How to Calculate Electric Field Strength Between Parallel Plates

A voltmeter measures the potential of one plate as \(+420\text{ V}\) and the other as \(+180\text{ V}\), both relative to earth. The plates are separated by \(3.0\text{ mm}\).

A student calculates \(420 \div 0.003\) and gets \(1.4\times10^5\text{ V m}^{-1}\). Is that correct?

No. The second plate is not at \(0\text{ V}\), so the potential difference across the gap is only \(240\text{ V}\).

The correct calculation is:

\[ E=\frac{420-180}{0.003} \]

\[ \boxed{E=8.0\times10^4\text{ V m}^{-1}} \]

This is the electric field strength in the approximately uniform central region between the plates.

The method is straightforward: find the potential difference between the plates, convert their perpendicular separation into metres, and divide. The important part is knowing exactly which voltage and distance belong in the calculation.

01What are you actually measuring?

Imagine two large, flat metal plates facing each other. A small gap separates them, and a DC power supply maintains a potential difference between them.

The experimental arrangement has three important parts:

• The power supply connects to the two plates and maintains their potential difference.
• A voltmeter connects across the plate terminals, measuring the potential difference in volts.
• A ruler, vernier calipers, or a fixed spacer determines the perpendicular distance between the facing surfaces.

The voltmeter does not measure the electric field directly. It measures the potential difference, which we use alongside the plate separation to calculate the field strength.

There is a reason the voltage difference matters more than either individual plate potential.

Suppose both plates have their potentials increased by \(200\text{ V}\). Their readings become \(+620\text{ V}\) and \(+380\text{ V}\).

The potential difference is still:

\[ \Delta V=620-380=240\text{ V} \]

Nothing about the potential difference across the plates has changed. Therefore, assuming the same geometry and surrounding conditions, the central electric field has not changed either.

This is an important experimental principle: the electric field depends on the potential difference, not the absolute potential relative to an arbitrary reference.

A first attempt that gives the wrong result

Imagine measuring each plate’s potential relative to earth, then using the higher reading directly in the field calculation.

That approach only works if the other plate is actually at \(0\text{ V}\).

The improved method is to either:

1. Connect the voltmeter directly across the two plates and read the potential difference.
2. Measure both plate potentials relative to the same reference and subtract them.

For the magnitude of the potential difference:

\[ \Delta V=|V_1-V_2| \]

The absolute value ensures that the voltage difference is positive when calculating field strength. We determine the field direction separately.

02Why does dividing voltage by distance give the electric field?

Before using the equation, consider what an electric field actually describes.

An electric field exists because of the arrangement of source charges. It describes the force that a small positive test charge would experience at a particular location.

Its definition is:

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

where:

• \(\vec E\) is the electric field strength in \(\text{N C}^{-1}\).
• \(\vec F\) is the electric force in newtons (\(\text{N}\)).
• \(q\) is the positive test charge in coulombs (\(\text{C}\)).

The test charge is used to define and investigate the field. It does not create the original field being investigated.

This distinction matters. Put a proton and an electron at the same location between the plates. They experience the same electric field, but their forces point in opposite directions.

Now consider what happens to electric potential as you move from one plate towards the other.

Electric potential tells us the electric potential energy per unit charge. A uniform electric field produces a constant rate of change of potential with perpendicular distance.

In other words, if you travel equal distances through the field, you experience equal changes in electric potential.

For a uniform field, the relationship is:

\[ |\Delta V|=Ed \]

Rearranging gives:

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

This is the equation used to calculate the electric field between parallel plates.

The equation makes physical sense. A large potential difference across a small gap produces a strong field. The same potential difference spread across a larger gap produces a weaker field.

The units also make sense:

\[ \frac{\text{V}}{\text{m}}=\text{V m}^{-1} \]

Electric field strength can therefore be measured in either:

• Volts per metre (\(\text{V m}^{-1}\)).
• Newtons per coulomb (\(\text{N C}^{-1}\)).

These units are equivalent.

