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Q. Two identical thin metal plates has charge $q_1$ and $q _2$ respectively such that $q _ 1 > q _2$. The plates were brought close to each other to form a parallel plate capacitor of capacitance $C$. The potential difference between them is :

JEE MainJEE Main 2022Electrostatic Potential and Capacitance

Solution:

Electric field between plates $E =\frac{ q _1- q _2}{2 A \epsilon_0}$
$V = Ed =\frac{ q _1- q _2}{2 A \in_0} d$
$V =\frac{ q _1- q _2}{2 C }$