A parallel plate capacitor has a capacitance of 6 x 10⁻¹² F and a charge of 0,3 x 10⁶ C on each one of the identical metal plates - NSC Technical Sciences - Question 8 - 2023 - Paper 1

The capacitance is directly proportional to the charge on the plates. This can be expressed mathematically as:

$C \propto Q$

where $C$ is the capacitance and $Q$ is the charge.

Using the formula for capacitance:

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

we can rearrange to find the potential difference $V$:

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

Substituting the values:

$V = \frac{0,3 \times 10^{-6} C}{6 \times 10^{-12} F} = 50,000 V = 50 kV$

Using the formula for capacitance in terms of area and distance:

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

Where:

• $C$ is the capacitance
• $\varepsilon_0$ (permittivity of free space) = $8.85 \times 10^{-12} \text{ F/m}$
• $A$ is the area of one plate
• $d$ is the separation between the plates (5 cm = 0.05 m)

Rearranging for $A$ gives:

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

Substituting the known values:

$A = 6 \times 10^{-12} F \cdot \frac{0.05 m}{8.85 \times 10^{-12} F/m} \approx 3.39 \times 10^{-2} m^2$