Question

Two infinite parallel metal planes, contain electric charges with charge densities $+\sigma$ and $-\sigma$ respectively and they are separated by a small distance in air. If the permittivity of air is ${\epsilon }_0$ then, the magnitude of the field between the two planes with its direction will be:

• A. $\sigma /{\epsilon }_0$ towards the positively charged plane
• B. $\sigma /{\epsilon }_0$ towards negatively charged plane
• C. $\sigma /\left(2{\epsilon }_0\right)$ towards the positively charged plane
• D. $0$ and towards any direction

Answer 1

The correct option is B $\sigma /{\epsilon }_0$ towards negatively charged plane

Electric field due to charged sheet is $E=\frac{\sigma }{2{\epsilon }_o}$

Electric field due to sheet having charge density $\sigma =\frac{\sigma }{2{\epsilon }_o}$ away from the sheet.

Electric field due to sheet having charge density $-\sigma =\frac{\sigma }{2{\epsilon }_o}$ towards the sheet.

So,

The net field $=\frac{\sigma }{2{\epsilon }_o}+\frac{\sigma }{2{\epsilon }_o}=\frac{\sigma }{{\epsilon }_o}$ towards the negatively charged plate.