Two capacitors, each of capacitance 2$\mu$F are connected in parallel. If the p.d across them is 120V, calculate the charge on each capacitor

Analyze the question: Two capacitors, each 2 μF, are connected in parallel across a 120 V supply. We need the charge on each capacitor.

Concept: In a parallel connection, every component has the same potential difference across it. The charge on a single capacitor is:

Q = CV

Solution:

For each capacitor:

• C = 2 μF = 2 × 10⁻⁶ F
• V = 120 V

Q = CV = (2 × 10⁻⁶)(120) = 240 × 10⁻⁶ C = 2.4 × 10⁻⁴ C

Check (total charge): The total capacitance in parallel is 2 + 2 = 4 μF, so the total charge is 4 × 10⁻⁶ × 120 = 4.8 × 10⁻⁴ C. This is option D, but it is the charge on both capacitors together, not on each one. Splitting it equally gives 2.4 × 10⁻⁴ C each.

Answer: The charge on each capacitor is 2.4 × 10⁻⁴ C, which is option C.

Exam tip: Always read the last line of the question carefully. WASSCE and JAMB often include the total charge as a distractor when the question asks for the charge on each capacitor. Remember: in parallel, the voltage is the same, so the charges add up. In series, the charge is the same, so the voltages add up.