Question

A capacitor consists of two closely spaced metal conductors of large area, separated by a thin insulating foil. It has an electrical capacity of 3800.0 μF and is charged to a potential difference of 78.0 V. Calculate the amount of energy stored in the capacitor. Tries 0/20 Calculate the charge on this capacitor when the electrical energy stored in the capacitor is 6.84 J. Tries 0/20 If the two plates of the capacitor have their separation increased by a factor of 4 while the charge on the plates remains constant, by what factor is the energy stored in the capacitor increased?

Answer 1

Answer:

- E = 11.55J

- Q = 0.17C

- E' = (1/4)E

Explanation:

- To calculate the amount of energy stored in the capacitor, you use the following formula:

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

C: capacitance = 3800.0*10^-6F

V: potential difference = 78.0V

$E=\frac{1}{2}(3800.0*10^{-6}C)(78.0V)^2=11.55J$

The energy stored in the capacitor is 11.55J

- If the electrical energy stored in the capacitor is 6.84J, the charge on the capacitor is:

$E=\frac{1}{2}QV\\\\Q=\frac{2E}{V}\\\\Q=\frac{2(6.84J)}{78.0V}=0.17C$

The charge on the capacitor is 0.17C

- If you take the capacitor as a parallel plate capacitor, you have that the energy stored on the capacitor is:

$E=\frac{1}{2}CV^2=\frac{1}{2}(\frac{\epsilon_oA}{d})V^2=\frac{1}{2}\frac{\epsilon_oAV^2}{d}$

A: area of the plates

d: distance between plates

If the distance between plates is increased by a factor of 4, you have:

$E'=\frac{1}{2}\frac{\epsilon_oAV^2}{(4d)}=\frac{1}{4}\frac{\epsilon_oAV^2}{2d}=\frac{1}{4}E$

Then, the stored energy in the capacitor is decreased by a a factor of (1/4)