---
title: Electric Field and Force
module: Electric Fields
moduleNumber: 1
lessonNumber: 3
order: 103
summary: >
  Rather than ask how one charge reaches across empty space to another, we credit
  the source with a field that fills the space and let a second charge respond to
  whatever field sits at its own location. Electric field is force per unit positive
  test charge, $\vec E=kq\hat r/r^2$ for a point source, and source fields add before
  any receiving charge is placed. We compute those fields and the force $\vec F=q\vec E$
  they exert, then follow a charge along its parabolic path through a uniform field
  and into nonuniform fields where the dynamics turn position-dependent.
topics: [Electric Fields]
draft: false
sources:
  - book: Tipler & Mosca
    ref: "Ch. 21 — The Electric Field I; §21-4 The Electric Field"
---

An electric field is defined operationally through the force on a sufficiently small
positive test charge. At position $\vec r$,

$$
\vec E(\vec r)=\lim_{q_0\to0}\frac{\vec F(\vec r)}{q_0}.
$$

The limiting condition prevents the probe from becoming part of the source
configuration. A large test charge can alter the distribution on nearby conductors or
polarize nearby matter, producing a force that no longer measures the original source
configuration. A source configuration determines a field vector at each position
independently of a later particle placed there.

A particle with charge $q$ responds according to

$$
\vec F=q\vec E,
\qquad m\vec a=q\vec E.
$$

Positive charge accelerates with the field. A negative charge experiences force and
acceleration opposite to the field, even though the field itself has not reversed.
The field has SI unit $\mathrm{N/C}$, equivalent to $\mathrm{V/m}$. Its direction is
defined by a positive test charge, not by electron motion in a conductor.

At $P$, the source configuration sets $\vec E$, and a probe experiences
$\vec F=q\vec E$.

$$
% caption: A positive source sets up a radial field. At $P$ the field $\vec E$
% points away from the source; a test charge placed there feels $\vec F=q\vec E$,
% whose size scales with $q$ but leaves the source field unchanged.
\begin{tikzpicture}[>=stealth, font=\footnotesize]
  \definecolor{acc}{HTML}{4A6FA5}
  \filldraw[draw=black, fill=black!8] (0,0) circle (0.24);
  \node[black!70] at (0,0) {+};
  \filldraw[draw=black, fill=black!8] (3.2,0) circle (0.09);
  \node[black, anchor=south] at (3.2,0.16) {P};
  \draw[->, acc, very thick] (0.28,0) -- (2.9,0) node[midway, above] {$E$};
  \draw[->, black, thick] (3.35,0) -- (4.5,0) node[right] {$F$};
\end{tikzpicture}
$$

## Probe limit and source-response separation

The definition uses a limiting probe because every charged probe exerts a reciprocal
force on the sources. A finite probe charge can move a light source particle,
redistribute charge on a conductor, or polarize an insulating body. The ratio
$\vec F/q_0$ then describes the combined source-plus-probe configuration rather
than the field prepared before the probe was introduced. The field at a point is
defined by the limiting configuration:

$$
\vec E(\vec r)
=\lim_{q_0\to0}
\frac{\vec F_{\rm sources\to probe}(\vec r)}{q_0}.
$$

The limit specifies the regime in which reducing the probe charge leaves the inferred
ratio unchanged within measurement uncertainty. A probe of charge $q_0$ measures a force
$\vec F=q_0\vec E$; halving $q_0$ should halve the force while preserving
$\vec F/q_0$ if source perturbation is negligible.

The field and the probe response are different physical quantities:

- **Field $\vec E$:** determined by the source configuration and the field point;
  its unit is $\mathrm{N/C}$.
- **Force $\vec F$:** determined by the field and the particular probe charge;
  its unit is $\mathrm N$.
- **Acceleration $\vec a$:** determined by force and probe mass; its unit is
  $\mathrm{m\,s^{-2}}$.

Changing the sign of the probe reverses the force and acceleration while leaving the
field unchanged. Changing the mass alters acceleration only. These distinctions
prevent a common error in which a field arrow is reversed when an electron is placed
at the point.

$$
% caption: The probe's sign flips its response, not the field. Along the same
% field $\vec E$, a positive charge is pushed with the field while an electron is
% pushed against it; the source-fixed field arrow is identical in both panels.
\begin{tikzpicture}[>=stealth, font=\footnotesize]
  \definecolor{acc}{HTML}{4A6FA5}
  \draw[->, acc, very thick] (0,0.4) -- (2.2,0.4) node[right] {$E$};
  \filldraw[draw=black, fill=black!8] (0.9,-0.5) circle (0.2);
  \node[black!70] at (0.9,-0.5) {+};
  \draw[->, black, thick] (1.16,-0.5) -- (2.15,-0.5) node[midway, below] {force};
  \draw[black, dashed] (3.3,-1.0) -- (3.3,0.9);
  \draw[->, acc, very thick] (4.2,0.4) -- (6.4,0.4) node[right] {$E$};
  \filldraw[draw=black, fill=black!8] (5.4,-0.5) circle (0.2);
  \node[black!70, font=\scriptsize] at (5.4,-0.5) {neg};
  \draw[->, black, thick] (5.14,-0.5) -- (4.15,-0.5) node[midway, below] {force};
\end{tikzpicture}
$$

## Field of one point charge

Let a source charge $q_i$ occupy position $\vec r_i$ and let $P$ be a field
point at $\vec r$. Coulomb's law gives the force on a positive test charge $q_0$:

$$
\vec F_{i\to0}
=kq_iq_0
\frac{\vec r-\vec r_i}{|\vec r-\vec r_i|^3}.
$$

Dividing by $q_0$ and taking the probe limit yields the point-charge field:

$$

\vec E_i(\vec r)
=kq_i
\frac{\vec r-\vec r_i}{|\vec r-\vec r_i|^3}
=k\frac{q_i}{r_{iP}^2}\hat r_{iP}.

$$

The unit vector points from the source location toward the field point. A positive
source gives a field in that direction; a negative source reverses it. The field
magnitude is

$$
E_i=k\frac{|q_i|}{r_{iP}^2}.
$$

The denominator has third power in the first vector expression because the
displacement vector contains one factor of distance. Its magnitude therefore reduces
to the inverse-square scalar form. The same algebra appeared in the vector form of
Coulomb's law, with the test-charge factor removed.

