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Question

(A) Let the side of the square be $a$. The electric field at corner $D$ due to charge at $A$ $(q/2)$ is $E_A = \frac{1}{4 \pi \epsilon_0} \frac{q/2}{a

Vedclass · Accepted Answer

$\frac{q}{4 \pi \epsilon_{0} a^{2}}\left(\frac{1}{\sqrt{2}}+\frac{1}{2}\right)$

Vedclass · Answer

(A) Let the side of the square be $a$. The electric field at corner $D$ due to charge at $A$ $(q/2)$ is $E_A = \frac{1}{4 \pi \epsilon_0} \frac{q/2}{a^2} = \frac{kq}{2a^2}$ (directed along $AD$).The electric field at corner $D$ due to charge at $C$ $(q/2)$ is $E_C = \frac{1}{4 \pi \epsilon_0} \frac{q/2}{a^2} = \frac{kq}{2a^2}$ (directed along $CD$).The resultant of $E_A$ and $E_C$ is $E_{AC} = \sqrt{E_A^2 + E_C^2} = \sqrt{(\frac{kq}{2a^2})^2 + (\frac{kq}{2a^2})^2} = \frac{kq}{2a^2} \sqrt{2} = \frac{kq}{\sqrt{2}a^2}$ (directed along the diagonal $BD$).The electric field at corner $D$ due to charge at $B$ $(q)$ is $E_B = \frac{1}{4 \pi \epsilon_0} \frac{q}{(\sqrt{2}a)^2} = \frac{kq}{2a^2}$ (directed along the diagonal $BD$).Since both $E_{AC}$ and $E_B$ are directed along the same diagonal $BD$,the net electric field at $D$ is $E_{net} = E_{AC} + E_B = \frac{kq}{\sqrt{2}a^2} + \frac{kq}{2a^2} = \frac{kq}{a^2} (\frac{1}{\sqrt{2}} + \frac{1}{2})$.Substituting $k = \frac{1}{4 \pi \epsilon_0}$,we get $E_{net} = \frac{q}{4 \pi \epsilon_0 a^2} (\frac{1}{\sqrt{2}} + \frac{1}{2})$.

The three charges $q/2, q$ and $q/2$ are placed at the corners $A, B$ and $C$ of a square of side '$a$' as shown in the figure. The magnitude of the electric field $(E)$ at the corner $D$ of the square is:

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