Question 11

A 600 pF capacitor is charged by a 200 V supply. It is then disconnected from the supply and is connected to another uncharged 600 pF capacitor. How much electrostatic energy is lost in the process?

Answer

Capacitance of the capacitor, *C* = 600 pF

Potential difference, *V* = 200 V

Electrostatic energy stored in the capacitor is given by,

If supply is disconnected from the capacitor and another capacitor of capacitance *C* = 600 pF is connected to it, then equivalent capacitance (*C*^{'}) of the combination is given by,

New electrostatic energy can be calculated as

Loss in electrostatic enegy = *E - E'*

= 1.2 x 10^{-5} - 0.6 x 10^{-5}

= 0.6 x 10^{-5}

= 6 x 10^{-6} J

Therefore, the electrostatic energy lost in the process is 6 x 10^{-6} J.

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A parallel plate capacitor with air between the plates has a capacitance of 8 pF (1pF = 10

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A point charge +10 μC is a distance 5 cm directly above the centre of a square of side 10 cm, as shown in Fig. 1.34. What is the magnitude of the electric flux through the square? (Hint: Think of the square as one face of a cube with edge 10 cm.)

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- Q:-
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Anjali Kashyap
2019-07-23 18:58:14

Thank you

Adithya
2019-06-23 19:28:03

common potential =c1c2\\c1+c2

Deepika
2019-06-14 11:05:07

How can we now that it is connected in series

Ameer Navas
2019-06-08 22:37:09

It is parallel not series.

Gode Basar
2019-05-10 07:07:36

How can we know that the capacitors are connected in series?

Nandini
2019-05-08 10:41:10

How can we conclude that capacitors are in series

Vibhor Singh
2018-05-23 00:58:49

In this question it is not written that uncharged capacitor is connected in series then how you take in series according to the solution??

vinay
2018-02-28 07:17:15

what if the unchanged capacitor is connected to parallel with it

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