Four positive point charges +q are kept at th | vedclass

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(B) Let the square be $ABCD$ with side length $l$. Let $P$ be the midpoint of side $AD$. The charges at $A$ and $D$ are at distance $l/2$ from $P$. Th

Vedclass · Accepted Answer

$\frac{16 k q}{5 \sqrt{5} l^2}$

Vedclass · Answer

(B) Let the square be $ABCD$ with side length $l$. Let $P$ be the midpoint of side $AD$. The charges at $A$ and $D$ are at distance $l/2$ from $P$. The electric field due to charge at $A$ $(E_A)$ is directed away from $A$ along $AD$,and the field due to charge at $D$ $(E_D)$ is directed away from $D$ along $DA$. Since $E_A = E_D = \frac{kq}{(l/2)^2} = \frac{4kq}{l^2}$,they cancel each other out.Now consider charges at $B$ and $C$. The distance from $B$ to $P$ is $r = \sqrt{l^2 + (l/2)^2} = \sqrt{5l^2/4} = \frac{\sqrt{5}l}{2}$.The electric field due to $B$ is $E_B = \frac{kq}{r^2} = \frac{kq}{5l^2/4} = \frac{4kq}{5l^2}$. Similarly,$E_C = \frac{4kq}{5l^2}$.Let $	heta$ be the angle that the line $BP$ makes with the horizontal. Then $\cos 	heta = \frac{l}{r} = \frac{l}{\sqrt{5}l/2} = \frac{2}{\sqrt{5}}$.The vertical components of $E_B$ and $E_C$ cancel out,while the horizontal components add up:$E_{net} = E_B \cos 	heta + E_C \cos 	heta = 2 \cdot \frac{4kq}{5l^2} \cdot \frac{2}{\sqrt{5}} = \frac{16kq}{5\sqrt{5}l^2}$.

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 (Take,$\frac{1}{4 \pi \varepsilon_0}=k$)

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