Why E = V/d Requires an Approximately Uniform Electric Field

A voltmeter measures a 10.0 V potential difference between two points 10.0 cm apart. Dividing gives \(100\ \text{V/m}\). But suppose the potential drops by just 0.5 V across the first 2.0 cm and by 4.3 V across the last 2.0 cm. Is the electric field really \(100\ \text{V/m}\) everywhere?

No. The potential is changing much faster in one region than the other.

This is why \(E=V/d\) needs an approximately uniform electric field. The equation gives the field strength at each point only when the field has approximately constant magnitude and direction over the distance being considered, with that distance measured along the field. Otherwise, the calculation describes an average potential gradient, not necessarily the local electric field.

The useful question is not simply whether you know the voltage and distance. It’s whether the potential changes at approximately the same rate throughout that distance.

01What does an electric field actually measure?

Imagine placing a tiny positive charge at a particular position between two charged objects.

It experiences an electric force. Move it somewhere else, and that force might change.

But here’s a complication: doubling the test charge would also double the force. Does that mean the electric field has doubled?

No. The field belongs to the arrangement of source charges, not to the small charge you use to investigate it.

Electric field is defined as force per unit positive test charge:

\[ \vec E=\frac{\vec F}{q_t} \]

where:

• \(\vec E\) is the electric field vector, measured in newtons per coulomb (N/C).
• \(\vec F\) is the electric force on the test charge, measured in newtons (N).
• \(q_t\) is the positive test charge, measured in coulombs (C).

We assume the test charge is sufficiently small that it does not significantly disturb the original field.

A negative charge experiences a force opposite to the field direction, but it does not change the direction used to define the field.

An electric field is uniform when its magnitude and direction are constant throughout the region being considered.

For example, in the central region between two large parallel conducting plates, the field can be approximately uniform.

Outside an isolated point charge, however, the field becomes weaker as distance increases. Its direction also changes depending on where you are around the charge.

Both situations contain electric fields. Only one generally allows us to use a single value of \(E=V/d\) to represent the local field throughout the region.

02An experiment that tests the uniform-field assumption

How could we investigate whether the electric field is actually uniform, rather than simply assuming it?

One approach is to measure electric potential at several positions instead of measuring only the total potential difference.

Setting up a voltage-mapping experiment

Consider a rectangular sheet of conductive paper, connected to a low-voltage DC power supply.

The apparatus is arranged as follows:

1. A straight metal strip touches the left side of the paper and connects to the positive supply terminal.
2. A second straight metal strip touches the right side and connects to the negative terminal.
3. The supply maintains the left strip at 10.0 V relative to the right strip, which we define as 0 V.
4. The strips are parallel and separated by 10.0 cm.
5. A high-input-resistance voltmeter has its negative lead fixed to the right strip. Its positive probe touches different positions along the central horizontal line of the paper.

The voltmeter measures electric potential relative to the right-hand electrode.

Moving the probe in 2.0 cm steps lets us investigate how potential changes with position.

This is a proposed measurement experiment, not a report of one that has been performed. The numerical readings below are illustrative model data.

There’s also an important physical limitation: the paper conducts a steady electric current, unlike the insulating space between capacitor plates. Its electric field is related to its potential gradient in the same way, but the precise field distribution depends on the conducting sheet and its electrode geometry.

What should a uniform field produce?

Take the horizontal coordinate \(x\) as increasing from left to right.

The left electrode is at higher potential, so the electric field should point to the right in the central region.

If that field is uniform, every equal horizontal distance should correspond to an equal potential drop.

For a total drop of 10.0 V across 10.0 cm, each 2.0 cm interval should account for a 2.0 V drop.

The predicted potentials are therefore:

The graph of potential against horizontal position would be a straight line.

The gradient of that line tells us how rapidly potential changes with distance.

03Change the electrode shape, keep the same voltage

Suppose we revise the apparatus.

Keep the left metal strip at 10.0 V, but replace the long right-hand strip with a small circular contact maintained at 0 V.

The central measurement line still runs horizontally between the electrodes, over a distance of 10.0 cm.

What changes?

The small contact concentrates the electric field into a different spatial arrangement. Near it, the potential can change much more rapidly with distance.

