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 ${{{q}}_{{1}}}{{ = }}{{{q}}_{{2}}}{{ = }}\dfrac{{{Q}}}{{{2}}}$. Charge density on the left side and the right side of the plate is ${{\sigma = }}\dfrac{{{Q}}}{{{{2A}}}}$. 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
${{E = }}\dfrac{{{\sigma }}}{{{{2}}{{{\varepsilon }}_{{0}}}}}$
Thus, the magnitudes of electric fields due to plate on each side is given by
$\Rightarrow \dfrac{{{{{\sigma }}_1}}}{{{{2}}{{{\varepsilon }}_{{0}}}}}$ and $\dfrac{{{{{\sigma }}_2}}}{{{{2}}{{{\varepsilon }}_{{0}}}}}$
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 ${{{q}}_{{1}}}{{ = }}{{{q}}_{{2}}}{{ = }}\dfrac{{{Q}}}{{{2}}}$
Therefore, the net surface charge density on each side will be $\dfrac{{{Q}}}{{{{2A}}}}$.

(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
${{\sigma = }}\dfrac{{{Q}}}{{{{2A}}}}$
Hence, the electric field is $\dfrac{{{Q}}}{{{{2A}}{{{\varepsilon }}_{{0}}}}}$

(c) Finding the electric field at a point in between the plates
The charge developed on the each side of the plate is ${{{q}}_{{1}}}{{ = }}{{{q}}_{{2}}}{{ = }}\dfrac{{{Q}}}{{{2}}}$
Electric field due to plate is given by
$\Rightarrow {{E = }}\dfrac{{{\sigma }}}{{{{{\varepsilon }}_{{0}}}}}$
$\Rightarrow {{E = }}\dfrac{{{\sigma }}}{{{{{\varepsilon }}_{{0}}}}}{{ = }}\dfrac{{{{(Q/2)/A}}}}{{{{{\varepsilon }}_{{0}}}}}$
On further solving, we get
$\Rightarrow {{E = }}\dfrac{{{Q}}}{{{{2A}}{{{\varepsilon }}_{{0}}}}}$
Thus, the electric field at a point in between the plates is $\dfrac{{{Q}}}{{{{2A}}{{{\varepsilon }}_{{0}}}}}$.

(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
${{\sigma = }}\dfrac{{{Q}}}{{{{2A}}}}$
Hence, the electric field is $\dfrac{{{Q}}}{{{{2A}}{{{\varepsilon }}_{{0}}}}}$

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 ${{Y}}$.