JEE MAIN - Physics (2022 - 28th June Morning Shift - No. 9)
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 figure. The magnitude of electric field (E) at the corner D of the square, is :
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$${q \over {4\pi { \in _0}{a^2}}}\left( {{1 \over {\sqrt 2 }} + {1 \over 2}} \right)$$
$${q \over {4\pi { \in _0}{a^2}}}\left( {1 + {1 \over {\sqrt 2 }}} \right)$$
$${q \over {4\pi { \in _0}{a^2}}}\left( {1 - {1 \over {\sqrt 2 }}} \right)$$
$${q \over {4\pi { \in _0}{a^2}}}\left( {{1 \over {\sqrt 2 }} - {1 \over 2}} \right)$$
Explanation
$$\left| {{E_0}} \right| = {{kq/2} \over {{a^2}}}\sqrt 2 + {{kq} \over {{{\left( {a\sqrt 2 } \right)}^2}}}$$
$$ = {{kq} \over {\sqrt 2 {a^2}}} + {{kq} \over {2{a^2}}}$$
$$ = {{kq} \over {{a^2}}}\left( {{1 \over {\sqrt 2 }} + {1 \over 2}} \right),\,k = {1 \over {4\pi {\varepsilon _0}}}$$
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