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Two identical metal plates are given the charges 0 1 and 02 (02 < 01) respectively. If they are now brought close together to form a parallel plate capacitor with capacitance C then what is the potential difference between them?

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Solution

$p a r a l l e l p l a t e c a p a c i t o r c o n t a i n s t h e c h a r g e s o f o p p o s i t e s i g n . S O n e t c h a r g e = Q_{01} - Q_{02} c i s t h e c a p a c i \tan c e o f t h e m p o t e n t i a l i s g i v e b y V = \frac{Q}{C} V = \frac{Q_{01} - Q_{02}}{C}$

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