Capacitance HCV Doubt Q30

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Capacitance HCV Doubt Q30

Post  Shohini on Wed Jun 11, 2008 11:14 am

A chare +2*10^-8C is placed on the positive plate and a charge of -1*10^-8C on the negative plate of a parallel plate capacitor of capacitance 1.2*10^-3microF.
Calculate the potential difference developed between the plates.

Ans:12.5V

Please explain

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Re: Capacitance HCV Doubt Q30

Post  Pinnaka on Thu Jun 12, 2008 3:16 am

once again charge re distribution takes place, as said in my other post at the following link

http://iitjeescience.bigforumpro.com/hc-verma-corner-f28/capacitance-hcv-doubt-q31-t26.htm?sid=2e5ff27f983e7292d0c159d929761fb1

charge distribute symmetrically on a plate when they are isolated. therefore on the positive plate the sides have 1*10^-8mc on them. the negative plate is having 0.5*10^-8mc on its sides. Now when they are brought together, the sides facing each other have un equal magnitudes of charges, now they influence the charge distribution on the other plate. the negatively charge plate induces a charge of +0.5*10^-8 C on the side facing it and 0.5*10^-8C on the side facing away from it, therefore the net charges on the side of the positively charges plate are as follows
0.5*10^-8C (facing away)
and 1.5 *10^-8C (facing towards negatively charged plate)

Similarly on the negatively charged plate there is induction which leads to the following distribution.
the side facing positive plate is -1.5*10^-8 C and the side facing away 0.5*10^-8 C
the charges (values) refer to the final distribution, once again as stated n my post in


http://iitjeescience.bigforumpro.com/hc-verma-corner-f28/capacitance-hcv-doubt-q31-t26.htm?sid=2e5ff27f983e7292d0c159d929761fb1
the effective charge is 1.5*10^-8 C

by V = q/c

we get the answer


Hope this helps
Pinnakas

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