Question

Four positive point charges $+q$ are kept at the four corners of a square of side $l$. The net electric field at the mid-point of any one side of the square is $\left(\right.$ Take, $\left.\frac{1}{4 \pi \varepsilon_{0}}=k\right)$

• A. $\frac{4kq}{l^{2}} $
• B. $\frac{16 kq}{5 \sqrt{5} l^{2}}$
• C. $\frac{8kq}{5 \sqrt{5} l^{2}}$
• D. $\frac{kq}{ l^{2}}$

Answer 1

Answer: (B)

The force at midpoint of any side of a square let's say at point $P$ which is midpoint of side $A B$ is due to charges at corners $A, B, C$ and $D$
$F_{P}=F_{A P}+F_{B P}+F_{C P}+F_{D P}$

$A P= D P =l$
$B P = C P =\sqrt{l^{2}+(l / 2)^{2}}=\sqrt{5 l^{2} / 4}$
Electric field due to charges at $A$ and $D$ will be equal in magnitude and opposite in direction, so will cancel each other.
Similarly, the vertical components of electric field due to charges at $B$ and $C$ will also cancel each other, but the horizontal components of these will add up to give resultant electric field.
$E=2 \cdot \frac{k q}{\left(\sqrt{5 l^{2} / 4}\right)^{2}} \cdot \cos \theta$
$=2 \times \frac{4 k q}{5 l^{2}} \times \frac{2}{5}=\frac{16 k q}{5 \sqrt{5} l^{2}}$