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(B) Let the square be $ABCD$ with side length $l$. Let the charges be at corners $A, B, C, D$. Consider the midpoint $M$ of side $CD$. The electric fi

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 the charges be at corners $A, B, C, D$. Consider the midpoint $M$ of side $CD$. The electric fields due to charges at $C$ and $D$ are $E_C$ and $E_D$. Since $MC = MD = l/2$,$E_C = k q / (l/2)^2$ directed away from $C$,and $E_D = k q / (l/2)^2$ directed away from $D$. These are equal and opposite,so they cancel out.Now,consider the fields due to charges at $A$ and $B$. The distance from $A$ to $M$ is $r = \sqrt{l^2 + (l/2)^2} = \sqrt{5l^2/4} = l\sqrt{5}/2$. The magnitude of the field is $E_A = E_B = k q / r^2 = k q / (5l^2/4) = 4kq / 5l^2$.The vertical components of $E_A$ and $E_B$ add up,while horizontal components cancel. Let $	heta$ be the angle the vector $AM$ makes with the vertical. Then $\cos 	heta = l / r = l / (l\sqrt{5}/2) = 2/\sqrt{5}$.The net electric field is $E_{net} = 2 E_A \cos 	heta = 2 	imes (4kq / 5l^2) 	imes (2/\sqrt{5}) = 16kq / 5\sqrt{5}l^2$.

Four positive point charges $+q$ are kept at the four corners of a square of side $l$. The net electric field at the midpoint of any one side of the square is (take $\frac{1}{4 \pi \epsilon_0}=k$ ).

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