Question

Two conducting plates $X$ and $Y$ , each having large surface area $A$ (on one side), are placed parallel to each other as shown in figure. The plate $X$ is given a charge $Q$ whereas the other is neutral. Find:

(a) the surface charge density at inner surface of the plate $X$
(b) the electric field at a point to the left of the plates
(c) the electric field at a point in between the plates and
(d) the electric field at a point to the right of the plates

Answer 1

Hint: For plate $X$, the net charge given to the plate will be equally distributed on both the sides and the charge developed on the each side will be $q_1=q_2={\displaystyle \frac{Q}{2}}$. Charge density on the left side and the right side of the plate is $\sigma ={\displaystyle \frac{Q}{2A}}$. Use these relations to find out the electric field at a point to the left, right and in between the plates.

Complete step by step solution:
Given: The surface area of conducting plates $X$ and $Y$ is A
Plate $X$ is charged while plate $Y$ is uncharged.
(a) Finding the surface charge density at inner surface of the plate $X$
Let us consider that the surface charge densities on both sides of the plate be ${\sigma }_{1}and{\sigma }_2$.
Electric field due to plate is given by
$E={\displaystyle \frac{\sigma }{2{\epsilon }_0}}$
Thus, the magnitudes of electric fields due to plate on each side is given by
$\Rightarrow {\displaystyle \frac{{\sigma }_1}{2{\epsilon }_0}}$ and ${\displaystyle \frac{{\sigma }_2}{2{\epsilon }_0}}$
The plate has two sides and the area of both the sides is $A$.
Hence, the net charge given to the plate will be equally distributed on both the sides.
The charge developed on the each side will be $q_1=q_2={\displaystyle \frac{Q}{2}}$
Therefore, the net surface charge density on each side will be ${\displaystyle \frac{Q}{2A}}$.

(b) Finding the electric field at a point to the left of the plates
Charge density on the left side of the plate is given by
$\sigma ={\displaystyle \frac{Q}{2A}}$
Hence, the electric field is ${\displaystyle \frac{Q}{2A{\epsilon }_0}}$

(c) Finding the electric field at a point in between the plates
The charge developed on the each side of the plate is $q_1=q_2={\displaystyle \frac{Q}{2}}$
Electric field due to plate is given by
$\Rightarrow E={\displaystyle \frac{\sigma }{{\epsilon }_0}}$
$\Rightarrow E={\displaystyle \frac{\sigma }{{\epsilon }_0}}={\displaystyle \frac{(Q/2)/A}{{\epsilon }_0}}$
On further solving, we get
$\Rightarrow E={\displaystyle \frac{Q}{2A{\epsilon }_0}}$
Thus, the electric field at a point in between the plates is ${\displaystyle \frac{Q}{2A{\epsilon }_0}}$.

(d) Finding the electric field at a point to the right of the plates
Charge density on the left side of the plate is given by
$\sigma ={\displaystyle \frac{Q}{2A}}$
Hence, the electric field is ${\displaystyle \frac{Q}{2A{\epsilon }_0}}$

Note: Electric field is defined as the electric force per unit charge. The plate $X$ is positively charged and plate $Y$ is neutral. Here, the charged plate acts as a source of electric field with positive in the inner side and a negative charge is induced on the inner side of plate $Y$.