How Plate Separation Changes Electric Field at Fixed Voltage

A power supply maintains a potential difference of 600 V between two parallel metal plates. You pull the plates further apart, doubling their separation, but the voltmeter still reads 600 V.

Has the electric field changed?

Yes. The electric field strength halves when the plate separation doubles at constant voltage. The same potential difference is now spread across twice the distance, so the potential changes less rapidly with position.

For example:

• At a separation of 10 mm, the field is \(60\,000\ \text{V m}^{-1}\).
• At a separation of 20 mm, the field is \(30\,000\ \text{V m}^{-1}\).

The voltage hasn’t changed, but the field has.

Before looking at the mathematics, predict something else: what happens to the force on a small test charge placed between the plates? It also halves, provided the charge stays the same. The reason is that electric force depends on both the charge and the electric field.

01What happens when plate separation increases?

Start with two large, flat, parallel conducting plates.

One plate is positively charged, and the other is negatively charged. Their potential difference is maintained by a connected power supply.

In the central region between the plates, the electric field is approximately uniform. This means its magnitude and direction are nearly constant from one position to another.

For this arrangement, the magnitude of the electric field is:

\[ E\approx\frac{\Delta V}{d} \]

where:

• \(E\) is electric field strength, measured in \(\text{V m}^{-1}\) or \(\text{N C}^{-1}\).
• \(\Delta V\) is the potential difference between the plates, measured in volts (V).
• \(d\) is the perpendicular separation between the plates, measured in metres (m).

The important part is what happens when voltage stays constant.

Rearranging the relationship conceptually:

\[ E\propto\frac{1}{d} \qquad \text{(constant voltage)} \]

This is an inverse relationship, not an inverse-square relationship.

The direction of the field does not change if the plate polarities stay the same. It still points from the positive plate towards the negative plate.

There is an important distinction here: a uniform field is constant across the central region of a particular arrangement. It does not mean the field strength stays constant when you change the arrangement itself.

For a fuller explanation of the underlying relationship, see Electric Fields Between Parallel Plates: Using E = V/d.

02Why does the field change if the voltage stays the same?

Think about what voltage tells you.

Potential difference measures the change in electric potential energy per unit charge between two locations. It tells you how much energy is transferred per coulomb of charge moving between those locations.

Electric field strength describes how rapidly electric potential changes with position.

In a uniform field, the magnitude of the field is the potential difference divided by the distance over which that change occurs.

Imagine two arrangements:

• Arrangement A: a potential difference of 600 V across 10 mm.
• Arrangement B: the same potential difference of 600 V across 20 mm.

In both arrangements, a positive test charge experiences the same total potential difference when travelling from one plate to the other.

However, in arrangement B, it travels twice as far to experience that same change.

The potential therefore changes less rapidly with distance.

This is why the field is weaker.

The potential gradient explains the result

There is a more precise way to express this relationship.

Take the positive \(x\)-direction to point from the positive plate towards the negative plate. Electric potential decreases in this direction.

For a uniform field:

\[ E_x=-\frac{\Delta V}{\Delta x} \]

The negative sign indicates that the electric field points towards decreasing electric potential.

For field strength, we use the magnitude:

\[ E=\left|\frac{\Delta V}{\Delta x}\right| \]

The field is therefore related to the slope of a potential-versus-position graph.

This gives you a useful way to interpret the change:

• Same voltage change over a shorter distance means a stronger field.
• Same voltage change over a longer distance means a weaker field.

Notice that the field depends on the change in potential with position, not simply the numerical potential at a particular point.

03Is electric field strength independent of the test charge?

Yes. This is an important distinction.

Electric field strength is a property of the source charges and their arrangement. It does not depend on which test charge you place in the field.

We define electric field strength as the force per unit positive test charge:

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

where:

• \(\vec E\) is the electric field vector.
• \(\vec F\) is the electric force acting on the test charge.
• \(q\) is the test charge, measured in coulombs (C).

