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  • Two identical thin metal plates has charge <img alt="\mathrm{q_{1} \: and \: q_{2}}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7Bq_%7B1%7D%20%5C%3A%20and%20%5C%3A%20q_%7B2%7D%7

Two identical thin metal plates has charge \mathrm{q_{1} \: and \: q_{2}} respectively such that \mathrm{q_{1}>q_{2}}. The plates were brought close to each other to form a parallel plate capacitor of capacitance \mathrm{ C}. The potential difference between them is :
 

Option: 1

\mathrm{\frac{\left(q_{1}+q_{2}\right)}{C}}


Option: 2

\mathrm{\frac{\left(q_{1}-q_{2}\right)}{C}}


Option: 3

\mathrm{\frac{\left(q_{1}-q_{2}\right)}{2 C}}


Option: 4

\mathrm{\frac{2\left(q_{1}-q_{2}\right)}{C}}


Answers (1)

best_answer

\mathrm{q=C(\Delta v)}

\mathrm{\Delta v=\frac{q}{c}=\frac{\left ( q_{1}-q_{2} \right )}{2c}}

The inner surface charge is taken into consideration as outer surface charges has no contribution in the potential difference between the plates of capacitor

Hence 3 is correct option

Posted by

Suraj Bhandari

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