A potential difference of 120 V is applied across two parallel plates of a capacitor - NSC Technical Sciences - Question 8 - 2021 - Paper 1

To calculate the capacitance (C) of the capacitor, we use the formula:

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

where:

• $\varepsilon_0$ (permittivity of free space) = $8.85 \times 10^{-12} \text{ F/m}$
• $A$ (area of the plates) = $2 \text{ m}^2$
• $d$ (distance between the plates) = $6 \text{ mm} = 6 \times 10^{-3} \text{ m}$

Plugging in the values:

$C = \frac{(8.85 \times 10^{-12}) \cdot (2)}{6 \times 10^{-3}}$ $C = 2.95 \times 10^{-9} \text{ F}$

The charge (Q) on each plate can be calculated using the relationship:

$Q = C \cdot V$

where:

• $V$ = voltage applied across the capacitor = 120 V
• $C$ = capacitance calculated previously = $2.95 \times 10^{-9} \text{ F}$

Substituting the values into the formula:

$Q = (2.95 \times 10^{-9}) \cdot 120$ $Q = 3.54 \times 10^{-7} \text{ C}$