At a fixed distance, doubling the source charge doubles the field magnitude. At a
fixed source charge, doubling distance reduces the field magnitude to one quarter:

$$
\frac{E(2r)}{E(r)}=\frac14.
$$

The field diverges at the mathematical location of a point source. A real charged
object has finite size and charge distribution, so the point-source expression is
used only outside the scale where its geometry is unresolved.

**Field measurement and units.**

The operational definition gives a direct measurement procedure. Place a small
positive probe at the selected point, measure the force vector, and divide by its
charge. Repeating the measurement with a different small positive probe checks
whether the ratio is independent of the probe. The field unit follows directly:

$$
[\vec E]
=\frac{\mathrm N}{\mathrm C}.
$$

The equivalent unit $\mathrm{V/m}$ follows from the later definition of electric
potential. The two unit forms represent the same dimensions, but $\mathrm{N/C}$
keeps the force-per-charge meaning visible at this stage.

> **Worked example (field from a measured force).** A probe of charge
> $q_0=5.0\ \mathrm{nC}$ feels force $\vec F=(2.0\times10^{-4}\ \mathrm N)\,\hat\imath$.
> The field at that point is
>
> $$
> \vec E=\frac{\vec F}{q_0}=\frac{2.0\times10^{-4}}{5.0\times10^{-9}}\,\hat\imath
> =4.0\times10^4\ \mathrm{N/C}\,\hat\imath.
> $$
>
> The force and charge each carry two significant figures, so the field does too.

A negative test charge of the same magnitude at that point would feel
$-2.0\times10^{-4}\ \mathrm N\,\hat\imath$, yet the inferred field is unchanged,
$4.0\times10^4\ \mathrm{N/C}\,\hat\imath$.

## Force and acceleration in a prescribed field

Once a field has been specified, the motion of a particle follows from

$$
m\vec a=q\vec E.
$$

In a uniform field $\vec E=E\hat x$, acceleration is constant:

$$
a_x=\frac{qE}{m},
\qquad
x(t)=x_0+v_{0x}t+\frac12\frac{qE}{m}t^2.
$$

The same kinematic equations used for uniform gravitational acceleration apply after
replacing $g$ with $qE/m$ and retaining the charge sign. A positive particle in a
positive $x$ field accelerates toward positive $x$. An electron has $q=-e$ and
accelerates toward negative $x$. Its acceleration magnitude can be large because the
electron mass is small.

The field is a property of the source arrangement, whereas particle trajectories
also depend on charge, mass, initial velocity, and boundaries. Two particles at the
same point can experience forces of opposite direction and accelerations of different
magnitude without implying different fields.

The point-charge field and the force-response equation describe electrostatic
configurations. A rapidly changing source arrangement requires the time-dependent
electromagnetic field description. For stationary or slowly varying source charges,
the electrostatic field gives the local force relation used throughout the following
lessons.

**Components of a point-charge field.**

Coordinate components follow from the source-to-field-point displacement before any
magnitude is rounded. For a source at the origin and a field point $P=(x,y)$,

$$
\vec r=x\hat\imath+y\hat\jmath,
\qquad
r=\sqrt{x^2+y^2}.
$$

Substitution into the point-charge field gives

$$
\vec E
=kq\frac{x\hat\imath+y\hat\jmath}{(x^2+y^2)^{3/2}},
$$

with components

$$
E_x=kq\frac{x}{(x^2+y^2)^{3/2}},
\qquad
E_y=kq\frac{y}{(x^2+y^2)^{3/2}}.
$$

The sign of $q$ remains in both components. A positive source produces component
signs that match the coordinate signs of the field point. A negative source reverses
both. This form retains direction algebraically throughout the calculation.

> **Worked example (field of a point charge).** A source $q=+6.0\ \mathrm{nC}$ sits
> at the origin; find $\vec E$ at $P=(0.30,0.40)\ \mathrm m$. The separation is a
> $3$–$4$–$5$ triangle,
>
> $$
> r=\sqrt{0.30^2+0.40^2}\ \mathrm m=0.50\ \mathrm m,
> $$
>
> so the magnitude is
>
> $$
> E=k\frac{q}{r^2}=(8.988\times10^9)\frac{6.0\times10^{-9}}{(0.50)^2}
> =2.16\times10^2\ \mathrm{N/C}.
> $$
>
> The unit vector to $P$ has components $(0.60,0.80)$, giving
>
> $$
> \vec E=(1.29\times10^2\,\hat\imath+1.73\times10^2\,\hat\jmath)\ \mathrm{N/C}.
> $$
>
> Checking, $\sqrt{E_x^2+E_y^2}=2.16\times10^2\ \mathrm{N/C}$ recovers the direct
> magnitude, and both components are positive — the field points away from the
> positive source.

$$
% caption: Components of a point-charge field. The field point $P$ lies
% $0.30\ \mathrm m$ in $x$ and $0.40\ \mathrm m$ in $y$ from the source, so
% $\vec E$ has positive $E_x$ and $E_y$ and a $0.50\ \mathrm m$ separation.
\begin{tikzpicture}[>=stealth, font=\footnotesize]
  \definecolor{acc}{HTML}{4A6FA5}
  \draw[->, black] (-0.3,0) -- (5.2,0) node[right] {$x$};
  \draw[->, black] (0,-0.3) -- (0,3.6) node[above] {$y$};
  \filldraw[draw=black, fill=black!8] (0.6,0.55) circle (0.2);
  \node[black!70] at (0.6,0.55) {+};
  \filldraw[draw=black, fill=black!8] (4.0,2.8) circle (0.09);
  \node[black, anchor=south west] at (4.12,2.86) {P};
  \draw[->, acc, very thick] (0.82,0.7) -- (3.82,2.68);
  \node[acc] at (1.95,1.95) {$E$};
  \draw[black, dashed] (0.6,0.55) -- (4.0,0.55) -- (4.0,2.8);
  \draw[->, black, thick] (0.85,0.2) -- (3.75,0.2) node[midway, below=3pt] {$E_x$};
  \draw[->, black, thick] (4.45,0.7) -- (4.45,2.7) node[midway, right=4pt] {$E_y$};
\end{tikzpicture}
$$

With an arbitrary source location $\vec r_s$, use

$$
\vec R=\vec r_P-\vec r_s,
\qquad
\vec E(P)=kq\frac{\vec R}{|\vec R|^3}.
$$

The subtraction order is source to field point. Reversing it gives the opposite
vector and would draw a positive-source field toward the source.