The original design showed what we expect for a nearly uniform central field. The revised design tests whether the same endpoint voltage and separation are enough to preserve that result.

They aren’t.

A comparison of potential readings

The following table contains illustrative, idealised values, not observations from a real experiment. The second set represents a plausible nonlinear potential profile along the centreline of the modified arrangement.

Both arrangements have the same 10.0 V potential difference across the same 10.0 cm separation.

Now look at the first and last intervals in the modified arrangement.

From 0 to 2.0 cm, the potential falls by just 0.5 V:

\[ \begin{aligned} E_{\text{first, avg}} &=\frac{0.5}{0.020}\\ &=25\ \text{V/m} \end{aligned} \]

From 8.0 to 10.0 cm, the potential falls by 4.3 V:

\[ \begin{aligned} E_{\text{last, avg}} &=\frac{4.3}{0.020}\\ &=215\ \text{V/m} \end{aligned} \]

These are average horizontal field components over their respective intervals. Along the centreline of a suitably symmetric arrangement, the field is directed horizontally, so they also provide useful estimates of field strength.

Yet the whole-gap calculation still gives:

\[ \frac{10.0}{0.100} =100\ \text{V/m} \]

The overall average has hidden substantial variation.

This is the central problem with applying \(E=V/d\) indiscriminately.

A fixed potential difference constrains the total potential change. It does not tell us how that change is distributed across the region.

There is another experimental limitation: changing the electrode shape may also create a field with a vertical component away from the symmetry line. A horizontal voltage scan cannot reveal that entire vector field.

To map it properly, we would need readings at multiple horizontal and vertical positions.

04The relationship that survives when the field changes

We need to distinguish three quantities:

• Electric potential, \(\phi\), measured in volts (V).
• Potential difference, \(\Delta\phi=\phi_B-\phi_A\), also measured in volts.
• Electric field, \(\vec E\), measured in V/m or N/C.

Here, \(\phi_A\) and \(\phi_B\) are the electric potentials at positions \(A\) and \(B\).

For the familiar magnitude equation \(E=V/d\), we use \(V=|\Delta\phi|\), the positive magnitude of the potential difference.

Keep the direction with a minus sign

Along a horizontal coordinate \(x\), an approximately constant field component can be found using:

\[ E_x\approx-\frac{\Delta\phi}{\Delta x} \]

where \(E_x\) is the horizontal component of electric field, and \(\Delta x=x_B-x_A\) is the signed horizontal displacement.

The minus sign tells us that electric field points towards decreasing potential.

For example, if potential decreases by 3.0 V as \(x\) increases by 0.010 m:

\[ \begin{aligned} E_x &\approx-\frac{-3.0}{0.010}\\ &=+300\ \text{V/m} \end{aligned} \]

The positive sign means the field points in the positive \(x\)-direction.

If the field changes over the interval, this finite-difference calculation gives the average horizontal component, rather than an exact value at one point.

The more general, exact one-dimensional relationship is:

\[ E_x=-\frac{d\phi}{dx} \]

This calculus notation describes the local gradient of potential. The important physical idea is accessible without differentiation: examine smaller intervals to find how rapidly potential changes near a particular position.

A further caution matters in two or three dimensions. A voltage change measured along one direction gives information about the field component along that direction. It does not necessarily give the full field magnitude.

05Why a point charge is different: the inverse-square field

Consider a small isolated positive source charge \(Q\).

The electric field points radially outwards, and its magnitude decreases as you move away.

For a point charge in vacuum,

\[ E(r)=\frac{k|Q|}{r^2} \]

where:

• \(E(r)\) is the electric field magnitude at distance \(r\), in N/C.
• \(Q\) is the source charge, in C.
• \(r\) is the distance from the charge, in m.
• \(k\) is Coulomb’s constant, approximately \(8.99\times10^9\ \text{N m}^2/\text{C}^2\).

This is an inverse-square relationship.

Double the distance from the charge, and the field becomes one-quarter as strong.

The electric potential of an isolated point charge, choosing zero potential at infinity, is:

\[ \phi(r)=\frac{kQ}{r} \]

Notice the different distance relationships.

Field strength follows \(1/r^2\), while potential follows \(1/r\).

