Question

a 1microfarad capacitor is charged to a potential difference of 220volts . it is the disconnected from battery its plate are then connected in parallel to another capacitor and it is found that , the potential difference fall to 100 volts . what is the capacitance of second capacitor

Answer 1

Answer:

43Pf

Explanation:

search-icon-header

Search for questions & chapters

search-icon-image

Question

Bookmark

A 100 pF capacitor is charged to a potential difference of 50 V, and the charging battery is disconnected. The capacitor is then connected in parallel with a second (initially uncharged) capacitor. If the potential difference across the first capacitor drops to 35 V, what is the capacitance of this second capacitor?

Hard

Solution

verified

Verified by Toppr

The charge initially on the charged capacitor is given by q=C

1

V

0

, where C

1

=100pF is the capacitance and V

0

=50V is the initial potential difference. After the battery is disconnected and the second capacitor wired in parallel to the first, the charge on the first capacitor is q

1

=C

1

V, where V = 35V is the new potential difference. Since charge is conserved in the process, the charge on the second capacitor is q

2

=q–q

1

, where C

2

is the capacitance of the second capacitor. Substituting C

1

V

0

for q and C

1

V for q

1

, we obtain q

2

=C

1

(V

0

–V). The potential difference across the second capacitor is also V , so the capacitance is :

C

2

=

V

q

2

=

V

V

0

−V

C

1

=

35V

50V−35V

(100pF)=43pF.