The test charge is assumed to be small enough that it does not significantly disturb the original electric field.

Consider placing an electron and a proton at the same point between the plates.

Both experience the same electric field because the field is determined by the plates.

However, their forces are different in direction because their charges have opposite signs.

The relationship is:

\[ \vec F=q\vec E \]

For a positive charge, the force is in the direction of the field.

For a negative charge, the force is opposite to the field.

This distinction matters when plate separation changes. The field strength changes because the plate arrangement has changed. The force on a particular test charge then changes accordingly.

04How could you investigate this experimentally?

Suppose you want to test whether doubling plate separation really halves the electric field.

You need to change the separation while keeping the potential difference constant, then measure how rapidly electric potential changes across the gap.

A useful investigation involves two parallel conducting plates, an adjustable support, a regulated power supply, and a potential probe.

Setting up the apparatus

Place two large, flat conducting plates facing each other.

Connect one plate to the positive terminal of a regulated DC power supply. Connect the other plate to the negative or grounded terminal.

A voltmeter connected across the plates checks that the potential difference remains constant.

Next, connect a second high-impedance voltmeter to a small potential probe. Use the negative plate as the reference connection, and move the probe along the central line between the plates.

A ruler or micrometer measures the perpendicular separation.

What should the investigation show?

Keep the power supply at 12.0 V.

First, set the plate separation to 10.0 mm.

Then increase the separation to 20.0 mm without disconnecting the power supply.

At each separation, record the potential at several positions along the central line.

The expected pattern for an ideal uniform field is:

These are illustrative ideal values, not claimed experimental measurements.

Notice that the potential difference remains 12.0 V in every trial.

However, the potential changes less per millimetre as the separation increases.

That is the evidence supporting the inverse relationship between field strength and separation at fixed voltage.

Turning a potential measurement into a field measurement

Consider the second trial, where the plates are separated by 20.0 mm.

Convert the distance into metres:

\[ d=20.0\times10^{-3}=0.0200\ \text{m} \]

Calculate the field:

\[ \begin{aligned} E&=\frac{\Delta V}{d}\\ &=\frac{12.0}{0.0200}\\ &=600\ \text{V m}^{-1} \end{aligned} \]

The field strength is \(600\ \text{V m}^{-1}\).

Compare this with the first trial, which produced \(1200\ \text{V m}^{-1}\).

Doubling the gap has halved the field.

What could go wrong with the investigation?

A voltmeter connected only across the plates is not enough to establish the electric field.

It confirms the potential difference, but it does not measure how that potential changes across the gap.

The improved method is to measure potential at several positions and calculate the gradient.

There are also practical limitations:

• The plates must be large compared with their separation.
• Measurements should be taken near the centre, away from the edges.
• The potential probe should be small and have a high-impedance input to minimise disturbance.
• The supply must remain connected so that the potential difference is maintained.

Real plates have finite dimensions, so their fields are not perfectly uniform. Near the edges, field lines curve outwards. This is called fringing.

For more detail about the limits of the uniform-field approximation, see Uniform Electric Fields Between Parallel Plates: HSC Physics.

05Don’t confuse inverse and inverse-square relationships

This is a common source of mistakes in HSC Physics.

You may have encountered another electric field relationship:

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

This describes the electric field strength around an isolated point charge.

Here:

• \(E\) is electric field strength.
• \(k\) is Coulomb’s constant, approximately \(9.0\times10^9\ \text{N m}^2\text{C}^{-2}\).
• \(Q\) is the magnitude of the source charge.
• \(r\) is the radial distance from the point charge.

For a fixed point charge, doubling the distance makes the electric field one-quarter as strong.

Why is this different?

A point charge produces a radial electric field. Its field spreads out in three dimensions as you move further away from the charge.

Parallel plates produce a different field arrangement. In their central region, the field is approximately uniform. The field strength is determined by the potential difference across the gap.

Compare the three situations:

The third situation is worth understanding.

