Question

A capacitor is charged until its stored energy is 7.54 J. A second capacitor is then connected to it in parallel. If the charge distributes equally, what is the total energy stored in the electric fields

Answer 1

Answer:

2 J

Explanation:

A charged capacitor of capacitance $C_1$ with energy of 7.54 J, is connected in parallel with another capacitor $C_2$ , so the charge is equally distributed between them.

(a) The energy stored in the capacitor before it being connected to the other capacitor is:

$U_O=q_0^2/2C_1=7.54 J$

The energy stored in the electric field is the sum of the energies of the two capacitors:

$U=U_1+U_2\\U=q_1^2/2C_1+q_2^2/2C_2$

since the charge equally distributed,  $q_1$ = $q_2$ = $q_o/2$. and since they are connected in parallel the potential difference on both of them is the same :

$V_1=V_2\\q_1/C_1=q_2/C_2\\q_0/C_1=q_0/C_2\\C_1=C_2=C_3$

hence,

$U=q_0^2/8C+q_0^2/8C\\U=q_0^2/4C\\U=2J$