The energy stored in a capacitor of capacitance 5μF is 40J. Calculate the voltage applied across its terminals?

  • A 4000V
  • B 200V
  • C 16V
  • D 6V
  • E 4V
waec 1995physics

The correct answer is A. 4000V

\(E = \frac{CV^{2}}{2}\)

\(V^{2} = \frac{2E}{C}\)

\(V^{2} = \frac{2\times 40}{5 \times 10^{-6}}\)

\(V^{2} = 16 \times 10^{6}\)

\(V = 4 \times 10^{3} = 4000v\)

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