Question

Two conducting plates $\text{A}$ and $\text{B}$ are placed parallel to each other. $\text{A}$ and $\text{B}$ are given charges $10\text{nC}$ and $12\text{nC}$ respectively . Find the distribution of charge on the inner surface of plate $\text{A}$.

• A. $-1\text{nC}$
• B. $-2\text{nC}$
• C. $-3\text{nC}$
• D. $-4\text{nC}$

Answer 1

The correct option is A $-1\text{nC}$


Consider a Gaussian surface as shown in figure $\left(a\right)$. Two faces of this closed surface lie completely inside the conductor where the electric field is zero. The flux through these faces is, therefore, zero.

The other two faces of the closed surface which are outside the conductor are parallel to the electric field and hence the flux through these parts is also zero.

The total flux of the electric field through the closed surface is, therefore, zero. From Gauss's law, the total charge inside this closed surface should also be zero.

i.e the charge on the inner surface of $\text{A}$ should be equal and apposite to that on the inner surface of $\text{B}$.

Let the distribution of charges be as shown in figure $\left(b\right)$.
To find the value of $q$, consider the field at a point $\text{P}$ inside the plate $\text{A}$.

Suppose, the surface area of the plate (one side) is $A$.
Due to the charge $10-q$, electric field at point $\text{P}$,
$E_1=\frac{10-q}{2A{ϵ}_0}$ $($downward$)$
Due to the charge $+q$,
$E_2=\frac{q}{2A{ϵ}_0}$ $($upward$)$
Due to the charge $-q$,
$E_2=\frac{q}{2A{ϵ}_0}$ $($downward$)$
Due to the charge $12+q$,
$E_4=\frac{12+q}{2A{ϵ}_0}$ $($upward$)$

As the point $\text{P}$ is inside the conductor, the net electric field at $\text{P}$ should be zero.
So, $\overrightarrow{E_1}+\overrightarrow{E_2}+\overrightarrow{E_3}+\overrightarrow{E_4}=0$

$\frac{10-q}{2A{ϵ}_0}-\frac{q}{2A{ϵ}_0}+\frac{q}{2A{ϵ}_0}-\frac{12+q}{2A{ϵ}_0}=0$

$\Rightarrow 10-q-12-q=0$

$\Rightarrow q=-1\text{nC}$

Hence, option (a) is the correct answer.