A 24-0μF capacitor is charged until the potential difference across it is 125 V - Scottish Highers Physics - Question 17 - 2018

To find the charge (Q) stored in a capacitor, we can use the formula: $Q = C imes V$ Where:

• Q is the charge in coulombs (C)
• C is the capacitance in farads (F)
• V is the potential difference in volts (V)

Given:

• Capacitance (C) = 24 μF = 24 \times 10^{-6} F
• Voltage (V) = 125 V

Substituting the values:

$$Q = (24 \times 10^{-6} \text{ F}) \times (125 \text{ V})$$

Calculating the above:

$$Q = 3.00 \times 10^{-3} \text{ C} = 3.00 \times 10^(-6) C$$

Therefore, the charge stored on the capacitor is:

• D 3.00 x 10^(-6) C