A capacitor is a device used in an electric circuit for storing electric charge - NSC Technical Sciences - Question 8 - 2024 - Paper 1

1. Distance between the plates: The capacitance decreases as the distance between the plates increases.

2. Type of dielectric material: Different dielectric materials affect the capacitance differently, with higher permittivity materials leading to greater capacitance.

3. Area of the plates: The larger the area of the plates, the greater the capacitance, as more charge can be stored.

To find the potential difference (V) across the capacitor plates, we first need to calculate the capacitance (C) using the formula:

$C = \frac{\epsilon_{0} \cdot A}{d}$

Where:

• $\epsilon_{0} = 8.85 \times 10^{-12} \ F/m$ (permittivity of free space)
• $A = 2 \times 10^{-4} \ m^{2}$ (area of the plates)
• $d = 2 \times 10^{-3} \ m$ (distance between the plates)

Substituting the values:

$C = \frac{(8.85 \times 10^{-12} \ F/m) \cdot (2 \times 10^{-4} \ m^{2})}{2 \times 10^{-3} \ m} = 8.85 \times 10^{-10} \ F$

Now, using the formula for capacitance: $C = \frac{Q}{V}$

Where:

• $Q = 4.5 \times 10^{-11} \ C$ (charge stored)

Rearranging for V gives: $V = \frac{Q}{C} = \frac{4.5 \times 10^{-11} \ C}{8.85 \times 10^{-10} \ F} \approx 50.85 \ V$

Thus, the potential difference across the capacitor plates is approximately $50.85 \ V$.