Question

You have a parallel-plate capacitor connected to a 12V battery. You charge it up such that it holds a certain amount of potential energy. You then replace the 12V battery with a 24V battery. What is one change you could make such that it will store the same amount of potential energy that the 12V battery did?

a. increase the separation between the plates by a factor of 2

b. increase the separation between the plates by a factor of 4

c. decrease the separation between the plates by a factor of 2

d. decrease the separation between the plates by a factor of 4

c. none of the above

I put none of the above, because given capacitance of a capacitor remains constant unless you change a physical parameter (i.e. add or remove dielectric material or distance between plates), the ratio of charge to voltage will remain the same (C=Q/V). Since U= 1/2(C)(V)^2=(Q^2)/C=1/2(Q)(V), none of the above options would correctly change the electric potential energy. Is that correct?

Answer 1

as we know

U = 0.5 C V^2

U = 0.5 (eo A / d) V^2

for constant value of energy

d is directly proportional to square of voltage

d2 / d1 = ( V2/v1) ^2

d2 = d * (24/12)^2

d2 = 4 * d

we need to increase the Seperation by factor of 4 in order to have same electric energy as before.

so option (b) is correct.

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