Question

In the figure shown the plates of a parallel capacitor have unequal charges. Its capacitance is ${}^{\prime }{C}^{\prime },P$ is a point outside the capacitor and close to the plate of charge $-Q$. The distance between the plates is ${}^{\prime }{d}^{\prime }.$

Choose the following correct statement(s)

• A. A point charge at point ${}^{\prime }{P}^{\prime }$ will experience electric force due to capacitor
• B. The potential difference between the plates will be $\frac{3Q}{2C}$
• C. The energy stored in the electric field in the region between the plates is $\frac{9Q^2}{8C}$
• D. The force on one plate due to the other plate is $\frac{Q^2}{2\pi {\epsilon }_0{d}^2}$

Answer 1

Answer: (A), (B), (C)

The energy stored in the electric field in the region between the plates is $\frac{9Q^2}{8C}$

Charge redistribution on the plates can be considered as


At point $P$ electric field of $\frac{3Q}{2}\text{and}\frac{-3Q}{2}$ will be cancelled but $\frac{Q}{2}\text{and}\frac{Q}{2}$ will result to non zero electric field at $P$
So, option $\left(a\right)$ is correct.
Now, potential difference between capacitors is
$\Rightarrow \Delta V=\frac{Q_{in}}{C}=\frac{\frac{3Q}{2}}{C}=\frac{3Q}{2C}....\left(b\right)$

Energy stored in the capacitor is
$\Rightarrow U=\frac{Q_{in}^2}{2C}=\frac{9Q^2}{4\times 2C}=\frac{9}{8}\frac{Q^2}{C}....\left(c\right)$
Force acting on one of the plates is
$\Rightarrow F=(\frac{\frac{3Q}{2A}}{2{\epsilon }_0}+\frac{\frac{Q}{2}}{2{\epsilon }_0})\left(Q\right)=\frac{Q^2}{A{\epsilon }_0}$