Question

Two parallel-plate capacitors, 2.6 μF each, are connected in parallel to a 28 V battery. One of the capacitors is then squeezed so that its plate separation is halved. Because of the squeezing, (a) how much additional charge is transferred to the capacitors by the battery and (b) what is the increase in the total charge stored on the capacitors?

Answer 1

a.
`72.8\mu C`

b.
`72.8\mu C`

Step-by-step explanation:

We are given that

Capacitance of one capacitor=
`2.6\mu F`

Capacitance of second capacitor=
`2.6\mu F`

Potential difference of battery=28 v

a.When one of the capacitor is squeezed so that plate separation is halved then we have to find the amount of additional charge is transferred to the capacitor by the battery.

Initial capacitance,
`C=(\epsilon_0A)/(d)`

If the separation is halved , then the capacitance will be doubled

`q=CV`

If the capacitance will be doubled then the charge will be doubled also.

Initial charge=
`2.6* 10^(-6)* 28=72.8\mu C`

Final charge on the capacitor

`q=2CV=2* 72.8\mu C=145.6\mu C`

Additional charge transmitted,
`q'=145.6-72.8=72.8\mu C`

b.We have to find the increase in the total charge stored on the capacitors

Initial total charge is given by

`q=(c_1+c_2)V=(2.6+2.6)* 28=145.6 \mu C`

Final total charge on the capacitor

`q=(2* 2.6+2.6)* 28=218.4\muC`

Increase in the charge=
`q'-q=218.4-145.6=72.8\mu C`