Capacitance HCV Doubt Q31

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Capacitance HCV Doubt Q31

Post  Shohini on Wed Jun 11, 2008 11:15 am

A charge of 20 microCis placed on the positive plate of an isolated parallel plate capacitor of capacitance 10 microF. Calculate the potential difference developed between the plates.

Ans:1V

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Post  Shohini on Wed Jun 11, 2008 10:40 pm

Plz give me the solution in some lucid language. I am new to these topics and thus have a little difficulty to understand. Smile

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Re: Capacitance HCV Doubt Q31

Post  Pinnaka on Thu Jun 12, 2008 3:15 am

Ok I can understand the trouble u are having, In the following solution I will give two methods to answer the question one s entirely conceptual .

Now consider the plate to which the charge is given to be free in space and the plate is charged to 20 micro coulombs (from now on I will refer charge given as mc and C as mF).
Then the charge on the plate symmetrically distributes on the both sides side of the plate. Therefore one side of the plate will be having 10 mc and the other 10mc. If ask why can’t the charge be inside the metal surface, it is not allowed. It is know fact that the charge contained in a metal (inside) is always zero under electrostatic conditions. Now we have charge given to the plate equally distributed on both sides of the plate. Now bring in the other plate of the capacitor. Now this plate is neutral and by the influence of the charged plate the 2nd plate is polarized. Once ones side of 1st plate is facing the 1 side of the 2nd plate, then by induction the 1st plate induces a charge of –10mc to the side of the 2nd plate which facing the 1st plate. Now the other side, which is facing away, gets 10mc. Now we have found out the charge distribution on the plates entirely by conceptual means.

Now we have 1st place with 10 and 10 mc on both sides, the second plate with –10mc and 10mc on two sides. Now what is the potential difference between the two plates.
Now the two side of the plates which are facing away from each other contain 10 mc which is equivalent to saying that they are having charge of equal polarization. Therefore we can treat them as having no charges. Now the effective charge of the capacitor is 10 mc and C is given to 10mF and V = q/c therefore V = 1V.

As a physics student I always like to try out the entirely conceptual ways before going inot the mathematics part. As u can see the answer is very simple. With enough practice u hardly need a pen and a paper to solve it

Hope this helps
Pinnaka

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