Which direction does the electric field point?

Electric potential decreases in the direction of the electric field.

Therefore, between parallel plates, the field points:

• Away from the higher-potential plate.
• Towards the lower-potential plate.

For example, if the left plate is at \(+500\text{ V}\) and the right plate is at \(+100\text{ V}\), the electric field points from left to right.

Notice that both plates have positive potentials relative to earth. That does not change the direction.

A negative test charge experiences a force opposite to the electric field, while a positive test charge experiences a force in the same direction.

03Uniform electric fields versus inverse-square fields

There is another important distinction to understand before using the equation.

Not every electric field behaves in the same way.

The electric field between large, closely spaced parallel plates is approximately uniform in the central region. Its magnitude and direction remain approximately constant as you move around within that region.

A point charge produces a different type of field.

The parallel-plate case

For large parallel plates, the central electric field is approximately:

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

If you move a test charge around within the central region, the field strength stays approximately constant.

There is an important distinction here.

If you double the plate separation while keeping the voltage constant, the calculated field halves.

However, that does not mean the field halves as you move from one position to another inside the original gap. The field is still approximately uniform throughout that central region.

Changing the apparatus is different from moving around within an existing field.

The point-charge case

An isolated point charge produces an electric field described by:

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

where:

• \(E\) is the electric field strength in \(\text{N C}^{-1}\).
• \(k\) is Coulomb’s constant, approximately \(8.99\times10^9\text{ N m}^2\text{ C}^{-2}\).
• \(Q\) is the source charge in coulombs.
• \(r\) is the radial distance from the point charge in metres.

This is an inverse-square relationship.

If you double the distance from a point charge, the field becomes one-quarter as strong:

\[ E_{\text{new}}=\frac{E_{\text{old}}}{4} \]

That is very different from the uniform field between parallel plates.

Do not use the inverse-square equation simply because a question mentions a distance. Equally, do not use \(E=\Delta V/d\) just because a question gives you a voltage and a distance.

Look at the physical arrangement first.

04When is the parallel-plate equation valid?

The equation \(E=\Delta V/d\) assumes that the field is uniform over the distance being considered.

For real parallel plates, this is a good approximation when:

• The plates are parallel.
• Their dimensions are much larger than their separation.
• The location being considered is well away from the edges.
• Nearby charged objects do not significantly distort the field.

Near the edges of finite plates, the electric field bends outwards. This is called fringing.

The field is no longer perfectly uniform. Its magnitude and direction can change with position.

This creates an important limitation.

The equation can still give a useful characteristic field strength for the plate arrangement, but it does not necessarily give the exact local field near an edge.

For a nonuniform field, dividing the total potential difference by the plate separation gives an average potential gradient along that direction, not necessarily the field strength at every individual point.

For a closer explanation of the field geometry and edge effects, see Uniform Electric Fields Between Parallel Plates: HSC Physics.

05Worked examples

06Questions and solutions

Question 1

Two large horizontal parallel plates are separated by \(5.0\text{ mm}\). The upper plate has a potential of \(+180\text{ V}\), and the lower plate has a potential of \(-20\text{ V}\).

Calculate the electric field strength in the central region and state its direction.

Solution 1

The electric field strength is \(4.0\times10^4\text{ V m}^{-1}\), directed downwards.

First, calculate the potential difference:

\[ \Delta V=|180-(-20)| \]

\[ \Delta V=200\text{ V} \]

Convert the separation into metres:

\[ d=5.0\times10^{-3}\text{ m} \]

Apply the parallel-plate equation:

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

\[ E=\frac{200}{5.0\times10^{-3}} \]

\[ \boxed{E=4.0\times10^4\text{ V m}^{-1}} \]

The electric field points from the higher-potential upper plate to the lower-potential bottom plate.

The important step is calculating the full potential difference. Using \(180\text{ V}\) alone would ignore the negative potential of the second plate.

Question 2

Two large vertical parallel plates are separated by \(7.5\text{ mm}\). Their potentials, measured relative to earth, are:

• Left plate: \(+0.92\text{ kV}\).
• Right plate: \(+0.32\text{ kV}\).

