Question

Each of two large conducting parallel plates has one sided surface area A. If one of the plates is given a charge Q whereas the other is neutral, then the electric field at a point in between the plates is given by:

• A. $\frac{Q}{A{\epsilon }_0}$
• B. $\frac{Q}{2A{\epsilon }_0}$
• C. $\frac{Q}{4A{\epsilon }_0}$
• D. Zero

Answer 1

Answer: (B)

$\frac{Q}{2A{\epsilon }_0}$

Each of two large conducting parallel plates has one sided.

Surface area $=A$

Charge $=Q$

electric field $=\frac{Q}{2A{ϵ}_0}$

because,

there is two large conducting parallel plate.

hence, $\frac{Q}{2}$ and $\lambda =\frac{Q}{2}\times \frac{1}{A{ϵ}_0}$ (where $\frac{Q}{{ϵ}_0}=$ using Gauss law)