JAMB - Physics (2024 - No. 80)
The energy stored in the above capacitor is
2.25J
6.25J
18.25J
24.25J
คำอธิบาย
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