A small charged bead with charge \(-3.0\text{ nC}\) is placed halfway between the plates.

a) Calculate the electric field strength and direction.

b) Calculate the magnitude and direction of the electric force on the bead.

c) Both plate potentials are then lowered by \(0.20\text{ kV}\), without changing their separation. Explain whether the electric field and force change.

Solution 2

The electric field is \(8.0\times10^4\text{ V m}^{-1}\) to the right, and the bead experiences a force of \(2.4\times10^{-4}\text{ N}\) to the left.

Part a: Electric field strength

Calculate the potential difference:

\[ \Delta V=920-320 \]

\[ \Delta V=600\text{ V} \]

Convert the separation:

\[ d=7.5\times10^{-3}\text{ m} \]

Calculate the field:

\[ E=\frac{600}{7.5\times10^{-3}} \]

\[ \boxed{E=8.0\times10^4\text{ V m}^{-1}} \]

The field points from the left plate towards the right plate because the left plate has the higher potential.

Part b: Electric force

Use:

\[ F=|q|E \]

Convert the bead’s charge:

\[ q=-3.0\times10^{-9}\text{ C} \]

Substitute:

\[ F=(3.0\times10^{-9})(8.0\times10^4) \]

\[ \boxed{F=2.4\times10^{-4}\text{ N}} \]

The bead is negatively charged, so its force acts opposite to the electric field.

Therefore, the force is directed to the left, towards the higher-potential plate.

Part c: Changing both plate potentials

The new plate potentials are:

\[ V_L=+0.72\text{ kV} \]

\[ V_R=+0.12\text{ kV} \]

Their potential difference is:

\[ \Delta V=720-120=600\text{ V} \]

The potential difference has not changed, and neither has the plate separation.

Therefore, the electric field remains:

\[ E=8.0\times10^4\text{ V m}^{-1} \]

The bead’s charge is also unchanged, so its electric force remains \(2.4\times10^{-4}\text{ N}\) to the left.

The tempting mistake is to assume that lowering both absolute potentials must weaken the field. It doesn’t. Only the potential difference across the plates determines the field in this model.

Question 3

Two large parallel plates have a potential difference of \(360\text{ V}\) and a separation of \(4.0\text{ mm}\).

An isolated positive point charge of \(+1.6\text{ μC}\) is placed in a separate experiment.

a) Calculate the electric field strength between the plates.

b) Calculate the radial distance from the point charge at which its electric field has the same magnitude as the field between the plates.

c) If the distance from the point charge is doubled, explain how its field strength compares with the original plate field.

Use \(k=8.99\times10^9\text{ N m}^2\text{ C}^{-2}\).

Solution 3

The field between the plates is \(9.0\times10^4\text{ V m}^{-1}\). The point charge produces the same field strength approximately \(0.40\text{ m}\) from its centre.

Part a: Field between the plates

Convert the separation:

\[ d=4.0\times10^{-3}\text{ m} \]

Calculate the field:

\[ E=\frac{360}{4.0\times10^{-3}} \]

\[ \boxed{E=9.0\times10^4\text{ V m}^{-1}} \]

Part b: Distance from the point charge

The point charge produces an inverse-square field, so use:

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

We want the field to equal the field between the plates.

Rearrange:

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

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

Substitute:

\[ r=\sqrt{ \frac{(8.99\times10^9)(1.6\times10^{-6})} {9.0\times10^4} } \]

\[ r=\sqrt{0.1598} \]

\[ \boxed{r\approx0.40\text{ m}} \]

At approximately \(0.40\text{ m}\), the point charge produces the same field magnitude as the parallel plates.

Part c: Doubling the distance

For a point charge:

\[ E\propto\frac{1}{r^2} \]

Doubling the radial distance reduces the field to one-quarter of its original value.

