Question

A charge $Q$ is placed at the centre of a square. If electric field intensity due to the charge at the comers of the square is $E_{1}$ and the intensity at the mid point of the side of 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

Answer: (C)

$E_{1}=\frac{1}{4 \pi \varepsilon_{0}} \frac{Q^{2}}{l^{2}}\,\,\,...(i)$
$E_{2}=\frac{1}{4 \pi \varepsilon_{0}} \frac{Q^{4}}{l^{2}}$
$E_{2}=2 E_{1}$
$\frac{E_{1}}{E_{2}}=\frac{1}{2}$