A pair of parallel plates, forming a capacitor, are charged. The plates are pulled apart to double the original separation, the charges on the plates remaining the same. What is the ratio of the final energy stored to the original energy stored?

Question

Physics

Adriel Leonard

12 April, 15:57

A pair of parallel plates, forming a capacitor, are charged. The plates are pulled apart to double the original separation, the charges on the plates remaining the same. What is the ratio of the final energy stored to the original energy stored?

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Mark · Answer 1

The final energy will be halved of the original energy.

Explanation:

As we know that the capacitance,

C = εA/d

C = capacitance

A = area of plate

d = distance or separation between the plates

ε = permitivity of dielectric material

As you can see from the formula that the capacitance (ability of the capacitor to store charge) is inversaly proportional to the separation of the plates. If all the other factors kept constant then increasing the distance between the plates will decrease the ability of the capacitor to store energy. So we can say that by doubling the separation, the capacitance or energy stored will become half.