Two parallel plates of a capacitor are 3 x 10^-3 m apart and have an area of 1 x 10^-2 m^2 - NSC Technical Sciences - Question 8 - 2023 - Paper 1

To calculate the potential difference (V) across the plates of the capacitor, we can use the formula:

$C = \frac{Q}{V}$

Where:

• $C$ is the capacitance,
• $Q$ is the charge,
• $V$ is the potential difference.

First, we need to calculate the capacitance ($C$). The capacitance of a parallel plate capacitor can be calculated using the formula:

$C = \varepsilon_0 \frac{A}{d}$

Where:

• $\varepsilon_0 = 8.85 \times 10^{-12} F/m$ is the permittivity of free space,
• $A = 1 \times 10^{-2} m^2$ is the area of the plates,
• $d = 3 \times 10^{-3} m$ is the distance between the plates.

Calculating $C$:

$C = 8.85 \times 10^{-12} \frac{1 \times 10^{-2}}{3 \times 10^{-3}} = 2.95 \times 10^{-11} F$

Now we can substitute $C$ and $Q$ into the first equation to find $V$:

$2.95 \times 10^{-11} = \frac{7.08 \times 10^{-9}}{V}$

Rearranging gives:

$V = \frac{7.08 \times 10^{-9}}{2.95 \times 10^{-11}} \approx 240 V$