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Asked by Arushi Juyal 16th December 2017, 6:51 AM
Answered by Expert
This question is very strange to me.
At First, the capcitor is not air-filled/vaccum filled capacitor, because for a capacitor without any dielectric material between parallel plates, we do not have four charged surface as shown in the question. For such a capacitor, top surface of the plate which is connected to positive terminal of battery (or given some postive potential) is positively charged. Similarly the other surface connected to negative terminal has negative charges. Hence only two charged surfaces.
If the capacitor has dielectric material in between parallel plates, in addition to the two charged surfaces as mentioned above, there are two more surface charges due to electric field effect on dielectric material. The figure given below shows the four charged surfaces.
If V is the potential difference between plates and K is dielectric constant, then we have following relation
begin mathsize 12px style q apostrophe space equals space q open parentheses 1 minus 1 over K close parentheses fraction numerator q over denominator K epsilon subscript 0 A end fraction space equals space V over d space space o r space q over V equals space K fraction numerator epsilon subscript 0 A over denominator d end fraction space equals space C space space left parenthesis c a p a c i tan c e space i n space p r e s e n c e space o f space d i e l e c t r i c right parenthesis end style
(Refer any standard physics book like Resnick & Halliday )
A is area of the plate and d is the distance between plates.
if we assume the capacitor given in the question is filled with dielectric, then if we apply this theory given above, the answer to the question is Q1/C.
But this is not mentioned in the list of options.

Answered by Expert 18th December 2017, 10:12 PM
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