Question

A 4.55 nF parallel-plate capacitor contains 27.5 μJ of stored energy. By how many volts would you have to increase this potential difference in order for the capacitor to store 55.0 μJ of potential energy?
Express your answer in volts as an integer.

Answer 1

Answer:

$\Delta V=V_{2}-V_{1}=45.4V$

Explanation:

The energy, E, from a capacitor, with capacitance, C, and voltage V is:

$E=\frac{1}{2} CV^{2}$

$V=\sqrt{2E/C}$

If we increase the Voltage, the Energy increase also:

$V_{1}=\sqrt{2E_{1}/C}$

$V_{2}=\sqrt{2E_{2}/C}$

The voltage difference:

$V_{2}-V_{1}=\sqrt{2E_{2}/C}-\sqrt{2E_{1}/C}$

$V_{2}-V_{1}=\sqrt{2*55*10^{-6}/4.55*10^{-9}}-\sqrt{2*27.5*10^{-6}/4.55*10{-9}}=45.4V$