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 the 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

(a) For the surface change density of a single plate.

Let the surface charge density at both sides be ${\sigma }_1$ & ${\sigma }_2$

Now, electric field at both ends.

$=\frac{{\sigma }_1}{2{\epsilon }_0}$ & $\frac{{\sigma }_2}{2{\epsilon }_0}$

Due to a net balanced electric field on the plate $\frac{{\sigma }_1}{2{\epsilon }_0}$ & $\frac{{\sigma }_2}{2{\epsilon }_0}$

$\therefore {\sigma }_1={\sigma }_2$

So, $q_2=\frac{Q}{2}$

$\therefore$ Net surface charge density $=\frac{Q}{2A}$

(b) Electric field to the left of the plates $=\frac{\sigma }{{\epsilon }_0}$

Since $\sigma =\frac{Q}{2A}$

Hence electric field $=\frac{Q}{2A{\epsilon }_0}$

This must be directed towards left as '$X$' is the charged plate.

(c) and (d) Here ion both the cases the charged plate '$X$' acts as the only source of electric field, with (+ve) in the inner side and '$Y$' attracts towards it with (-ve) he in its inner side. So for the middle portion $E=\frac{Q}{2A{\epsilon }_0}$ towards right.

(d) Similarly, for extreme right, the outer side of the '$Y$' plate acts as positive and hence right with $E=\frac{Q}{2A{\epsilon }_0}$