If you disconnect a charged capacitor from its power supply before moving the plates, its charge is approximately fixed. You can no longer assume the voltage stays constant.

For ideal large plates with unchanged area, the field is approximately unchanged as the separation increases, while the potential difference increases.

This is a different physical condition from the question in this article.

Always identify what is being held constant before choosing an electric field relationship.

06Worked examples

07Questions and solutions

Question 1

Two large parallel-plate arrangements have the same potential difference of 300 V.

• Arrangement A has a separation of 5.0 mm.
• Arrangement B has a separation of 12.0 mm.

Calculate the electric field strength in each arrangement and determine which produces the stronger field. By what factor is it stronger?

Assume both arrangements have approximately uniform central fields.

Solution 1

Arrangement A produces the stronger field, with a field strength 2.4 times that of arrangement B.

Use the uniform-field relationship:

\[ E=\frac{\Delta V}{d} \]

First, convert the separations:

\[ d_A=0.0050\ \text{m} \]

\[ d_B=0.012\ \text{m} \]

Calculate the field in arrangement A:

\[ \begin{aligned} E_A&=\frac{300}{0.0050}\\ &=6.0\times10^4\ \text{V m}^{-1} \end{aligned} \]

Calculate the field in arrangement B:

\[ \begin{aligned} E_B&=\frac{300}{0.012}\\ &=2.5\times10^4\ \text{V m}^{-1} \end{aligned} \]

Now compare the field strengths:

\[ \frac{E_A}{E_B} =\frac{6.0\times10^4}{2.5\times10^4} =2.4 \]

Arrangement A has a field 2.4 times stronger than arrangement B.

Both have the same voltage, so the smaller separation produces the stronger field.

Question 2

Two parallel conducting plates maintain a potential difference of 360 V.

Initially, their separation is 6.0 mm. The separation is then increased to 15.0 mm while the power supply remains connected.

A proton and an electron are separately placed at the same point in the central region between the plates.

Calculate:

a) The electric field strength before and after increasing the separation.

b) The magnitude of the electric force on each particle before and after the change.

c) The direction of the force on each particle, given that the left plate is positive and the right plate is negative.

Use the magnitude of the elementary charge:

\[ e=1.60\times10^{-19}\ \text{C} \]

Solution 2

The electric field decreases from \(6.0\times10^4\) to \(2.4\times10^4\ \text{V m}^{-1}\). Both particles experience the same field, but their forces act in opposite directions.

Part a: Calculate the electric field.

Initially:

\[ \begin{aligned} E_1&=\frac{360}{6.0\times10^{-3}}\\ &=6.0\times10^4\ \text{V m}^{-1} \end{aligned} \]

After increasing the separation:

\[ \begin{aligned} E_2&=\frac{360}{15.0\times10^{-3}}\\ &=2.4\times10^4\ \text{V m}^{-1} \end{aligned} \]

The field becomes 40% of its original value.

Part b: Calculate the electric forces.

The magnitude of the charge on both particles is the same.

Use:

\[ F=|q|E \]

Initially:

\[ \begin{aligned} F_1&=(1.60\times10^{-19})(6.0\times10^4)\\ &=9.6\times10^{-15}\ \text{N} \end{aligned} \]

After the separation increases:

\[ \begin{aligned} F_2&=(1.60\times10^{-19})(2.4\times10^4)\\ &=3.84\times10^{-15}\ \text{N} \end{aligned} \]

Both particles experience the same force magnitude in each arrangement.

Part c: Determine the force directions.

The electric field points from the positive left plate towards the negative right plate.

• The proton is positively charged, so its force acts towards the right, in the direction of the field.
• The electron is negatively charged, so its force acts towards the left, opposite to the field.

The field itself is independent of the test particle. It is the sign of the particle’s charge that determines the direction of its force.

Question 3

An isolated positive point charge has a magnitude of \(5.0\ \text{nC}\).

Calculate the electric field strength at distances of 0.10 m and 0.20 m from the charge.