## Uniform fields and trajectory response

A uniform field has the same magnitude and direction throughout the region of
interest. Parallel plates with a large central region provide an approximate example;
edge regions have fringing fields and require a more detailed model. In a uniform
field $\vec E=E\hat y$, a particle has constant acceleration

$$
a_y=\frac{qE}{m}.
$$

If the particle enters with horizontal speed $v_x$ and zero initial vertical speed,
then

$$
x(t)=v_xt,
\qquad
y(t)=y_0+\frac12\frac{qE}{m}t^2.
$$

Eliminating time yields a parabolic trajectory:

$$
y(x)=y_0+\frac{qE}{2mv_x^2}x^2.
$$

The sign of $qE$ determines the direction of curvature. A positive particle curves
with the field. An electron curves against it. The field remains the same in both
cases; the charge changes the force relation.

$$
% caption: A charged particle crossing a uniform field. Between the plates the
% field is uniform (parallel arrows); a positive charge entering horizontally
% curves with the field, an electron against it, at the same entry speed.
\begin{tikzpicture}[>=stealth, font=\footnotesize]
  \definecolor{acc}{HTML}{4A6FA5}
  \draw[black, line width=1pt] (0,1.2) -- (6.3,1.2);
  \draw[black, line width=1pt] (0,-1.2) -- (6.3,-1.2);
  \foreach \x in {0.7,1.5,2.3,3.1,3.9,4.7,5.5}{\draw[->, acc, thick] (\x,0.95) -- (\x,-0.6);}
  \filldraw[draw=black, fill=black!8] (0.4,0.5) circle (0.14);
  \node[black!70, font=\scriptsize] at (0.4,0.5) {+};
  \draw[black, thick] plot[smooth] coordinates
    {(0.56,0.5) (1.8,0.36) (3.05,0.0) (4.2,-0.5) (5.3,-1.05)};
  \filldraw[draw=black, fill=black!8] (0.4,-0.5) circle (0.14);
  \node[black!70, font=\scriptsize] at (0.4,-0.5) {neg};
  \draw[black, thick, dashed] plot[smooth] coordinates
    {(0.56,-0.5) (1.8,-0.36) (3.05,0.0) (4.2,0.5) (5.3,1.05)};
\end{tikzpicture}
$$

A particle deflected by plates of length $L$ has transit time
$t=L/v_x$. The vertical exit speed is

$$
v_y=\frac{qE}{m}\frac{L}{v_x},
$$

and the deflection at the exit is

$$
y_{\rm exit}
=\frac{qEL^2}{2mv_x^2}.
$$

The dependence on $v_x^{-2}$ gives a strong speed-selection effect. Doubling the
entry speed reduces the deflection to one quarter. A measured deflection therefore
cannot be interpreted as a field value unless the charge, mass, entry speed, and
field-region length are also known.

**Field magnitude scales and source changes.**

Point-charge fields span a wide range. A field of order $10^2\ \mathrm{N/C}$ occurs
in ordinary atmospheric conditions, while fields in an atomic environment can be
many orders of magnitude larger. The magnitude alone does not determine the force
on a particle: a $1\ \mathrm{nC}$ probe in a field of
$10^5\ \mathrm{N/C}$ experiences $10^{-4}\ \mathrm N$, whereas an electron in the
same field experiences a force magnitude $eE$.

The inverse-square dependence gives a local scale estimate. A source charge of
$1.0\ \mathrm{nC}$ at $0.10\ \mathrm m$ produces

$$
E\approx
(9.0\times10^9)
\frac{1.0\times10^{-9}}{(0.10)^2}
=9.0\times10^2\ \mathrm{N/C}.
$$

Moving to $0.20\ \mathrm m$ reduces the field to
$2.25\times10^2\ \mathrm{N/C}$. Such estimates identify whether a stated
instrument sensitivity or particle deflection is plausible before detailed numerical
work begins.

The electrostatic field model assumes a source arrangement that is fixed over the
observation interval. If a source charge is moved, the field at distant points does
not change instantaneously. Electromagnetic changes propagate at finite speed. The
static point-charge formula remains the appropriate local result once the new source
configuration has settled and the observation time is long compared with the
propagation time across the apparatus.

## Direction, sign, and local vector measurements

An electric field is a vector at each observation point, so a magnitude without a
direction is incomplete information. For one positive point source, the field points
away from the source at every surrounding point. For one negative point source, the
field points toward the source. The reversal belongs to the source charge in

$$
\vec E=kq_s\frac{\vec R}{R^3},
\qquad
\vec R=\vec r_P-\vec r_s,
$$

not to a later test charge. The sign of a test charge enters only after the field is
known, through $\vec F=q_0\vec E$. Keeping those two signs in separate steps
prevents source--probe sign reversals in diagrams that contain both a source and a
particle whose force is being found.

For example, place a negative source at the origin and examine a point on the
positive $x$ axis. The displacement vector $\vec R$ points in the positive $x$
direction, whereas the factor $q_s<0$ reverses the field. Thus $E_x<0$ at that
point. A positive test charge placed there feels a negative-$x$ force, directed
toward the source. An electron placed at the same point feels a positive-$x$ force,
directed away from the source. The source field has not changed between the two
experiments.

$$
% caption: Source sign and probe sign are separate choices. To the right of a
% negative source the field points back toward the source; a positive probe is
% pushed that way, a negative probe the opposite way — the field itself does not
% change.
\begin{tikzpicture}[>=stealth, font=\footnotesize]
  \definecolor{acc}{HTML}{4A6FA5}
  \filldraw[draw=black, fill=black!8] (0,0) circle (0.28);
  \node[black!70, font=\scriptsize] at (0,0) {neg};
  \node[black, anchor=north] at (0,-0.34) {source};
  \filldraw[draw=black, fill=black!8] (3.1,0.6) circle (0.16);
  \node[black!70, font=\scriptsize] at (3.1,0.6) {+};
  \filldraw[draw=black, fill=black!8] (3.1,-0.6) circle (0.16);
  \node[black!70, font=\scriptsize] at (3.1,-0.6) {neg};
  \draw[->, acc, very thick] (2.75,0.05) -- (0.5,0.05) node[midway, above] {$E$};
  \draw[->, black, thick] (2.9,0.6) -- (1.9,0.6) node[midway, above] {force};
  \draw[->, black, thick] (3.3,-0.6) -- (4.3,-0.6) node[midway, below] {force};
\end{tikzpicture}
$$