That difference is exactly why you cannot generally take the potential difference between two radii, divide by their separation, and claim the result is the electric field at both locations.

06When can you actually use \(E=V/d\)?

The geometry of the source charges matters as much as the voltage.

For parallel conducting plates, the central field becomes a good uniform-field approximation when the plate dimensions are large compared with the gap, the plates are parallel, and nearby charged objects do not significantly disturb the field.

Near the edges of finite plates, field lines bend. This is called fringing.

The distinction isn’t that \(E=V/d\) suddenly becomes mathematically meaningless near an edge.

Rather, the same voltage and separation can no longer identify the full local electric field there.

Use the following decision process.

For standard uniform-field plate calculations, Electric Fields Between Parallel Plates: Using E = V/d develops the usual plate model.

Here, the important skill is recognising when that model no longer answers the question being asked.

A practical test is to ask:

Would dividing the region into smaller intervals give approximately the same field component each time?

If yes, the constant-gradient model may be useful.

If not, a single \(V/d\) value is concealing a changing field.

And if the arrangement is two-dimensional or three-dimensional, remember to check field direction as well as magnitude.

07Questions and solutions

Question 1

Two parallel conducting plates are separated horizontally by 6.0 cm. Their central electric field may be treated as uniform.

The left plate is at 9.0 V, and the right plate is at 3.0 V.

A voltmeter also reads 6.0 V at a point halfway between them.

Calculate:

a) The electric field magnitude in the central region.

b) The direction of the electric field.

c) The potential difference between the midpoint and the right plate, and explain whether it agrees with the uniform-field model.

Solution 1

The electric field is \(100\ \text{V/m}\), directed to the right. The midpoint reading is consistent with a uniform field.

a) Field magnitude

The potential difference is:

\[ V=9.0-3.0=6.0\ \text{V} \]

The plate separation is \(0.060\ \text{m}\).

\[ \begin{aligned} E&=\frac{V}{d}\\ &=\frac{6.0}{0.060}\\ &=100\ \text{V/m} \end{aligned} \]

b) Field direction

Electric field points towards decreasing potential.

The left plate is at 9.0 V and the right plate is at 3.0 V, so the field points right.

c) Midpoint comparison

The potential difference from the midpoint to the right plate is:

\[ 6.0-3.0=3.0\ \text{V} \]

Across the 0.030 m interval:

\[ \frac{3.0}{0.030} =100\ \text{V/m} \]

This agrees with the whole-gap calculation.

The tempting mistake is to treat agreement at three positions as proof of perfect uniformity. It supports the stated approximation, but measurements at only three positions cannot establish the complete field distribution.

Question 2

An isolated positive point charge creates the following electric potentials:

• At point \(A\), 3.0 cm from the charge: 90 V.
• At point \(B\), 9.0 cm from the charge: 30 V.

Both points lie along the same radial line.

A student calculates \(60/0.060=1000\ \text{V/m}\) and claims this is the electric field at the midpoint, 6.0 cm from the source.

Assume an isolated point charge in vacuum, with potential zero at infinity.

a) Calculate the actual field magnitude at the midpoint.

b) Calculate the field magnitude at \(A\) and \(B\).

c) Find the distance from the source where the local field magnitude really equals \(1000\ \text{V/m}\).

d) Explain why that distance is not the midpoint.

Solution 2

The student’s midpoint claim is incorrect. The field at the midpoint is \(750\ \text{V/m}\), and the local field equals \(1000\ \text{V/m}\) at approximately 5.2 cm from the charge.

a) Midpoint field

For a positive point charge:

\[ \phi=\frac{kQ}{r} \]

At \(A\):

\[ \begin{aligned} kQ&=\phi_A r_A\\ &=(90)(0.030)\\ &=2.7\ \text{V m} \end{aligned} \]

The midpoint is at \(r=0.060\ \text{m}\).

Using the inverse-square field equation:

\[ \begin{aligned} E_{\text{mid}} &=\frac{kQ}{r^2}\\ &=\frac{2.7}{(0.060)^2}\\ &=750\ \text{V/m} \end{aligned} \]

b) Endpoint fields

At \(A\):

\[ \begin{aligned} E_A&=\frac{2.7}{(0.030)^2}\\ &=3000\ \text{V/m} \end{aligned} \]

At \(B\):

\[ \begin{aligned} E_B&=\frac{2.7}{(0.090)^2}\\ &\approx333\ \text{V/m} \end{aligned} \]

The field is nine times stronger at \(A\), because \(B\) is three times farther from the source.

c) Where is the field \(1000\ \text{V/m}\)?

