A 20 µF capacitor is connected to a 12 V d.c - Scottish Highers Physics - Question 20 - 2016

To find the maximum charge (Q) stored in a capacitor, we use the formula:

$Q = C imes V$

where:

• $C$ is the capacitance in farads (F), and
• $V$ is the voltage in volts (V).

In this case, we have:

• $C = 20 \, \mu F = 20 \times 10^{-6} \ F$
• $V = 12 \ V$

Plugging in the values:

$Q = (20 \times 10^{-6}) \times 12 = 240 \times 10^{-6} \ C = 2.4 \times 10^{-4} \ C$

Thus, the maximum charge stored on the capacitor is B 2.4 × 10⁻⁴ C.