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Suppose a charge +Q1 is given to the positive plate and a charge -Q2 to the negative plate of a capacitor. What is the "charge on the capacitor" ?
Dear Nikhil,By applying gauss law it can easily be proven that inner faces of plates shall have equal charge of opposite magnitudes, say q0 and -q0. Field on the outside of plates will be (Q1-Q2)/2ε0 and this gives charge of (Q1-Q2)/2 on the outer plates from gauss law. This implies inner sides of plates shall have q0 = (Q1+Q2)/2 and -q0 = -(Q1+Q2)/2. Charge on the capacitor hence is (Q1+Q2)/2.Please feel free to post as many doubts on our discussion forum as you can.we will get you the answer and detailed solution very quickly. All the best.Regards,Askiitians Experts,Gokul Joshi
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