# A 4 micro farad capacitor charged by 200 volt supply. It is then dissconnected from the supply and is connected to anothe uncharged 2 micro farad capacitor. How much electrostatic energy of first capacitor is lost in the form of heat or e. m radiations?

Here,

C1 = 4 μF = 4 ×10-6F

V1 = 200 volt

C2 = 2 μF = 2 ×10-6F

V2 = 0

Initial Electrostatic energy stored in the capacitors will be

Ui = ½ C1V12 + ½ C2V22

=>Ui = ½ ×4 ×10-6 × (200)2 + 0 = 8×10-2J

After connecting the two capacitors, the total capacitance becomes

C = C1 + C2 = 4 ×10-6 + 2 ×10-6 = 6×10-6 F

Charge in the first capacitor

Q = C1V1

=> Q = 4 ×10-6×200 = 8×10-4 C

Since, the 2nd capacitor is uncharged so total charge on the system will be Q = 8×10-4 C

Let V be the common potential of the capacitors after connection

V = Q/C

=> V = 8×10-4/6×10-6 = 4/3×102 volt

Now final electrostatic energy of the combination will be

Uf = ½CV2 = ½×6×10-6×(4/3×102)2  = 4×10-2 J = 5.33×10-2

Now loss in energy in form of heat etc will be

Ui - Uf = (8-5.33) ×10-2 J = 2.67 ×10-2J

• 62

U1 = (C1V^2)/2

when its is connent to the second capacitor , the net charge will reamain same and the net capacitace will become 4/3 micro farad .... so here by Q=CV   we can find the new V . and we already know the new capacitance , so U2 =(c2v^2)/2

loss in energy = U1-U2

• 6

ooh k thnx a lot vidit,, il try it...!!:)

• -5
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