Question

A charge $Q$ is placed at the center of a square. If electric field intensity due to the charge at the corners of the square is $E_1$ and the intensity at the mid point of the side square is $E_2$ then the ratio of $\frac{E_1}{E_2}$ will be

• A. $\frac{1}{2\sqrt{2}}$
• B. $\sqrt{2}$
• C. $\frac{1}{2}$
• D. $2$

Answer 1

The correct option is C $\frac{1}{2}$

$\text{Step 1: Calculate all the distances[Ref. fig]}$

Distance of charge $Q$ from corner $D$ :

$r_1=OD=\frac{BD}{2}=\frac{\sqrt{a^2+a^2}}{2}=\frac{a}{\sqrt{2}}$

Distance of charge $Q$ from midpoint $E$ of the side $AD$:

$r_2=OE=\frac{a}{2}$

$\text{Step 2: Electric field at corner D}$

$E_1=\frac{1}{4\pi {ϵ}_0}\frac{Q}{r_1^2}=\frac{1}{4\pi {ϵ}_0}\frac{Q}{(a/\sqrt{2}{)}^2}=\frac{2Q}{\left(4\pi {ϵ}_0\right)a^2}$ $....\left(1\right)$

$\text{Step 3: Electric field at midpoint E of side AD}$

$E_2=\frac{1}{4\pi {ϵ}_0}\frac{Q}{r_2^2}=\frac{1}{4\pi {ϵ}_0}\frac{Q}{(a/2{)}^2}$ $=\frac{4Q}{\left(4\pi {ϵ}_0\right)a^2}$ $....\left(2\right)$

$\text{Step 4: Ratio of}{E}_1\&E_2$

Dividing eq $\left(1\right)$ and $\left(2\right)$

$\frac{E_1}{E_2}=\frac{1}{2}$

Hence, Option (C) is correct.