Now consider a separate parallel-plate arrangement with a fixed potential difference of 500 V.

Its separation increases from 5.0 mm to 10.0 mm.

Calculate the field strength before and after the change.

Finally, explain why doubling the distance produces different changes in the two electric fields.

Use:

\[ k=9.0\times10^9\ \text{N m}^2\text{C}^{-2} \]

Solution 3

Doubling the distance from the point charge reduces its field to one-quarter, whereas doubling the plate separation at fixed voltage halves the field.

These results follow from two different physical models.

Part a: Electric field from the point charge.

Use the inverse-square relationship:

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

Convert the charge:

\[ Q=5.0\times10^{-9}\ \text{C} \]

At \(r=0.10\ \text{m}\):

\[ \begin{aligned} E_1&=\frac{(9.0\times10^9)(5.0\times10^{-9})}{(0.10)^2}\\ &=\frac{45}{0.010}\\ &=4.5\times10^3\ \text{N C}^{-1} \end{aligned} \]

At \(r=0.20\ \text{m}\):

\[ \begin{aligned} E_2&=\frac{45}{(0.20)^2}\\ &=\frac{45}{0.040}\\ &=1.125\times10^3\ \text{N C}^{-1} \end{aligned} \]

The field becomes one-quarter of its original value.

Part b: Electric field between parallel plates.

For the first arrangement:

\[ \begin{aligned} E_1&=\frac{500}{5.0\times10^{-3}}\\ &=1.0\times10^5\ \text{V m}^{-1} \end{aligned} \]

For the second arrangement:

\[ \begin{aligned} E_2&=\frac{500}{10.0\times10^{-3}}\\ &=5.0\times10^4\ \text{V m}^{-1} \end{aligned} \]

The field becomes half its original value.

Part c: Explain the difference.

The point-charge field spreads radially outwards. Its strength decreases with the square of the distance from the source charge.

The field between parallel plates is approximately uniform in the central region. When the voltage is fixed, increasing the separation spreads the same potential difference over a longer distance.

Therefore:

• Point charge: \(E\propto 1/r^2\).
• Parallel plates at fixed voltage: \(E\propto 1/d\).

Do not apply the inverse-square equation to the separation between parallel plates. The two arrangements have different field geometries.

Question 4

A student investigates the electric field between parallel plates using a regulated power supply and a high-impedance potential probe.

The supply maintains a constant potential difference of 12.0 V.

The probe measures electric potential at several positions along the central line between the plates.

The following illustrative readings are obtained. The distance \(x\) is measured from the positive plate.

A student argues:

“The potential at the midpoint is 6.0 V in both trials, so the electric field must be the same.”

a) Calculate the average electric field strength in each trial using the first and last probe readings.

b) Explain why the student’s conclusion is incorrect.

c) Identify one experimental limitation and explain how it could affect the results.

Solution 4

The student’s conclusion is incorrect. Trial A has an average field of \(1.2\times10^3\ \text{V m}^{-1}\), while trial B has an average field of \(6.0\times10^2\ \text{V m}^{-1}\).

The potential at one position does not tell us the electric field strength. We need to know how rapidly potential changes with distance.

Part a: Calculate the field in trial A.

The potential changes from 9.6 V to 2.4 V.

Therefore:

\[ \Delta V=9.6-2.4=7.2\ \text{V} \]

The distance between the two measurement positions is:

\[ \Delta x=8.0-2.0=6.0\ \text{mm} \]

Convert this into metres:

\[ \Delta x=0.0060\ \text{m} \]

Calculate the field magnitude:

\[ \begin{aligned} E_A&=\frac{\Delta V}{\Delta x}\\ &=\frac{7.2}{0.0060}\\ &=1.2\times10^3\ \text{V m}^{-1} \end{aligned} \]

Part b: Calculate the field in trial B.

