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

2 J

Step-by-step 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`