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Electricity

New kid on the Block

 Joined: 11 Jun 2009 Post: 23
30 Jul 2009 22:16:37 IST
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Work to charge a capacitor
Engineering Entrance , Medical Entrance , NEET , JEE Main , AIIMS , JEE Main & Advanced , Physics , Electricity

When a capacitor is charged the potential energy stored in it is 1/2 CV^2. Isn't the work done to charge a capacitor CV^2 as half the work done is dissipated as heat?(efficiciency of a capacitor is 50% isn't).

Blazing goIITian

Joined: 4 Apr 2009
Posts: 2068
30 Jul 2009 22:21:33 IST
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No dude.. the work done is actually 1/2 CV^2

Blazing goIITian

Joined: 4 Apr 2009
Posts: 2068
30 Jul 2009 22:24:11 IST
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On any plate of the capacitor , the force is Q*r / 2 E0    instead of   Q*r /  E   .. where r is surface charge density

so the work done is 1/2 CV^2

and the statement is wrong that efficiency of a capacitor is 50%

New kid on the Block

Joined: 11 Jun 2009
Posts: 23
31 Jul 2009 09:04:00 IST
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Well check the refference books and you will see that work done is V*CV but energy stored is half of that and the rest is liberated as heat

Blazing goIITian

Joined: 4 Apr 2009
Posts: 2068
31 Jul 2009 09:49:57 IST
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NO dude..go and check the lectures of MIT , then come and tell ..!! the concept is a little HATKE and the reference books have many mistakes ..!!

kbhi time mile to samjha doonga ..!!

Hot goIITian

Joined: 28 Jul 2009
Posts: 165
31 Jul 2009 12:02:01 IST
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No dude, Suppose that q is the charge on the capacitor at some instant during the charging process. At the same instant, the potential difference across the capacitor is V=q/C. We know that the work necessary to transfer an increment of charge dq from the plate carrying charge -q to the plate carrying charge q (which is at the higher electric potential) isdW= Vdq =q dq / CThe total work required to charge the capacitorfrom q=0 to some final charge q=Q isW = integral of (q dq / C) from limit 0 to Q which is equals to = Q^2 / 2CThe work done in charging the capacitor appears as electric potential energy Ustored in the capacitor.

New kid on the Block

Joined: 11 Jun 2009
Posts: 23
31 Jul 2009 17:56:54 IST
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How come that's true. the potential or work donebring a unit charge is 1 V. then the work done to charge CV coloumb's must be CV*V, ryt.

Blazing goIITian

Joined: 24 Jun 2009
Posts: 391
31 Jul 2009 18:02:26 IST
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WORK DONE IN CHARGING IS CV^2 :HALF IS STORED AS POT. ENERGY AND HALF IS DISSIPATED AS HEAT.AND I M SURE THAT THIS IS RIGHT.

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