Rearrange the inverse-square relationship:

\[ r=\sqrt{\frac{kQ}{E}} \]

Substitute:

\[ \begin{aligned} r&=\sqrt{\frac{2.7}{1000}}\ \text{m}\\ &\approx0.052\ \text{m}\\ &=5.2\ \text{cm} \end{aligned} \]

d) Why not the midpoint?

The field decreases according to \(1/r^2\), not linearly with distance.

The midpoint of two positions is therefore not necessarily the position where the local field equals its average value over the interval.

The original calculation of \(1000\ \text{V/m}\) was valid as an average outward radial field component from \(A\) to \(B\). The error was assigning that average to a particular location without checking the field’s distance relationship.

Question 3

A student investigates an electric field between two differently shaped electrodes maintained at 10.0 V on the left and 0 V on the right.

The electrodes are 10.0 cm apart horizontally. Because their shapes are asymmetric, the field may have horizontal and vertical components.

Potential measurements are taken along two horizontal lines:

• A middle line at \(y=0\).
• An upper line at \(y=2.0\ \text{cm}\).

The coordinate \(x\) increases to the right, and \(y\) increases upwards.

The measurements are:

a) Calculate the average horizontal electric field component between \(x=0\) and \(x=5.0\ \text{cm}\) along each measurement line.

b) At \(x=5.0\ \text{cm}\), estimate the average vertical field component between the two measurement lines. State its direction.

c) Explain why the student cannot conclude that the electric field everywhere is \(100\ \text{V/m}\) directed horizontally.

d) Calculate the work done by the electric field when a positive \(2.0\ \text{nC}\) test charge moves from the left electrode to the right electrode. Explain whether the result depends on which route it takes, assuming a stationary electrostatic field.

Solution 3

The field is not purely horizontal and uniform throughout the measured region. The data reveal different horizontal gradients and a downward average vertical component, although the work done between the electrodes is still \(2.0\times10^{-8}\ \text{J}\).

a) Horizontal components

For the middle line:

\[ \begin{aligned} E_{x,\text{mid,avg}} &=-\frac{5.0-10.0}{0.050}\\ &=+100\ \text{V/m} \end{aligned} \]

For the upper line:

\[ \begin{aligned} E_{x,\text{upper,avg}} &=-\frac{6.5-10.0}{0.050}\\ &=+70\ \text{V/m} \end{aligned} \]

Both average components point to the right, but their magnitudes differ.

b) Vertical component

At \(x=5.0\ \text{cm}\), the potential increases from 5.0 V to 6.5 V as we move upwards through 0.020 m.

\[ \begin{aligned} E_{y,\text{avg}} &=-\frac{6.5-5.0}{0.020}\\ &=-75\ \text{V/m} \end{aligned} \]

The negative result means the average vertical field component points downwards.

This is an average between the measurement lines, not necessarily the exact vertical field at either endpoint.

c) Why the uniform-field conclusion fails

The whole-gap voltage divided by horizontal separation is:

\[ \frac{10.0}{0.100} =100\ \text{V/m} \]

That gives the average horizontal component across the full gap.

But the upper line has an average horizontal component of only \(70\ \text{V/m}\) over its first half, and the vertical measurements reveal another field component.

A purely horizontal, uniform field could not produce those differences.

The tempting route is to use the electrode voltages alone. The missing constraint is the field’s spatial distribution, which depends on the electrode geometry.

We also cannot combine these finite-difference averages to obtain an exact local field magnitude at one point. They describe averages over different intervals.

d) Work done

For a charge moving between two potentials, the work done by the electrostatic field is:

\[ W_{\text{field}}=q(\phi_A-\phi_B) \]

Here:

\[ q=2.0\ \text{nC} =2.0\times10^{-9}\ \text{C} \]

Therefore:

\[ \begin{aligned} W_{\text{field}} &=(2.0\times10^{-9})(10.0-0)\\ &=2.0\times10^{-8}\ \text{J} \end{aligned} \]

The work is positive because the positive charge moves from higher to lower potential.

