The energy stored in the above capacitor is
Attached Figure / Diagram
A
2.25J
B
6.25J
C
18.25J
D
24.25J
Verified Correct Answer: A.
Physical Principles & Numerical Solution
The energy stored in the capacitor = \(\frac{1}{2}\)\(\frac{q^2}{C}\)
Where C = 2F, q = 3C
= \(\frac{1}{2}\)\(\frac{3^2}{2}\) = \(\frac{9}{4}\) = 2.25J
Source: WAEC / JAMB / NECO Classroom Archive (Record #70518)