a 600pF capacitor is charged by 2oov supply. it is then disconnected from the supply and is connected to another uncharged 600pF capacitor.how much electrostatic energy is lost in the process

Asked by Manisha Mehrotra | 29th Sep, 2010, 11:54: AM

Expert Answer:

Dear Student

In this question, first you need to calculate the energy of the first capacitor initially and then you need to calculate the energies of both the capacitors finally, and the difference in the energy of initial capacitor and the combined final energies of the 2 capacitors will give you the energy loss.

So, initially the energy of the capacitor is given by (1/2)C*V*V, where C is the capacitance and V is the voltage between the terminals of the capacitor.

So, the initial energy , let us say E1 = (1/2)C*V*V =(1/2)*600pf*200*200 = 12000000 pJ = 12*10-6J          as 1pf = 10-12 farad.

so, E1 = 12*10-6J

Now when we connect that capacitor to the uncharged capacitor 2 things happen

1- charge goes from charged to uncharged capacitor but the total charge will remain fixed as charge can not escape from anywhere.

2- the voltage between the terminals of both the capacitotrs will be same as both terminals are now connected.

so let us say the final common voltage is V

so total final charge = initial charge

and charge of a capacitor is given by q= CV

so, v(C1+c2)= 600*200pC

or, v (1200pf)= 120000pf

so V= 100V

so final energy = 1/2(c1+c2)*V*V = 1/2 * 1200 *100*100 = 6000000pJ = 6 *10-6 J

so differnece in energy = (12-6)* 10-6 J= 6 *10-6 J

this is the energy lost in the process.

We hope that clarifies your query.

Regards

Team

TopperLearning

Answered by  | 1st Oct, 2010, 04:20: PM

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