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Two identical conducting plates of surface area A each are fixed parallel to each other with a small separation d between them. The plates are given total charge +3Q and -Q. The potential difference between the plates is $X (\frac{Qd}{ε_{0} A})$ . The value of X is ___________.

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answer is 2.

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Detailed Solution

Electric field between the plates, $E = \frac{Q_{1} - Q_{2}}{2 ε_{0} A} = \frac{2 Q}{ε_{0} A}$

Therefore, the potential difference, $Δ V = E d = \frac{2 Q d}{ε_{0} A}$

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