It does not depend on which route is taken between the same endpoints in a stationary electrostatic field.

This is an important contrast: the local electric force can vary along different routes, even though the total work between the same endpoints is unchanged.

Question 4

An investigator measures electric potential inside a 4.0 cm horizontal gap.

The left boundary is at 12.0 V, the right boundary is at 0 V, and the midpoint is at 6.0 V.

Two models could explain these readings.

Model A: A uniform electric field directed right, with magnitude \(300\ \text{V/m}\).

Model B: A one-dimensional field directed right, but with different strengths in different sections. The first 1.0 cm and final 1.0 cm each have a uniform field of \(100\ \text{V/m}\). The central 2.0 cm has an unknown uniform field magnitude \(E_c\).

For Model B, assume thin, fixed, charged sheets at \(x=1.0\ \text{cm}\) and \(x=3.0\ \text{cm}\) maintain the changes in field. Ignore their thickness.

a) Calculate \(E_c\) so Model B matches the total potential difference.

b) Show that Model B also predicts 6.0 V at the midpoint.

c) Determine the potential at \(x=1.0\ \text{cm}\) in each model. Explain how this measurement could distinguish them.

d) A positive test charge of \(3.0\ \text{nC}\) is placed at \(x=0.50\ \text{cm}\). Compare the electric forces in the two models.

e) Calculate the work done by the field when this charge moves from the left boundary to the right boundary in each model.

Solution 4

Both models reproduce all three original voltage readings, even though their local forces differ. An additional measurement at \(x=1.0\ \text{cm}\) would distinguish them.

a) Central field in Model B

The total potential drop is 12.0 V.

Each outer section contributes:

\[ \Delta\phi_{\text{outer}} =(100)(0.010)=1.0\ \text{V} \]

The two outer sections account for 2.0 V altogether.

Therefore, the central section must account for:

\[ 12.0-2.0=10.0\ \text{V} \]

Across 0.020 m:

\[ \begin{aligned} E_c&=\frac{10.0}{0.020}\\ &=500\ \text{V/m} \end{aligned} \]

b) Midpoint potential

From the left boundary to the midpoint, Model B has:

• A 1.0 V drop through the first outer section.
• A 5.0 V drop through the first 1.0 cm of the central section.

The total drop is:

\[ 1.0+5.0=6.0\ \text{V} \]

Hence:

\[ \phi_{\text{mid}}=12.0-6.0=6.0\ \text{V} \]

Model B matches the midpoint reading.

This is the first surprise: agreement with the total voltage and midpoint voltage does not establish uniformity.

c) An additional measurement

In Model A, the potential drop over the first 1.0 cm is:

\[ \Delta\phi=(300)(0.010)=3.0\ \text{V} \]

So:

\[ \phi_A(1.0\ \text{cm})=9.0\ \text{V} \]

In Model B, the first section loses only 1.0 V:

\[ \phi_B(1.0\ \text{cm})=11.0\ \text{V} \]

A measurement at this position distinguishes the models:

• Model A predicts 9.0 V.
• Model B predicts 11.0 V.

The tempting assumption is that three points on a straight potential-versus-position line force the entire graph to be straight. They don’t. Different potential profiles can pass through the same measured points.

d) Force on the test charge

The field is:

\[ F=qE \]

With \(q=3.0\times10^{-9}\ \text{C}\), Model A gives:

\[ \begin{aligned} F_A&=(3.0\times10^{-9})(300)\\ &=9.0\times10^{-7}\ \text{N} \end{aligned} \]

Model B has \(100\ \text{V/m}\) at the specified position, so:

\[ \begin{aligned} F_B&=(3.0\times10^{-9})(100)\\ &=3.0\times10^{-7}\ \text{N} \end{aligned} \]

Both forces point right, but Model A produces three times the force at this position.

e) Work done across the gap

Both models have the same endpoint potential difference of 12.0 V.

Therefore:

\[ \begin{aligned} W_{\text{field}} &=q(\phi_A-\phi_B)\\ &=(3.0\times10^{-9})(12.0)\\ &=3.6\times10^{-8}\ \text{J} \end{aligned} \]

The work done is identical in both models.

