Question

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

Answer 1

Answer:

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.  

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

Answer:

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

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)

$\\W_{energy}=\frac{1}{2}CV^{2}$

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

$\\W_{A}=\frac{1}{2}CV^{2}$

and for capacitor B, we have $\\W_{B}=\frac{1}{2}C[2V]^{2}$

$\\ W_{B}=2CV^{2}$

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