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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.

The energy in the spark produced by dis-
charging the second capacitor is
1. more energetic than the discharge spark
of the first capacitor.
2. the same as the discharge spark of the
first capacitor.
3. less energetic than the discharge spark of
the first capacitor.

1 Answer

5 votes

Answer:

The energy in the spark produced by discharging the second capacitor is more energetic than the discharge spark of the first capacitor.

Step-by-step explanation:

Given that the two isolated parallel plate have the same plate area A and

are given the same charge Q.

Energy stored in the capacitor
= (Q^2)/(2C)


E = (Q^2)/(2C)\\\\Q = √(2EC) \\\\√(2E_1C_1) = √(2E_2C_2)\\\\√(E_1C_1) = √(E_2C_2)

Where;

C is the capacitance and it is given as;


C =(\epsilon A)/(d)

where;

ε is a constant known as permittivity of free space

substitute C in the above equation;


\sqrt{E_1*(A)/(d)} = \sqrt{E_2*(A)/(D)}\\\\\sqrt{(E_1)/(d)} = \sqrt{(E_2)/(D)} \\\\E_2 =(D*E_1)/(d), \ D >d, \ thus \ E_2>E_1

Therefore, the energy in the spark produced by discharging the second capacitor is more energetic than the discharge spark of the first capacitor.

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