This is possible because work depends on the total potential difference, while the instantaneous force depends on the local electric field.

The fixed charged sheets in Model B are an essential assumption. They allow the one-dimensional field to change between sections. Without those extra sources, an ideal charge-free parallel-plate arrangement would have additional constraints on its field.

Question 5

Extension challenge – a zero potential difference that hides two opposing fields

A positive point charge \(+Q\) is fixed at the origin of a horizontal \(x\)-axis.

A second, uniform electric field is applied throughout the region, directed towards the negative \(x\)-direction.

For positions on the positive \(x\)-axis, the two electric fields therefore point in opposite directions.

Define two points:

• \(A\) is at \(x=r=2.0\ \text{cm}\).
• \(C\) is at \(x=4r=8.0\ \text{cm}\).

The potential due to the point charge alone is 12.0 V at \(A\). Use the isolated point-charge potential convention for this contribution.

When the uniform field is also present, the total electric potentials at \(A\) and \(C\) are equal.

Assume the source charge and uniform background field remain fixed. A tiny positive test charge \(+q\) does not disturb them, and all motion is non-relativistic.

a) Calculate the magnitude of the uniform background field.

b) Find the point \(B\), between \(A\) and \(C\), where the net electric field is zero.

c) A positive test charge is released from rest at \(A\) and is constrained to move only along the \(x\)-axis. Does it reach \(C\)? Does it stop at \(B\)? Explain using both electric force and conservation of energy.

d) Now place the positive test charge at rest exactly at \(B\), without the one-dimensional constraint. If it is displaced slightly upwards and released from rest, will the electric force return it towards \(B\)? Explain.

Solution 5

The uniform background field has magnitude \(150\ \text{V/m}\), the net field is zero at \(x=4.0\ \text{cm}\), and a positive charge released from rest at \(A\) can reach \(C\), passing through \(B\) with its maximum speed. However, the equilibrium at \(B\) is unstable to small sideways displacements.

The crucial distinction is that equal endpoint potentials do not imply zero electric field between them.

a) Magnitude of the background field

For the positive point charge:

\[ \phi_Q=\frac{kQ}{r} \]

At \(A\), its potential contribution is 12.0 V.

At \(C\), which is four times farther away:

\[ \phi_Q(C)=\frac{12.0}{4}=3.0\ \text{V} \]

The point-charge potential therefore decreases by 9.0 V between \(A\) and \(C\).

The uniform background field points left, so its potential increases as \(x\) increases.

For the total potentials at \(A\) and \(C\) to be equal, that increase must be exactly 9.0 V.

The separation is:

\[ \begin{aligned} \Delta x&=4r-r=3r\\ &=0.060\ \text{m} \end{aligned} \]

Hence the background field magnitude is:

\[ \begin{aligned} E_0&=\frac{9.0}{0.060}\\ &=150\ \text{V/m} \end{aligned} \]

Its direction is left, as stated in the question.

b) Position where the net field is zero

For the point charge, the field at \(A\) is:

\[ \begin{aligned} E_Q(A)&=\frac{\phi_Q(A)}{r}\\ &=\frac{12.0}{0.020}\\ &=600\ \text{V/m} \end{aligned} \]

This field points right.

At distance \(x\), the inverse-square relationship gives:

\[ E_Q(x)=600\left(\frac{0.020}{x}\right)^2 \]

The two fields cancel when:

\[ E_Q(x)=150\ \text{V/m} \]

Therefore:

\[ 600\left(\frac{0.020}{x}\right)^2=150 \]

Simplifying:

\[ \begin{aligned} \left(\frac{0.020}{x}\right)^2&=\frac14\\ x&=0.040\ \text{m}\\ &=4.0\ \text{cm} \end{aligned} \]

The cancellation point \(B\) is at \(x=2r\), halfway between the origin and \(C\), but not halfway between \(A\) and \(C\).

At \(B\), the rightward point-charge field and the leftward uniform field have equal magnitudes.

c) Motion of a positive charge released from rest at A

The charge reaches \(C\), where it momentarily stops before reversing direction. It does not stop at \(B\).

First examine the forces.

