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Question

# Three electric charges +q each are placed at the three corners of a square of side d. The intensity of electric field at the fourth corner is :

A
14πϵ0qd2(2+12)
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B
14πϵ0qd2(2+12)
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C
14πϵ02Qd2
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D
14πϵ02qd2
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Solution

## The correct option is D 14πϵ0qd2(√2+12)Now E due to charge q at A is kqd2(−^y).E due to charge q at C is kqd2(−^x)E due charge q at B is kq(√2d)2cos450(−^x)+kq(√2d)2sin450(−^y)=kq2√2d2(−^x)+kq2√2d2(−^y)Now total E=kqd2(1+12√2)(−^x)+kqd2(1+12√2)(−^y)Magnitude of E=√Ex2+Ey2E=kqd2(√2+12)

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