The potential change is again:

\[ \Delta V=9.6-2.4=7.2\ \text{V} \]

However, the measurement positions are further apart:

\[ \Delta x=16.0-4.0=12.0\ \text{mm} \]

Convert into metres:

\[ \Delta x=0.0120\ \text{m} \]

Therefore:

\[ \begin{aligned} E_B&=\frac{7.2}{0.0120}\\ &=6.0\times10^2\ \text{V m}^{-1} \end{aligned} \]

The field in trial B is half the field in trial A.

Part c: Explain the student’s mistake.

The student has confused electric potential with electric field strength.

Both trials have a potential of 6.0 V at the midpoint because the midpoint is halfway across the potential difference in both arrangements.

However, the field depends on the potential gradient, not the potential at one particular point.

In trial B, the same potential change occurs across twice the distance. This produces half the field.

One experimental limitation is that the potential probe may disturb the electric field slightly. Using a small probe with a high-impedance voltmeter reduces this effect.

Measurements should also be taken away from the plate edges, where fringing makes the field nonuniform.

Question 5

Two parallel-plate arrangements, A and B, are connected to regulated power supplies maintaining the same potential difference of 480 V.

Their plate separations are:

• Arrangement A: 12 mm.
• Arrangement B: 24 mm.

A proton is released from rest in each arrangement. In both cases, its initial position is exactly 3.0 mm from the positive plate.

The proton moves towards the negative plate.

Determine:

a) The electric field strength in each arrangement.

b) The acceleration of the proton in each arrangement.

c) The potential difference through which the proton moves before reaching the negative plate.

d) The final speed of the proton in each arrangement.

e) The time taken for the proton to reach the negative plate in each arrangement.

Finally, explain why the proton in the wider-gap arrangement can reach a greater final speed despite experiencing a weaker electric field.

Use:

\[ e=1.60\times10^{-19}\ \text{C} \]

\[ m_p=1.67\times10^{-27}\ \text{kg} \]

Assume:

• Both plate arrangements produce uniform fields.
• The power supplies maintain 480 V throughout.
• Gravity and other forces are negligible.
• The proton remains non-relativistic.

Solution 5

The proton in arrangement B reaches a greater final speed, approximately \(2.84\times10^5\ \text{m s}^{-1}\), compared with \(2.63\times10^5\ \text{m s}^{-1}\) in arrangement A. However, it takes longer to arrive.

This is the most interesting part of the problem.

A weaker electric field means a smaller acceleration. It is tempting to conclude that the proton must also reach a lower final speed.

That conclusion ignores how far the proton travels and how much potential difference it experiences.

Part a: Calculate the electric field strength.

For arrangement A:

\[ \begin{aligned} E_A&=\frac{480}{12\times10^{-3}}\\ &=4.0\times10^4\ \text{V m}^{-1} \end{aligned} \]

For arrangement B:

\[ \begin{aligned} E_B&=\frac{480}{24\times10^{-3}}\\ &=2.0\times10^4\ \text{V m}^{-1} \end{aligned} \]

The field in arrangement B is half as strong.

Part b: Calculate the proton’s acceleration.

Use the electric force relationship:

\[ F=qE \]

Then apply Newton’s second law:

\[ F=ma \]

Combining these:

\[ a=\frac{qE}{m} \]

For arrangement A:

\[ \begin{aligned} a_A&=\frac{(1.60\times10^{-19})(4.0\times10^4)} {1.67\times10^{-27}}\\ &=3.83\times10^{12}\ \text{m s}^{-2} \end{aligned} \]

For arrangement B:

\[ \begin{aligned} a_B&=\frac{(1.60\times10^{-19})(2.0\times10^4)} {1.67\times10^{-27}}\\ &=1.92\times10^{12}\ \text{m s}^{-2} \end{aligned} \]

The proton’s acceleration is also halved.

Part c: Calculate the distance travelled and potential difference experienced.

This is where the important detail appears.

The proton starts 3.0 mm from the positive plate in both arrangements.

