Physics 2002 JAMB Past Questions - Question 30: The energy stored in a capacitor of capacitance 10μF carrying a charge of 100μC

Correct Answer

B

Explanation

Energy =1/2CV2and v=Q/CEnergy =1/2 Q2/C =1/2 *(100*10-6)2/10*10-6=5*10-4joulesThe energy stored in a capacitor can be calculated using the formula:

\[U = \frac{1}{2} \cdot C \cdot V^2\]

Where:
U = energy stored in the capacitor (in joules)
C = capacitance of the capacitor (in farads)
V = voltage across the capacitor (in volts)

In your case, you have a capacitor with a capacitance of 10μF (microfarads) and a charge of 100μC (microcoulombs). To find the energy stored, you first need to calculate the voltage across the capacitor using the formula:

\[V = \frac{Q}{C}\]

Where:
Q = charge on the capacitor (in coulombs)

In your case, Q = 100μC and C = 10μF, so:

\[V = \frac{100 \times 10^{-6} C}{10 \times 10^{-6} F} = 10 \text{ volts}\]

Now that you know the voltage, you can calculate the energy stored in the capacitor:

\[U = \frac{1}{2} \cdot 10 \times 10^{-6} F \cdot (10 \text{ volts})^2\]

\[U = \frac{1}{2} \cdot 10 \times 10^{-6} F \cdot 100 \text{ V} = 5 \times 10^{-4} \text{ joules} = 0.0005 \text{ joules}\]

So, the energy stored in the capacitor is 0.0005 joules.