The local vector can be measured by resolving force into coordinate components.
Suppose a small positive probe of charge $q_0$ experiences

$$
\vec F=(F_x\hat\imath+F_y\hat\jmath+F_z\hat k).
$$

Then

$$
E_x=\frac{F_x}{q_0},
\qquad
E_y=\frac{F_y}{q_0},
\qquad
E_z=\frac{F_z}{q_0}.
$$

Component signs are obtained from the force components on the positive probe. A
negative $F_y$ produces a negative $E_y$; it does not mean that the magnitude of the
field is negative. The magnitude remains

$$
E=\sqrt{E_x^2+E_y^2+E_z^2}.
$$

At a point where $\vec E=(300\hat\imath-400\hat\jmath)\ \mathrm{N/C}$,
the magnitude is $500\ \mathrm{N/C}$. A $+2.0\ \mathrm{nC}$ probe has force

$$
\vec F
=(6.0\hat\imath-8.0\hat\jmath)\times10^{-7}\ \mathrm N,
$$

and an equal-magnitude negative probe has the opposite force vector. The calculation
also checks units: multiplying charge in coulombs by field in newtons per coulomb
leaves newtons.

Component form retains signs when a field point lies left of, below, or behind the
source. The signs follow directly from the displacement vector and the source charge,
whereas a scalar inverse-square calculation returns only a
magnitude. In three dimensions, retain all three components until the final magnitude
or direction is needed; projecting a result onto a drawing plane can hide a nonzero
third component.

**Distance scaling and the point-source approximation.**

Holding source charge fixed eliminates Coulomb's constant from a field-magnitude
comparison. If the field is measured at distances $r_1$ and $r_2$,

$$
\frac{E_2}{E_1}=\left(\frac{r_1}{r_2}\right)^2.
$$

Moving from $r$ to $3r$ reduces the magnitude to $E/9$; moving from $r$ to $r/2$
increases it to $4E$. Direction at corresponding radial locations remains unchanged
for a single source. The comparison is valid only when the same source can be
treated as a point at both observation distances.

Consider a $+4.0\ \mathrm{nC}$ charge. At $0.20\ \mathrm m$ the point-charge
model gives

$$
E_1=(8.988\times10^9)
\frac{4.0\times10^{-9}}{(0.20)^2}
=8.99\times10^2\ \mathrm{N/C}.
$$

At $0.60\ \mathrm m$, the ratio calculation gives $E_2=E_1/9$, or
$9.99\times10^1\ \mathrm{N/C}$. Substitution into the inverse-square expression
gives the same value. A ratio calculation exposes an incorrect distance exponent
because that error changes the result by a large factor.

The point-source model requires the source size to be small compared with the
source-to-observer distance and with the spatial resolution of the calculation. A
charged metal sphere with radius $a$ has
the exact external field of a point charge at its center when the charge distribution
is spherically symmetric and the field point lies outside the sphere. A charged rod,
disk, or irregular object generally needs a distributed-charge calculation when the
observer is near enough to distinguish different parts of the source. Those
calculations are developed in the next module.

Far from any bounded charge distribution, the total charge provides the leading
field term. Suppose several charges occupy a region of size $a$ and the field point
is at distance $R$ with $R\gg a$. Replacing the collection by one point charge equal
to its total charge gives the dominant radial field. The approximation loses the
internal geometry: two equal and opposite charges have total charge zero, so their
far field is much smaller than either individual field but is not identically zero.
That residual structure belongs to the dipole treatment rather than the one-point
source formula.

$$
% caption: Near and far views of a finite charged object. From a far point the
% rays from its parts are nearly parallel and the total charge suffices; from a
% near point they arrive from visibly different directions and the distribution
% matters.
\begin{tikzpicture}[>=stealth, font=\footnotesize]
  \definecolor{acc}{HTML}{4A6FA5}
  \fill[black!8] (0,0) circle (0.6);
  \draw[black, thick] (0,0) circle (0.6);
  \foreach \p in {(0.18,0.2),(-0.26,0.15),(0.05,-0.28)}{\filldraw[draw=black, fill=black!8] \p circle (0.06);}
  \node[black, anchor=north] at (0,-0.72) {size a};
  \filldraw[draw=black, fill=black!8] (1.5,0.7) circle (0.08);
  \node[black, anchor=south] at (1.5,0.82) {near P};
  \draw[black] (0.3,0.42) -- (1.5,0.7);
  \draw[black] (-0.28,0.42) -- (1.5,0.7);
  \draw[black] (0.1,-0.42) -- (1.5,0.7);
  \filldraw[draw=black, fill=black!8] (5.4,0) circle (0.08);
  \node[black, anchor=south] at (5.4,0.14) {far P};
  \draw[->, acc, thick] (0.66,0.24) -- (5.15,0.05);
  \draw[->, acc, thick] (0.62,0.0) -- (5.15,0.0);
  \draw[->, acc, thick] (0.6,-0.24) -- (5.15,-0.05);
  \node[black, anchor=north] at (5.4,-0.2) {distance R};
\end{tikzpicture}
$$

Use the point-source expression only for an exterior point sufficiently far from a
compact, effectively symmetric source. Other geometries require continuous-charge
or symmetry methods.

**Finite uniform-field regions.**

The uniform-field model applies to a specified region of an apparatus. Between two
large oppositely charged parallel conducting plates, the central region is well
approximated by parallel field vectors normal to the surfaces. Near an edge, field
lines spread outward and the field develops a horizontal component. A charged particle
can be treated with constant acceleration only while it remains in the central region
and while its displacement is small enough that the field variation is negligible.

