A parallel plate capacitor with an air dielectric is attached to a voltage sourc

ID: 2016144 • Letter: A

Question

A parallel plate capacitor with an air dielectric is attached to a voltage source and charged.
The voltage source is removed, and then the plates are brought closer together to half their
previous distance. What happens to the electric field between the plates when they are brought
together?
A. It is diminished by a factor of 4.
B. It halves.
C. It doubles.
D. It stays the same.
E. It quadruples.

explain please, it says it stays the same but i got the equation E = vBL , so if L halves then doesn't E also half. What equation do you use for parallel plate capacitors?

Explanation / Answer

Data: Let E = Electric field C = Capacitance d = Distance between the plates V = Potential difference between the plates Q = Charge on each plate of the capacitor Solution: We have, Q = C V As charge remains constant, C V = constant C1 V1 = C2 V2 V1 / V2 = C2 / C1 ...(1) Capacitance of parallel plate capacitor, C = o A / d Hence, C1 / C2 = d2 / d1 ...(2) By substtuting (2) in (1): V1 / V2 = d1 / d2 ...(3) But, electric field, E = V / d So, E2 / E1 = ( V2 / V1 ) * ( d1 / d2 ) = ( d2 / d1 ) * ( d1 / d2 ) [ since V2 / V1 = d2 / d1 ] = 1 This results, E2 = E1 Ans: Option-D: It stays the same

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