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:
[tex]E=\frac{1}{2}CV^2[/tex]
C: capacitance = 3800.0*10^-6F
V: potential difference = 78.0V
[tex]E=\frac{1}{2}(3800.0*10^{-6}C)(78.0V)^2=11.55J[/tex]
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:
[tex]E=\frac{1}{2}QV\\\\Q=\frac{2E}{V}\\\\Q=\frac{2(6.84J)}{78.0V}=0.17C[/tex]
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:
[tex]E=\frac{1}{2}CV^2=\frac{1}{2}(\frac{\epsilon_oA}{d})V^2=\frac{1}{2}\frac{\epsilon_oAV^2}{d}\\\\[/tex]
A: area of the plates
d: distance between plates
If the distance between plates is increased by a factor of 4, you have:
[tex]E'=\frac{1}{2}\frac{\epsilon_oAV^2}{(4d)}=\frac{1}{4}\frac{\epsilon_oAV^2}{2d}=\frac{1}{4}E[/tex]
Then, the stored energy in the capacitor is decreased by a a factor of (1/4)