For arrangement A, the remaining distance is:

\[ s_A=12-3=9.0\ \text{mm} \]

For arrangement B:

\[ s_B=24-3=21.0\ \text{mm} \]

The proton travels a much greater distance in arrangement B.

Because the field is uniform, the potential drop along the direction of travel is proportional to the distance travelled.

For arrangement A:

\[ \begin{aligned} \Delta V_A&=E_As_A\\ &=(4.0\times10^4)(9.0\times10^{-3})\\ &=360\ \text{V} \end{aligned} \]

For arrangement B:

\[ \begin{aligned} \Delta V_B&=E_Bs_B\\ &=(2.0\times10^4)(21.0\times10^{-3})\\ &=420\ \text{V} \end{aligned} \]

The proton experiences a larger potential drop in arrangement B.

This happens because the starting point is the same absolute distance from the positive plate, not the same fraction of the gap.

Part d: Calculate the final speed.

The proton starts from rest.

The work done by the electric field becomes kinetic energy:

\[ q\Delta V=\frac12 mv^2 \]

Rearranging:

\[ v=\sqrt{\frac{2q\Delta V}{m}} \]

For arrangement A:

\[ \begin{aligned} v_A&=\sqrt{ \frac{2(1.60\times10^{-19})(360)} {1.67\times10^{-27}} }\\ &=2.63\times10^5\ \text{m s}^{-1} \end{aligned} \]

For arrangement B:

\[ \begin{aligned} v_B&=\sqrt{ \frac{2(1.60\times10^{-19})(420)} {1.67\times10^{-27}} }\\ &=2.84\times10^5\ \text{m s}^{-1} \end{aligned} \]

The proton reaches a greater final speed in arrangement B.

The reason is that it experiences a larger potential drop, so the electric field transfers more energy to it.

Part e: Calculate the time taken to reach the negative plate.

The acceleration is constant in each arrangement, and the proton starts from rest.

Use the constant-acceleration equation:

\[ s=ut+\frac12at^2 \]

Since \(u=0\):

\[ s=\frac12at^2 \]

Rearrange:

\[ t=\sqrt{\frac{2s}{a}} \]

For arrangement A:

\[ \begin{aligned} t_A&=\sqrt{\frac{2(0.0090)}{3.83\times10^{12}}}\\ &=6.85\times10^{-8}\ \text{s} \end{aligned} \]

For arrangement B:

\[ \begin{aligned} t_B&=\sqrt{\frac{2(0.021)}{1.92\times10^{12}}}\\ &=1.48\times10^{-7}\ \text{s} \end{aligned} \]

Arrangement B takes approximately 2.16 times as long.

Why is this result surprising?

The tempting prediction is that a weaker electric field must produce a lower final speed.

But acceleration and final speed are not the same thing.

The wider gap produces a weaker field and therefore a smaller acceleration. However, the proton travels further and experiences a larger potential drop.

Since the energy gained is determined by the potential difference travelled through, the proton in arrangement B gains more kinetic energy.

There is another useful check.

If the proton had started at the midpoint of each gap instead, it would have experienced half the total voltage in both arrangements: 240 V.

Its final speed would then have been the same in both cases, even though the wider-gap arrangement would still take longer.

That is why the starting position matters. The same voltage, the same field relationship, and a different starting position can produce a different final result.

07What to remember when solving HSC questions

When a question changes the separation between parallel plates, work through these decisions:

1. Check what stays constant. If the power supply remains connected, the potential difference is fixed.

2. Identify the field geometry. Large parallel plates produce an approximately uniform field in their central region.

3. Use the correct relationship. At fixed voltage, \(E\propto1/d\), not \(1/d^2\).

4. Separate field from force. The field is independent of the test charge, but the force depends on its magnitude and sign.

5. Check the starting and finishing positions. If a charged particle moves through the field, its change in kinetic energy depends on the potential difference it actually experiences, not just the field strength.

Once you can distinguish these ideas, the next step is combining electric fields with Newton’s second law to predict the motion of charged particles between parallel plates.