Let the particle enter the central region at $x=0$, $y=y_0$ with velocity components
$v_{0x}$ and $v_{0y}$. With a field $\vec E=E\hat y$,

$$
x(t)=v_{0x}t,
\qquad
y(t)=y_0+v_{0y}t+\frac{qE}{2m}t^2,
$$

and

$$
v_x(t)=v_{0x},
\qquad
v_y(t)=v_{0y}+\frac{qE}{m}t.
$$

The horizontal velocity is constant because the field has no $x$ component in this
model. A nonzero entry value $v_{0y}$ changes both the exit height and the exit angle;
the familiar simple parabola with horizontal entry is the special case $v_{0y}=0$.
With plate length $L$ and $v_{0x}>0$, the particle spends

$$
t_L=\frac{L}{v_{0x}}
$$

inside the modeled field. Its exit height and vertical velocity are therefore

$$
y_L=y_0+\frac{v_{0y}L}{v_{0x}}
+\frac{qEL^2}{2mv_{0x}^2},
\qquad
v_{yL}=v_{0y}+\frac{qEL}{mv_{0x}}.
$$

These equations give two separate observables. The position $y_L$ depends on the
full transit history, whereas $v_{yL}$ records the accumulated impulse. A detector
placed beyond the plates receives a particle moving in a straight line at the exit
velocity if no other force acts in that later region.

The geometry imposes a further condition. If the plate separation is $d$ and the
midplane is $y=0$, the particle reaches a plate before the exit whenever

$$
|y(t)|\geq\frac d2
$$

for some $0<t<t_L$. The formula for $y_L$ alone is insufficient when the trajectory
first moves toward a plate and then turns around, because the maximum displacement
can occur before the exit. Solve $v_y(t)=0$ for a turning time when $qE$ and $v_{0y}$
have opposite signs, then evaluate $y(t)$ there. The same check appears in a beam
deflection tube, an electrostatic steering element, or a charged-particle analyzer.

For horizontal entry, the curvature has the sign of $qE$. A positive field in the
negative $y$ direction bends a positive ion downward and an electron upward. Reversing
the plate charge reverses $\vec E$ and therefore reverses both trajectories. The
initial horizontal speed also matters strongly: a particle traveling twice as fast
spends half as long in the field and acquires one quarter of the vertical displacement.

An electric field calculation must identify the source region and particle
observation point. Extending a uniform field beyond the plate edges is an
idealization that changes the predicted flight time, exit direction, and detector
position. Keeping the boundaries in the diagram avoids that hidden assumption.

**Electric force together with gravity.**

The equation $\vec F=q\vec E$ gives one force contribution. A particle in a
laboratory also has weight, contact forces, drag, or forces from other fields. Newton's
second law uses their vector sum. For a particle of mass $m$ in a vertical electric
field, take upward as the positive $y$ direction. If the field is upward, then

$$
\sum F_y=qE-mg.
$$

A positive charge experiences an upward electric force; a negative charge experiences
a downward one. A suspended positive charge requires $qE=mg$, so the field must point
upward. A suspended negative charge requires a downward field. The field direction
can therefore be inferred from a known charge sign and an observed force balance.

The balance condition determines a ratio rather than either property alone:

$$
\frac qm=\frac gE.
$$

> **Worked example (charge suspended against gravity).** A droplet of mass
> $m=3.0\times10^{-15}\ \mathrm{kg}$ carries charge $q=+6.0\times10^{-18}\ \mathrm C$
> in an upward field $E=4.9\times10^3\ \mathrm{N/C}$. The electric force is
>
> $$
> qE=(6.0\times10^{-18})(4.9\times10^3)=2.94\times10^{-14}\ \mathrm N\ (\text{up}).
> $$
>
> Its weight is $mg=(3.0\times10^{-15})(9.8)=2.94\times10^{-14}\ \mathrm N$ (down), so
> $\sum F_y=qE-mg=0$ and the droplet floats. Raise the field by $10\%$ and the net
> upward force becomes $0.10\,mg$, giving acceleration $0.10\,g=0.98\ \mathrm{m/s^2}$
> upward. The sign of the net force, not the equality of magnitudes alone, decides the
> motion.

$$
% caption: Vertical balance of a charged droplet. In an upward field the electric
% force $qE$ acts up and the weight $mg$ down; equal magnitudes give zero net force
% and the droplet floats.
\begin{tikzpicture}[>=stealth, font=\footnotesize]
  \definecolor{acc}{HTML}{4A6FA5}
  \foreach \x in {0.4,1.2,2.8,3.6}{\draw[->, acc, thick] (\x,-1.1) -- (\x,1.1);}
  \filldraw[draw=black, fill=black!8] (2.0,0) circle (0.22);
  \node[black!70] at (2.0,0) {+};
  \draw[->, black!70, very thick] (2.0,0.24) -- (2.0,1.25) node[anchor=south east] {$qE$};
  \draw[->, black!70, very thick] (2.0,-0.24) -- (2.0,-1.25) node[anchor=north] {$mg$};
  \node[acc, anchor=west] at (3.9,0.7) {applied E};
  \node[black, anchor=west] at (3.9,-0.7) {a = 0};
\end{tikzpicture}
$$

For many elementary charged-particle calculations, gravity is negligible. The
comparison is quantitative. An electron in a field of only $1\ \mathrm{N/C}$ has
electric-force magnitude $eE=1.60\times10^{-19}\ \mathrm N$, while its weight is
about $8.9\times10^{-30}\ \mathrm N$. Their ratio is approximately
$1.8\times10^{10}$. For a macroscopic charged bead with a much larger mass and a
small net charge, the ratio may be near unity instead. Dropping gravity without a
scale comparison is justified for some particles and wrong for others.

The vector equation also distinguishes static equilibrium from motion at constant
speed. When the net force vanishes, acceleration is zero; the particle may remain at
rest or continue with whatever velocity it already has. A particle launched upward in
an exact electric-gravitational balance travels upward at constant speed until another
force or boundary changes its motion. Conversely, a particle at rest in an imbalance
immediately begins accelerating even though its initial velocity is zero.

When the field is horizontal and gravity is vertical, the motions separate into two
components:

$$
a_x=\frac{qE}{m},
\qquad
a_y=-g.
$$

The resulting trajectory is determined by both accelerations. A calculation that
uses $qE/m$ as the entire acceleration implicitly assumes that all other forces are
either negligible or included in a separately stated direction. That assumption must
be tied to the physical setup, particularly for slowly moving charged droplets and
charged grains.

**Impulse, momentum change, and charge-to-mass ratio.**

During a finite interval, a prescribed electric field changes momentum through the
impulse relation

$$
\Delta\vec p
=\int_{t_i}^{t_f}q\vec E\,\d t.
$$

If both $q$ and $\vec E$ are constant over an interval $\Delta t$, this reduces to

$$
\Delta\vec p=q\vec E\,\Delta t.
$$

The momentum form compares equal impulses before particle mass is specified. Two
particles with the same charge passing through the same field for the same time
receive equal electric impulses. Their velocity changes differ because $\Delta\vec v=
\Delta\vec p/m$. A light particle therefore responds with a larger acceleration
than a heavy particle under identical field conditions.