Therefore:

\[ E_{\text{new}}=\frac{9.0\times10^4}{4} \]

\[ \boxed{E_{\text{new}}=2.25\times10^4\text{ N C}^{-1}} \]

The field between the plates, however, remains approximately uniform throughout the central region.

This comparison highlights why choosing the correct field equation matters. The same field strength can occur in two very different arrangements, but the way each field changes with position is not the same.

Question 4

A student investigates the electric field between two finite parallel plates.

The plates have dimensions of \(200\text{ mm}\times120\text{ mm}\), are separated by \(10\text{ mm}\), and have a potential difference of \(480\text{ V}\).

The student places a field-measuring probe \(2\text{ mm}\) from the right-hand edge of the plates.

They calculate:

\[ E=\frac{480}{0.010} =4.8\times10^4\text{ V m}^{-1} \]

The student concludes that the local electric field at the probe must be exactly \(4.8\times10^4\text{ V m}^{-1}\).

Evaluate this conclusion. Explain what the calculation tells us and why it may not describe the field at the probe.

Solution 4

The calculation gives a characteristic central-field estimate of \(4.8\times10^4\text{ V m}^{-1}\), but it does not establish the exact local field at the probe.

The calculation itself is correct for the uniform-field approximation:

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

\[ E\approx\frac{480}{0.010} \]

\[ E\approx4.8\times10^4\text{ V m}^{-1} \]

However, the probe is only \(2\text{ mm}\) from the edge.

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

As a result:

• The field direction may no longer be perpendicular to the plate surfaces.
• The field magnitude may vary with position.
• A single value of \(\Delta V/d\) cannot describe the exact local field.

The student’s mistake is treating an approximate uniform-field calculation as an exact description of every point between the plates.

To determine the field at the probe, more information about the local potential variation or the actual field distribution would be needed.

If the probe were positioned well inside a sufficiently uniform central region, the calculated value would be a much more appropriate estimate.

Question 5

Two large parallel plates are separated by \(12\text{ mm}\). The higher-potential plate is at \(+500\text{ V}\), and the lower-potential plate is at \(+140\text{ V}\).

A small positive test charge of \(+2.0\text{ nC}\) is carried from point A on the higher-potential plate to point B on the lower-potential plate.

Point B is \(16\text{ mm}\) sideways from point A, measured parallel to the plate surfaces. The charge travels along a straight diagonal path between the points.

Assume both endpoints and the entire path remain well inside the central uniform-field region.

Answer the following:

a) Calculate the magnitude and direction of the electric field between the plates.

b) Calculate the diagonal distance travelled by the charge. Explain why dividing the potential difference by this diagonal distance does not give the magnitude of the electric field.

c) Calculate the work done by the electric field on the positive test charge as it moves from A to B.

d) Both plate potentials are increased by \(500\text{ V}\), without changing their separation. Explain what happens to the electric field and the work done by the field.

e) The positive test charge is replaced by a negative charge of \(-2.0\text{ nC}\), which is still carried from A to B. Explain what changes and what remains the same.

Solution 5

The electric field is \(3.0\times10^4\text{ V m}^{-1}\), directed from A towards B across the gap. The diagonal path does not change the field magnitude, and the work done by the field on the positive charge is \(7.2\times10^{-7}\text{ J}\).

Part a: Calculate the electric field

First, find the potential difference:

\[ \Delta V=500-140 \]

\[ \Delta V=360\text{ V} \]

Convert the perpendicular separation:

\[ d=12\times10^{-3}\text{ m} \]

Calculate:

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

\[ E=\frac{360}{0.012} \]

\[ \boxed{E=3.0\times10^4\text{ V m}^{-1}} \]

The field points from the higher-potential left plate towards the lower-potential right plate.

Part b: Calculate the diagonal distance

The sideways displacement is \(16\text{ mm}\), and the perpendicular displacement is \(12\text{ mm}\).

These form a right-angled triangle.

