JAMB Physics 2017 Paper

Concept:Energy stored in a capacitor depends on the capacitance and the voltage across the plates.Explanation:From the diagram, the capacitance is $C = 8\,\mu F$ and the supplied voltage is $V = 10\,V$.Convert the capacitance from microfarads to farads, since $1\,\mu F = 10^{-6}\,F$.Thus, $C = 8 \times 10^{-6}\,F$.The energy stored in a capacitor is given by $E = \frac{1}{2} C V^2$.Substitute the values into the formula: $E = \frac{1}{2} \times (8 \times 10^{-6}) \times (10)^2$.Simplify the square of voltage: $E = \frac{1}{2} \times (8 \times 10^{-6}) \times 100$.Complete the multiplication: $\frac{1}{2} \times 8 \times 100 = 400$, and include $10^{-6}$.This gives $E = 400 \times 10^{-6}\,J = 4.0 \times 10^{-4}\,J$.Answer:The energy stored in the capacitor is $4.0 \times 10^{-4}\,J$.Therefore, the correct option is B.