In a uniform field of length $L$ traversed with constant horizontal speed $v_x$, the
time in the field is $L/v_x$. With the field along $y$,

$$
\Delta p_y=qE\frac{L}{v_x},
\qquad
\Delta v_y=\frac{qEL}{mv_x}.
$$

The ratio $q/m$ controls the velocity deflection. Suppose two singly charged positive
ions enter the same horizontal field region at the same speed. The smaller-mass ion
has a greater vertical exit velocity and a larger deflection. The sign of $q$ fixes
the direction; the mass affects only the magnitude of the acceleration for a fixed
charge.

$$
% caption: Equal impulse, unequal deflection. Two positive charges spend the same
% time in the same downward field, so both gain the same downward momentum
% $qE\,\Delta t$; the lighter particle gains more speed and curves more sharply.
\begin{tikzpicture}[>=stealth, font=\footnotesize]
  \definecolor{acc}{HTML}{4A6FA5}
  \foreach \x in {0.6,1.5,2.4,3.3,4.2,5.1}{\draw[->, acc, thick] (\x,1.2) -- (\x,-1.2);}
  \filldraw[draw=black, fill=black!8] (0.34,0.55) circle (0.14);
  \node[black!70, font=\scriptsize] at (0.34,0.55) {m};
  \filldraw[draw=black, fill=black!8] (0.34,-0.5) circle (0.2);
  \node[black!70, font=\scriptsize] at (0.34,-0.5) {M};
  \draw[black, thick] plot[smooth] coordinates
    {(0.52,0.55) (1.7,0.4) (2.8,-0.02) (3.75,-0.62) (4.45,-1.15)};
  \draw[black, thick, dashed] plot[smooth] coordinates
    {(0.56,-0.5) (1.8,-0.54) (3.0,-0.66) (4.1,-0.86) (5.05,-1.14)};
  \node[black, anchor=west] at (5.35,0.5) {light m};
  \node[black, anchor=west] at (5.35,-0.55) {heavy M};
  \node[acc, anchor=west] at (5.35,-0.02) {applied E};
\end{tikzpicture}
$$

A measured deflection can determine $q/m$ when the field, geometry, and initial
velocity are independently known. For horizontal entry with $v_{0y}=0$,

$$
y_L=\frac{qEL^2}{2mv_x^2}
$$

gives

$$
\frac qm=\frac{2y_Lv_x^2}{EL^2}.
$$

The sign of $y_L$ must be retained. When the positive $y$ axis is chosen along the
field, a positive measured $y_L$ indicates a positive $q/m$; a negative deflection
indicates a negative charge. Using only the absolute displacement yields the
magnitude $|q|/m$ and discards that information.

A numerical check can use a particle entering a $5.0\times10^3\ \mathrm{N/C}$
field at $v_x=2.0\times10^6\ \mathrm{m/s}$ through plates of length
$L=4.0\times10^{-2}\ \mathrm m$. If its measured exit displacement is
$y_L=1.0\times10^{-3}\ \mathrm m$, then

$$
\frac qm
=\frac{2(1.0\times10^{-3})(2.0\times10^6)^2}
{(5.0\times10^3)(4.0\times10^{-2})^2}
=1.0\times10^9\ \mathrm{C/kg}.
$$

The value is positive because the deflection has been taken in the field direction.
Before trusting a result like this, verify that the displacement is much smaller than
the plate separation, that the trajectory did not strike a plate, and that the
horizontal speed remains approximately constant in the central field. Each condition
belongs in the model before the numerical result is interpreted.

The momentum form also helps compare a short, strong field pulse with a weak field
acting for a long time. Equal values of $E\Delta t$ give equal impulse per unit
charge when the field direction is the same. The detailed trajectory differs if the
particle moves through a nonuniform region during the interval, because the field
must then be integrated along the actual path rather than replaced by one constant
vector.

## Measuring a field without changing its source

The test-charge definition requires a probe small enough to measure the prepared
field without significantly altering the source configuration. Its charge must also
produce a force above the sensor's resolution. The measurement range balances two
conditions:

$$
\text{source disturbance is negligible},
\qquad
|q_0|E\text{ is measurable}.
$$

Repeated readings with progressively smaller probes provide a practical test. If the
inferred ratio $\vec F/q_0$ remains unchanged within uncertainty, the probe is in
the test-charge regime. If the ratio drifts as the charge is reduced, the larger probes
were polarizing, displacing, or otherwise modifying the source arrangement. A charged
conducting source is particularly sensitive because an external probe redistributes
its surface charge.

At one location, record the force components on a calibrated positive probe. For
example,

$$
q_0=(2.00\pm0.02)\ \mathrm{nC},
\qquad
F_x=(6.00\pm0.06)\times10^{-6}\ \mathrm N.
$$

The inferred horizontal field is

$$
E_x=\frac{F_x}{q_0}=3.00\times10^3\ \mathrm{N/C}.
$$

For independent small relative uncertainties, the fractional uncertainty estimate is

$$
\frac{\delta E_x}{|E_x|}
\approx
\sqrt{
\left(\frac{\delta F_x}{|F_x|}\right)^2+
\left(\frac{\delta q_0}{|q_0|}\right)^2}.
$$

Here each relative input uncertainty is $1\%$, so the combined estimate is about
$1.4\%$. The field should be reported with an uncertainty consistent with that
precision, such as $(3.00\pm0.04)\times10^3\ \mathrm{N/C}$. The sign of $E_x$
follows from the measured force direction on the positive probe; it is independent
of the uncertainty convention.

Mapping a field requires repeating this component measurement at a series of fixed
positions. The resulting data set is a list of vectors, not a single scalar profile.
A point source gives equal field magnitudes at samples on a circle of constant radius
within experimental uncertainty, while their directions rotate with
the radius. Sample points at increasing radius should follow the inverse-square trend
when the source is sufficiently compact. These comparisons test the model using
measured geometry rather than a drawing alone.

An experimental map also distinguishes a field measurement from the subsequent motion
of one freely released particle. A held probe permits force components to be measured
at a specified location. A released particle has a changing position, and it samples
different field vectors as it moves through a nonuniform region. Treating its initial
force as a constant force requires an additional uniform-field approximation.

