2022 A/L Physics Past Paper · Question 11 (English Medium)

Plate separation for a given potential difference

A uniform electric field of intensity 200 V m−1200\ \text{V m}^{-1} is applied in between a pair of parallel metallic plates. What should be the separation between the plates in order to produce a potential difference of 10 V10\ \text{V}?

  1. (1) 20 mm20\ \text{mm}
  2. (2) 30 mm30\ \text{mm}
  3. (3) 50 mm50\ \text{mm}
  4. (4) 20 m20\ \text{m}
  5. (5) 30 m30\ \text{m}
Answer and explanation

Answer: (3) 50 mm50\ \text{mm}

Given information

We know:

The electric field is uniform, with intensity E=200 V m−1E = 200\ \text{V m}^{-1}.

The potential difference between the plates is V=10 VV = 10\ \text{V}.

What the question asks

We want: the separation dd between the plates.

What each letter means:

EE (V m−1\text{V m}^{-1}): electric field intensity between the plates.

VV (V\text{V}): potential difference between the plates.

dd (m\text{m}): separation between the plates.

Relevant theory

the electric field intensity at any point in an electric field is equal to the potential gradient at that point.

For example, think of a ramp that drops a certain height.

The steeper the ramp, the more height it drops in each metre. The longer the ramp, the more height it drops in all.

The field is like the steepness. The separation is like the length. The potential difference is like the total drop.

In a uniform field the steepness is the same everywhere, so the drop is just the steepness times the length.

Required equations

E=VdE = \dfrac{V}{d}

d=VEd = \dfrac{V}{E}

Applying it to this question

Every quantity is in SI units already: volts, volts per metre. So dd comes out in metres.

d=10200d = \dfrac{10}{200}

0.050.05

d=0.05d = 0.05

5050

d=0.05×1000d = 0.05 \times 1000

Reasoning and calculation

The size of the field is the potential difference divided by the separation.

So the separation is 10 V200 V m−1=0.05 m\dfrac{10\ \text{V}}{200\ \text{V m}^{-1}} = 0.05\ \text{m}.

0.05 m0.05\ \text{m} is 50 mm50\ \text{mm}. It is option (3).

Checks

1000∗v/e1000 * v / e

The unit: VV m−1=m\dfrac{\text{V}}{\text{V m}^{-1}} = \text{m}. A length, as it should be.

Does it make sense?

Put the answer back: Ed=200×0.05=10 VE d = 200 \times 0.05 = 10\ \text{V}. That is what the question asks.

A separation of 5 cm5\ \text{cm} is a sensible size for a pair of plates in a lab. A separation of 20 m20\ \text{m} is not.

Conclusion

d=50 mmd = 50\ \text{mm}

The separation is d=VE=10200=0.05 m=50 mmd = \dfrac{V}{E} = \dfrac{10}{200} = 0.05\ \text{m} = 50\ \text{mm}.

The answer is (3).

In a uniform field, E=VdE = \dfrac{V}{d}. The separation is the potential difference divided by the field.

Do not turn it upside down. Check the answer by putting it back: EdE d should give the potential difference.

Exam tip

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