A 400, pF capacitor is charged with a 100, V battery. After disconnecting battery this capacitor is

English

Gujarati

Hindi

2. Electric Potential and Capacitance

medium

A $400\, pF$ capacitor is charged with a $100\, V$ battery. After disconnecting battery this capacitor is connected with another $400\, pF$ capacitor. Then find out energy loss.

A

$1\times10^{-6}\, J$

B

$2\times10^{-6}\, J$

C

$3\times10^{-6}\, J$

D

$4\times10^{-6}\, J$

Solution

Heat loss $ = \frac{{{C_1}{C_2}}}{{2\left( {{C_1} + {C_2}} \right)}}{\left( {{V_1} – {V_2}} \right)^2}$

Standard 12

Physics

Share

Similar Questions

The separation between the plates of a isolated charged parallel plate capacitor is increased. Which of the following quantities will change?

medium

If charge on each plate of a parallel plate capacitor is $Q$ and the magnitude of electric field between the plates is $E$ then force on each plate of a parallel plate capacitor will be

hard

A capacitor is charged by a battery. The battery is removed and another identical uncharged capacitor is connected in parallel. The total electrostatic energy of resulting system

medium

(NEET-2017)

In a uniform electric field, a cube of side $1\ cm$ is placed. The total energy stored in the cube is $8.85\ \mu J.$ The electric field is parallel to four of the faces of the cube. The electric flux through any one of the remaining two faces is.

medium

A parallel plate capacitor after charging is kept connected to a battery and the plates are pulled apart with the help of insulating handles. Now which of the following quantities will decrease?

medium