Systematic checks can expose a sign or calibration error. Reversing the probe charge
reverses the measured force and leaves $\vec F/q_0$ unchanged. Rotating the
apparatus through $180$ degrees reverses the coordinate components assigned to a fixed
physical field. Repeating measurements at the same marked position checks drift; a
different mark changes the observation point and constitutes a different observation.
These controls keep the measured field tied to a particular source configuration and
coordinate system.

## Complete vector calculation at one field point

A point-field calculation keeps four quantities distinct: source position, field
position, displacement vector, and probe response. Combining them too early often
reverses a direction or incorrectly includes the probe charge in the field. Let a source charge
$q_s$ be at $(x_s,y_s,z_s)$ and a field point be at $(x_P,y_P,z_P)$. The required
displacement is

$$
\vec R=(x_P-x_s)\hat\imath
+(y_P-y_s)\hat\jmath
+(z_P-z_s)\hat k.
$$

Its length is

$$
R=\sqrt{(x_P-x_s)^2+(y_P-y_s)^2+(z_P-z_s)^2}.
$$

The source field follows directly:

$$
\vec E(P)=kq_s\frac{\vec R}{R^3}.
$$

Only after that step is a probe charge introduced through $\vec F=q_0\vec E$.
The component formula preserves all signs without a separate verbal direction rule.

Consider a source $q_s=-3.0\ \mathrm{nC}$ at
$(0.10,-0.20)\ \mathrm m$ and a field point $P=(0.40,0.20)\ \mathrm m$. The
source-to-point displacement is

$$
\vec R=(0.30\hat\imath+0.40\hat\jmath)\ \mathrm m,
\qquad R=0.50\ \mathrm m.
$$

The field is

$$
\begin{aligned}
\vec E(P)
&=(8.988\times10^9)(-3.0\times10^{-9})
\frac{0.30\hat\imath+0.40\hat\jmath}{(0.50)^3}\\
&=(-64.7\hat\imath-86.3\hat\jmath)\ \mathrm{N/C}.
\end{aligned}
$$

Its magnitude is $108\ \mathrm{N/C}$, directed down and left toward the negative
source. A positive $2.0\ \mathrm{nC}$ probe at $P$ experiences

$$
\vec F
=(-1.29\hat\imath-1.73\hat\jmath)\times10^{-7}\ \mathrm N.
$$

The force direction agrees with the field direction because the probe charge is
positive. Replacing that probe with a $-2.0\ \mathrm{nC}$ electron-scale charge
reverses the force components while preserving the field components just calculated.

Check that $R>0$ and that each field component has units of newtons per coulomb.
The expression $kq_sR/R^3$ has units

$$
\left(\frac{\mathrm{N\,m^2}}{\mathrm{C^2}}\right)
\frac{\mathrm C\,\mathrm m}{\mathrm{m^3}}
=\frac{\mathrm N}{\mathrm C}.
$$

The field must point away from a positive source and toward a negative source. Its
magnitude must decrease if the field point is moved farther away along the same ray.
For the example, the ratio $|E_y|/|E_x|$ equals $0.40/0.30$, so the component ratio
matches the displacement geometry. A component sign that breaks any of these checks
usually indicates a reversed displacement or an omitted source-charge sign.

Coordinates can be translated without changing the physical field. Shifting both
source and field point by the same vector leaves $\vec R$ unchanged, hence leaves
$\vec E$ unchanged. Rotating the coordinate axes changes the numerical components
but not the field magnitude or its direction in physical space. These invariances
separate a true physical prediction from a bookkeeping choice.

**Spatial variation and the local-field approximation.**

The inverse-square field changes continuously with position. Along a radial line from
a positive point source,

$$
E(r)=\frac{kq_s}{r^2},
\qquad
\frac{\d E}{\d r}=-\frac{2kq_s}{r^3}=-\frac{2E}{r}.
$$

The derivative gives the local rate at which field magnitude changes with radial
distance. A small outward displacement $\Delta r$ produces the approximation

$$
\Delta E\approx-\frac{2E}{r}\Delta r,
\qquad
\frac{\Delta E}{E}\approx-2\frac{\Delta r}{r}.
$$

At $r=1.0\ \mathrm m$, a radial sensor motion of $1.0\ \mathrm{cm}$ changes the
field by about $2\%$. At $r=0.10\ \mathrm m$, the same motion changes the field by
about $20\%$. Thus the phrase "field at a point" is an approximation whenever a
probe has finite size or a particle travels through an extended region. The variation
must be small over the relevant distance before a single constant vector can represent
the field there.

The exact comparison across a finite radial interval is

$$
\frac{E(r+\Delta r)}{E(r)}
=\left(\frac r{r+\Delta r}\right)^2.
$$

For $r=0.50\ \mathrm m$ and $\Delta r=0.050\ \mathrm m$, the exact ratio is
$(0.50/0.55)^2=0.826$. The field falls by $17.4\%$. The derivative estimate predicts
$-2(0.050/0.50)=-20\%$, close enough to show the scale but not exact because the
displacement is no longer extremely small compared with $r$. The exact form should be
used for a stated finite interval; the derivative is a local approximation.

A small test charge has a finite spatial extent, and a force sensor can hold it only
over a finite position tolerance. The reported vector is therefore an average over a
small region. When the field changes little across that region, the average agrees
closely with the field at its center. When the field changes rapidly, the result
depends on the probe shape, orientation, and exact center location. Finite probe size
makes a point-charge model delicate close to a source even if the source itself is
approximately compact.

Particle motion through a nonuniform field requires the same caution. The force is

$$
\vec F(t)=q\vec E(\vec r(t)),
$$

so the acceleration changes as the particle changes location. Replacing it by
$q\vec E(\vec r_0)/m$ gives a local, short-time approximation around the
starting point $\vec r_0$. A finite trajectory requires either a field that is
uniform over the path or a solution of the position-dependent equation of motion. The
inverse-square field of a point charge is a standard case in which acceleration is
radial and changes strongly with radius.

The local approximation has a clear geometric criterion. If a particle travels a
distance $\ell$ near radius $r$ in a point field, then $2\ell/r$ estimates the
fractional field change along a radial path. Values much smaller than one support a
constant-field approximation over that segment. Values comparable with one require
the position dependence to remain in the force law. Tangential motion changes field
direction even when the radius remains nearly constant, so a vector-field check must
consider both magnitude and direction.