At \(A\), the net field is:

\[ E_{\text{net}}(A)=600-150=450\ \text{V/m} \]

It points right, so the positive charge accelerates towards \(B\).

At \(B\), the net field is zero.

But zero force means zero instantaneous acceleration, not necessarily zero velocity. A particle already moving can continue through a zero-force point.

To find what happens to its speed, consider electric potential.

Choose a convenient additive reference so that the background potential contribution is \(E_0x\). The total potential along the positive \(x\)-axis is then:

\[ \phi(x)=\frac{kQ}{x}+E_0x \]

At \(A\), the two contributions are:

\[ \begin{aligned} \phi_Q(A)&=12.0\ \text{V}\\ \phi_0(A)&=(150)(0.020)=3.0\ \text{V} \end{aligned} \]

So:

\[ \phi(A)=15.0\ \text{V} \]

At \(B=0.040\ \text{m}\):

\[ \begin{aligned} \phi_Q(B)&=6.0\ \text{V}\\ \phi_0(B)&=6.0\ \text{V} \end{aligned} \]

Thus:

\[ \phi(B)=12.0\ \text{V} \]

At \(C=0.080\ \text{m}\):

\[ \begin{aligned} \phi_Q(C)&=3.0\ \text{V}\\ \phi_0(C)&=12.0\ \text{V} \end{aligned} \]

Therefore:

\[ \phi(C)=15.0\ \text{V} \]

The potential decreases from 15.0 V at \(A\) to 12.0 V at \(B\), then increases to 15.0 V at \(C\).

For a positive charge, electric potential energy is \(U=q\phi\). As the charge moves from \(A\) to \(B\), it loses potential energy and gains kinetic energy.

Its maximum kinetic energy at \(B\) is:

\[ \begin{aligned} K_B&=q[\phi(A)-\phi(B)]\\ &=q(3.0\ \text{V}) \end{aligned} \]

Since \(q\) is measured in coulombs, this expression gives kinetic energy in joules.

The particle has its greatest speed at \(B\), even though the net force is zero there.

Beyond \(B\), the point-charge field weakens below \(150\ \text{V/m}\). The background field dominates, producing a net leftward force.

The particle continues right but slows down.

At \(C\), its potential energy has returned to its original value at \(A\). With no energy losses:

\[ K_A+q\phi(A)=K_C+q\phi(C) \]

Because \(K_A=0\) and \(\phi(A)=\phi(C)\):

\[ K_C=0 \]

The particle reaches \(C\) with zero speed and then accelerates back towards the left.

The tempting conclusion is that zero potential difference means no motion. The missing detail is the changing electric field and potential between the endpoints.

d) Stability of the equilibrium point

A small upward displacement does not produce a restoring force. The positive charge is pushed farther away from the axis.

At \(B\), the horizontal components cancel exactly.

Now move the test charge a tiny distance upwards.

The uniform background field still points horizontally left. It has no upward or downward component.

But the electric field from \(+Q\) points radially away from the source. At the displaced position, it has an upward component.

The positive test charge therefore experiences an upward force, in the same direction as its displacement.

It moves farther from \(B\), rather than returning.

There is a subtle distinction here.

Along the positive \(x\)-axis alone, \(B\) is a stable equilibrium for a positive test charge: small horizontal displacements produce restoring forces.

In three-dimensional space, it is not stable because sideways displacements produce forces away from the axis.

This is the conceptual high point of the problem. The same location can have zero net electric field and be stable for motion in one direction, while being unstable for motion in another.

Zero field at a point does not establish uniformity, zero potential, zero velocity, or stability in every direction.

08What to do when \(E=V/d\) is not enough

When you meet a new electric-field problem, identify the source geometry before selecting an equation.

Large parallel plates may justify a nearly uniform field in their central region. A point charge produces an inverse-square field. Several charges or electrodes may produce a field that changes in both magnitude and direction.

Potential differences remain useful even when the field is non-uniform. They tell you about changes in electric potential energy and the total work done by an electrostatic field between two positions.

What they do not automatically tell you is the local force on a charge at every point between those positions.

The next useful step is electric-field superposition: calculating how individual fields combine, including where they reinforce or oppose one another. Once you can find the net local field, you can connect it to force using \(\vec F=q\vec E\), while using potential differences separately to understand energy and motion.