Question

Two parallel plane sheets 1 and 2 carry uniform charge densities ${\sigma }_1$ and ${\sigma }_2$ (see Fig. 20.8). The electric field in the region marked I is $({\sigma }_1>{\sigma }_2)$. (Consider the right direction as the positive direction)

• A. $-\frac{{\sigma }_1}{2{ϵ}_0}$
• B. $-\frac{{\sigma }_2}{2{ϵ}_0}$
• C. $-\frac{({\sigma }_1+{\sigma }_2)}{2{ϵ}_0}$
• D. $-\frac{({\sigma }_1-{\sigma }_2)}{2{ϵ}_0}$

Answer 1

The correct option is C

$-\frac{({\sigma }_1+{\sigma }_2)}{2{ϵ}_0}$

The left sheet 1 produces a field of magnitude $E_1=\frac{{\sigma }_1}{2{ϵ}_0}$ which points away from it, to the left in region I and to the right in regions II and III. The right sheet 2 produces a field of magnitude $E_2=\frac{{\sigma }_2}{2{ϵ}_0}$ which points away from it, to the left in regions II and I and to the right in region III. If ${\sigma }_1>{\sigma }_2>0$ the field in each region from the principle of superposition will be as follows:

Region I : $E=\frac{1}{2{ϵ}_0}({\sigma }_1+{\sigma }_2)$ towards left.
Region II: $E=\frac{1}{2{ϵ}_0}({\sigma }_1-{\sigma }_2)$ towards right.
Region III: $E=\frac{1}{2{ϵ}_0}({\sigma }_1+{\sigma }_2)$ towards right.