## Domain of the electrostatic point-field model

The point-charge expression is an electrostatic model. Its source charge is treated as
fixed while the field is evaluated, and the observation point is distinct from the
source location. At the source location itself, the expression contains $R=0$ and is
undefined. The divergence belongs to the ideal mathematical point charge. Finite
charged objects have size, structure, and charge distributions that replace the
singular model in their immediate vicinity.

An isolated conducting sphere at electrostatic equilibrium has an exterior field
that is the same as that of a point charge equal to the sphere's total charge located at its
center. Only exterior points and spherical symmetry meet those conditions. A charged
rod observed close to one end requires its actual charge distribution because its
field depends strongly on the positions of individual charge elements. Those cases
need the field integrated over a continuous charge distribution, together with
symmetry arguments.

The model also assumes that the sources do not change appreciably during the
observation interval. Moving a source charge, closing a circuit, or redistributing
charge on a conductor creates time-dependent electromagnetic fields. The effect of
that change reaches a distant observation point after a finite propagation time. A
static formula describes the field before the change arrives and after the new
arrangement has settled. Time-dependent fields are developed later with induction and
electromagnetic waves.

The distinction between source field and response remains valid in every regime. A
prescribed field at $\vec r$ determines the force $q_0\vec E(\vec r)$ on a
small probe located there. A probe large enough to alter the source distribution
requires a coupled calculation in which its own field is part of the configuration.
A statement of “electric field due to the source” must specify which charges are held
fixed and which material responses are included.

Several limiting behaviors provide compact checks on a result:

- **Vanishing source charge.** As $q_s\to0$, every component of the source field
  tends to zero.
- **Large distance.** As $R\to\infty$, the point-source field tends to zero as
  $1/R^2$.
- **Positive versus negative source.** Replacing $q_s$ by $-q_s$ reverses every field
  component at the same field point.
- **Positive versus negative probe.** Replacing $q_0$ by $-q_0$ leaves the field
  unchanged and reverses the force.
- **Mass change.** Changing probe mass leaves field and force unchanged, while
  changing acceleration through $\vec a=q_0\vec E/m$.

Each limit follows from a different physical part of the model. Source charge sets
the field; field-point geometry sets its spatial dependence; probe charge converts a
field into force; probe mass converts force into acceleration. Keeping these roles
separate is the basis for the superposition, continuous-distribution, and particle
motion calculations that follow.

**Units, numerical scale, and reporting a vector field.**

The SI unit $\mathrm{N/C}$ follows directly from the operational definition. A field
of $1\ \mathrm{N/C}$ exerts a $1\ \mathrm N$ force on a $1\ \mathrm C$ positive
test charge. Coulomb is a large laboratory charge, so field-force calculations often
combine smaller prefixes. A field of $2.5\times10^4\ \mathrm{N/C}$ acting on a
$4.0\ \mathrm{nC}$ probe produces

$$
F=qE=(4.0\times10^{-9})(2.5\times10^4)
=1.0\times10^{-4}\ \mathrm N.
$$

The conversion of nanocoulombs to coulombs introduces nine powers of ten. Omitting that
conversion changes the force by a factor of $10^9$, a common scale error in otherwise
correct algebra. Scientific notation keeps the charge, field, and force exponents
visible until the final rounding step.

Vector reports require both components and a coordinate convention. The statement

$$
\vec E=(-2.6\hat\imath+1.4\hat\jmath)\times10^3\ \mathrm{N/C}
$$

specifies a field completely in a two-dimensional coordinate system. Its magnitude is
$3.0\times10^3\ \mathrm{N/C}$ to two significant figures, but the magnitude alone
does not specify the force direction on a charged particle. A diagram should state
which axis is positive whenever components are read from geometry; a different axis
orientation changes the signs of the components while leaving the physical vector
unchanged.

Precision should follow the data. If a source-field distance is measured to three
significant figures, the inverse-square calculation has roughly twice the fractional
distance uncertainty. For a distance $r\pm\delta r$,

$$
\frac{\delta E}{E}\approx2\frac{\delta r}{r}
$$

when source-charge uncertainty is negligible. A $1\%$ position uncertainty therefore
creates about a $2\%$ field uncertainty for a point-source measurement. Reporting six
digits from Coulomb's constant cannot compensate for a poorly known source-to-point
distance. Reported precision belongs to the complete physical measurement, not to
one constant in the formula.

Use the reciprocal relationship between field and force as a final numerical check.
Dividing a force by the known test charge must recover the original field vector;
multiplying that field by a negative test charge must reverse every force component.
Those two calculations test signs, units, and powers of ten without introducing a new
model.

## From a field measurement to a particle calculation

A measured or calculated field becomes a motion model only after a particle and a
region of validity are specified. Along one coordinate, the governing equation is

$$
m\frac{\d^2 x}{\d t^2}=qE_x(x,t).
$$

In a uniform static field, $E_x$ is constant and the familiar constant-acceleration
equations follow. For a point source, $E_x$ varies with position, and a particle
moving toward or away from the source experiences changing acceleration. At the
particle position, evaluate the source field and use
$\vec a=q\vec E/m$.

A short numerical approximation keeps that interface explicit. At time $t_n$, with
position $\vec r_n$ and velocity $\vec v_n$, evaluate

$$
\vec a_n=\frac qm\vec E(\vec r_n),
$$

then update over a sufficiently short interval $\Delta t$ by

$$
\vec v_{n+1}\approx\vec v_n+\vec a_n\Delta t,
\qquad
\vec r_{n+1}\approx\vec r_n+\vec v_n\Delta t.
$$

Reducing $\Delta t$ should make the calculated trajectory converge. A large time step
can move the particle across a region where the field direction or magnitude changes
substantially, replacing a varying force with one outdated value. The fractional
field-change estimate from the preceding section sets a physical step-size criterion
near a point source.

The charge sign remains in every update. A negative particle in a prescribed field
has acceleration opposite the local field vector at each step. Its path does not
reverse the field arrows; it samples the same vector map with the opposite force law.
Mass enters only through $q/m$, so particles with equal charge magnitude but different
masses trace different paths through the same field. These distinctions remain
necessary when later lessons add superposed fields, continuous charge distributions, magnetic
forces, and time-dependent sources.