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 :
Answer & explanation
Correct answer: option 4
Electric field with the plates $\vec{E}=\vec{E}_{Q_1}+\vec{E}_{Q_2}$
$E =E_1-E_2$
$=\frac{Q_1}{2 A \varepsilon_0}-\frac{Q_2}{2 A \varepsilon_0}$
E = $\frac{Q_1-Q_2}{2 A \varepsilon_0}$
∴ Potential difference between the plates
$V_A-V_B =E . d=\left(\frac{Q_1-Q_2}{2 A \varepsilon_0}\right) d$
$=\frac{Q_1-Q_2}{2\left(\frac{A \varepsilon_0}{d}\right)}$
$=\frac{Q_1-Q_2}{2 C}$