2023 A/L Physics Past Paper · Question 11 (English Medium)

Answer: (3) $2000\ \text{V m}^{-1}$

Given information

We know:

The potential of one plate is $-10\ \text{V}$.

The potential of the other plate is $30\ \text{V}$.

The separation between the plates is $2\ \text{cm}$.

What the question asks

We want: the electric field intensity $E$ between the plates.

What each letter means:

$V_1$ ($\text{V}$): the potential of the first plate.

$V_2$ ($\text{V}$): the potential of the second plate.

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

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

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

Relevant theory

Electric field around two parallel plates charged with opposite charges.

Hence according to the above expression, the electric field intensity at any point in an electric field is equal to the potential gradient at that point.

The first quote shows the field between two parallel plates. The second quote links the field to the potential.

For example, think of a playground slide. A long gentle slide and a short steep slide can drop the same height from the top to the ground.

The steepness is the drop divided by the length of the slide.

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

Required equations

$V = V_2 - V_1$

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

Applying it to this question

The separation is in centimetres, but a field in volts per metre needs metres. So change $2\ \text{cm}$ to $0.02\ \text{m}$ first.

$40$

$V = 30 - (-10)$

$0.02$

$d = \dfrac{2}{100}$

$2000$

$E = \dfrac{V}{d} = \dfrac{40}{0.02}$

Reasoning and calculation

The field is the potential difference divided by the separation.

The potential changes by $40\ \text{V}$ across $0.02\ \text{m}$.

So the potential changes by $\dfrac{40}{0.02} = 2000\ \text{V}$ across each metre. That is $2000\ \text{V m}^{-1}$, option (3).

Checks

$v / d$

The unit: $\dfrac{\text{V}}{\text{m}} = \text{V m}^{-1}$. It is a field intensity, as it should be.

Does it make sense?

Put the answer back: $E d = 2000 \times 0.02 = 40\ \text{V}$. That is the difference between $30\ \text{V}$ and $-10\ \text{V}$.

The potential difference is not $20\ \text{V}$ or $30\ \text{V}$. A potential of $-10\ \text{V}$ is $10\ \text{V}$ below zero, so the climb from it to $30\ \text{V}$ is $40\ \text{V}$.

Conclusion

$E = 2000\ \text{V m}^{-1}$

The field is $E = \dfrac{V}{d} = \dfrac{40}{0.02} = 2000\ \text{V m}^{-1}$.

The answer is (3).

Find the potential difference first. Subtract the smaller potential from the bigger one, and watch the signs: $30 - (-10) = 40$.

Then divide by the separation in metres. Check the answer by putting it back: $E d$ should give the potential difference.

Exam tip