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

- E = 11.55J

- Q = 0.17C

- E' = (1/4)E

Step-by-step explanation:

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

`E=(1)/(2)CV^2`

C: capacitance = 3800.0*10^-6F

V: potential difference = 78.0V

`E=(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=(1)/(2)QV\\\\Q=(2E)/(V)\\\\Q=(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=(1)/(2)CV^2=(1)/(2)((\epsilon_oA)/(d))V^2=(1)/(2)(\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'=(1)/(2)(\epsilon_oAV^2)/((4d))=(1)/(4)(\epsilon_oAV^2)/(2d)=(1)/(4)E`

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