Using Pythagoras:

\[ L=\sqrt{12^2+16^2} \]

\[ L=\sqrt{400} \]

\[ \boxed{L=20\text{ mm}} \]

A tempting calculation would be:

\[ E=\frac{360}{0.020} \]

\[ E=1.8\times10^4\text{ V m}^{-1} \]

But this is not the magnitude of the electric field.

Why?

Because the electric field is perpendicular to the plate surfaces, not parallel to the diagonal path.

The potential changes only with the displacement in the field direction. Moving sideways along the plates does not produce an additional potential change in an ideal uniform field.

The calculated value of \(1.8\times10^4\text{ V m}^{-1}\) is actually the component of the electric field along the diagonal path.

The relationship is:

\[ E_{\parallel}=E\cos\theta \]

where \(\theta\) is the angle between the path and the electric field.

Here:

\[ \cos\theta=\frac{12}{20}=0.60 \]

Therefore:

\[ E_{\parallel}=(3.0\times10^4)(0.60) \]

\[ E_{\parallel}=1.8\times10^4\text{ V m}^{-1} \]

The student’s calculation finds the field component along the path, not the full field magnitude.

Part c: Work done by the electric field

Work done by the electric field when a charge moves between two potentials is:

\[ W_{\text{field}}=q(V_A-V_B) \]

The positive charge is:

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

The potential difference is:

\[ V_A-V_B=360\text{ V} \]

Substitute:

\[ W_{\text{field}}=(2.0\times10^{-9})(360) \]

\[ \boxed{W_{\text{field}}=7.2\times10^{-7}\text{ J}} \]

The work is positive because the positive charge moves from higher potential to lower potential, in the direction of the electric force.

We can also check this result using the force and displacement.

The force magnitude is:

\[ F=qE \]

\[ F=(2.0\times10^{-9})(3.0\times10^4) \]

\[ F=6.0\times10^{-5}\text{ N} \]

Only the displacement component parallel to the force contributes to the work:

\[ W=F d \]

\[ W=(6.0\times10^{-5})(0.012) \]

\[ W=7.2\times10^{-7}\text{ J} \]

Both methods give the same result.

This is a useful consistency check: the field does not perform work based on the full diagonal distance because the force is not parallel to the diagonal displacement.

Part d: Increasing both plate potentials

The new potentials are:

\[ V_A=+1000\text{ V} \]

\[ V_B=+640\text{ V} \]

The new potential difference is:

\[ \Delta V=1000-640=360\text{ V} \]

The potential difference is unchanged.

Therefore, the electric field remains:

\[ \boxed{E=3.0\times10^4\text{ V m}^{-1}} \]

The work done on the positive charge also remains:

\[ \boxed{W_{\text{field}}=7.2\times10^{-7}\text{ J}} \]

Adding the same amount to both potentials changes their absolute values, but not the potential difference between them.

Part e: Replacing the positive charge with a negative charge

The electric field does not change.

It is determined by the plate arrangement and potential difference, not by the test charge.

However, the electric force reverses direction because the charge is now negative.

Its magnitude remains:

\[ F=|q|E \]

\[ F=(2.0\times10^{-9})(3.0\times10^4) \]

\[ F=6.0\times10^{-5}\text{ N} \]

The force now acts towards the higher-potential plate, opposite to the direction of the electric field.

The work done by the field also reverses sign:

\[ W_{\text{field}}=q(V_A-V_B) \]

\[ W_{\text{field}}=(-2.0\times10^{-9})(360) \]

\[ \boxed{W_{\text{field}}=-7.2\times10^{-7}\text{ J}} \]

The field does negative work because the negative charge is being carried against the direction of the electric force.

The key distinction is that changing the test charge changes the force and work, but not the electric field itself.

07What to do next

Once you can calculate the electric field from voltage and separation, the next step is predicting how a charged particle moves through that field.

Use:

\[ F=qE \]

to calculate the electric force, followed by:

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

to determine the particle’s acceleration.

The sign of the charge determines the force direction, while the mass determines how strongly the particle accelerates.

This connects the calculation of electric field strength to charged-particle motion, which is an important application of uniform electric fields in HSC Physics.