Question

Two metallic bodies separated by a distance $20\text{cm}$, are given equal and opposite charges of magnitude $0.88\mu \text{C}$. The component of electric field along the line $\text{AB}$, between the plates, varies as $E_x=3x^2+0.4\text{N/C}$, where $x$(in meters) is the distance from one body with respect to the other body as shown below.

• A. The capacitance of the system is $10\mu \text{F}$
• B. The capacitance of the system is $20\mu \text{F}$
• C. The potential difference between A and C is $0.088\text{V}$.
• D. The potential difference between A and C cannot be detrmined from the given data.

Answer 1

Answer: (A), (C)

The potential difference between A and C is $0.088\text{V}$.

Given, $d=20\text{cm}=0.2\text{m}$

$Q=0.88\mu \text{C}$

$E_{AB\left(x\right)}=3x^2+0.9\text{N/C}$

Since, $E$ is only a function of $x$, By using $E=-\frac{dV}{dx}$

${\int }_0^{V}dV=-\underset{0}{\overset{d}{\int }}Edx$

$\Rightarrow V=-\underset{0}{\overset{d}{\int }}(3x^2+0.4)dx$

$\Rightarrow V=-[\frac{3x^3}{3}+0.4x]{|}_0^d$

$\Rightarrow V=-(d^3+0.4d)$

$\Rightarrow V=-\left(\right(0.2{)}^3+\left(0.4\right)\left(0.2\right))$

$\Rightarrow V=-0.088\text{Volts}$

Potential difference between $\text{A}$ and $\text{B}$ is $0.088\text{V}$

Capacitance of the system $C=\frac{Q}{V}$

$\Rightarrow C=\frac{0.88}{0.088}=10\mu \text{F}$

$\text{B}$ and $\text{C}$ are equipotential because it is a metallic plate

Hence, options (a) and (c) are the correct answers.