Question

A battery charges capacitor A until the potential difference between the two conductors of the capacitor is V.

A second, identical capacitor, labeled B, is charged by a different battery until the potential difference of capacitor B is 2V.

How does the stored energy of capacitor B compare to that of capacitor A?

Answer 1

The stored energy (Potential difference) of capacitor B which is equal to 2V is twice as much as capacitor A which has a stored capacity of 1V or V. Increased voltage also implies a reduction in the current passing through it.

Step-by-step explanation:

Potential Difference is simply the work that has to be done in transferring a unit positive charge from one point to the other. It is also referred to as the "Electric Potential".

Also, the potential difference between points A and B, V(B) - V(A), is the change in Potential energy of a charge (q) moved from A to B, divided by the charge. Its units are joules per coulomb which is also equal to volt (V).

Note: the relationship between Potential energy and Potential difference is expressed thus: Change in Potential Difference = Change in Potential Energy / Charge.

Answer 2

The energy stored in capacitor B is four times the energy stored in capacitor A

Step-by-step explanation:

First let the write-out the formula for the energy stored in a capacitor in-terms of the capacitance(C)and the voltage(V)

`\\W_(energy)=(1)/(2)CV^(2)\\`

For capacitor A charged to a voltage of V, The energy stored is

`\\W_(A)=(1)/(2)CV^(2)\\`

and for capacitor B, we have
`\\W_(B)=(1)/(2)C[2V]^(2)\\`

`\\ W_(B)=2CV^(2)\\`

From the energy stored in Capacitor A we can arrive at

`2W_(A)=CV^(2)\\`

if we substitute this value into the energy for capacitor B we arrive at

`\\W_(B)=2(2W_(A) )\\W_(B)=4W_(A)\\`

Hence we can conclude that the energy stored in capacitor B is four times the energy stored in capacitor A