8.1 A 3 V battery is connected to a circuit, as shown below - NSC Technical Sciences - Question 8 - 2022 - Paper 1

The capacitor is connected to a battery, which causes it to charge.

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

$C = \frac{\epsilon_0 A}{d}$

Where:

• $\epsilon_0 = 8.85 \times 10^{-12} \ F/m$ (permittivity of free space)
• $A = 10.2 \ cm^2 = 10.2 \times 10^{-4} \ m^2$ (area)
• $d = 2.38 \ mm = 2.38 \times 10^{-3} \ m$ (distance between plates)

Substituting the values:

$C = \frac{(8.85 \times 10^{-12}) (10.2 \times 10^{-4})}{2.38 \times 10^{-3}}$

Calculating gives:

$C = 3.79 \times 10^{-12} \ F$

To calculate the potential difference (V), we use the formula:

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

Where:

• $Q = 0.345 \ nC = 0.345 \times 10^{-9} \ C$ (charge)
• $C = 3.79 \times 10^{-12} \ F$

Substituting the values:

$V = \frac{0.345 \times 10^{-9}}{3.79 \times 10^{-12}}$

Calculating gives:

$V \approx 91.03 \ V$