Complete Question
The two isolated parallel plate capacitors be- low, one with plate separation d and the other with D > d, have the same plate area A and
are given the same charge Q.
What is the energy in the spark produced by discharging the second capacitor?
1. The same as the discharge spark of the first capacitor
2. More energetic than the discharge spark of the first capacitor
3. Less energetic than the discharge spark of the first capacitor
Answer:
The correct option is 2
Explanation:
The formula for the energy stored in the capacitor is
[tex]U = \frac{Q^2}{2C}[/tex]
And generally the formula for finding the capacitance of a capacitor is
[tex]C = \frac{\epsilon_oA}{d}[/tex]
We can denote the capacitance of the first capacitor as [tex]C_1 = \frac{\epsilon_oA}{d}[/tex]
and denote the capacitance of the second capacitor as [tex]C_2 = \frac{\epsilon_oA}{D}[/tex]
Looking at this formula we can see that C varies inversely with d
as D > d it means that [tex]C_1 > C_2[/tex]
Since the charge is constant
[tex]U\ \alpha\ \frac{1}{C}[/tex] i.e U varies inversely with C
So [tex]C_1 > C_2[/tex] => [tex]U_2 >U_1[/tex]
This means that the energy of spark would be more for capacitor two compared to capacitor one
The correct option is 2
