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  4. Two identical metal plates are given positive charges Q1 and Q2 (< Q1) respectively. If they are now brought close together to form a parallel plate capacitor with capacitance C,the potential difference between them is

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Q. Two identical metal plates are given positive charges $Q_1 $ and $Q_2 (< Q_1) $ respectively. If they are now brought close together to form a parallel plate capacitor with capacitance C,the potential difference between them is

IIT JEEIIT JEE 1999Electrostatic Potential and Capacitance

Solution:

Electric field within the plates E $= E_{Q_1}+E_{Q_2} $
$\, \, \, \, E=E_1-E_2= \frac {Q_1}{2A \varepsilon_0}- \frac {Q_2}{2A\varepsilon_0},E= \frac {Q_1-Q_2}{2A\varepsilon_0} $
$\therefore $ Potential difference between the plates
$V_A-V_B=ED= \bigg (\frac {Q_1-Q_2}{2A \varepsilon_0}\bigg )d= \frac {Q_1-Q_2}{2 \bigg (\frac {A \varepsilon_0}{d}\bigg )}= \frac {Q_1-Q_2}{2C} $

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