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Question

Three point charges of 3×10−9C are situated at each of three corners of a square, whose side is 15 cm. The magnitude and direction of the electric field at the vacant corner of the square are:

A
2130 V/m along the diagonal
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B
9622 V/m along the diagonal
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C
22.0 V/m along the diagonal
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D
zero
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Solution

The correct option is A 2130 V/m along the diagonalEγAB=KQr2=9×109×3×10−915×10−2×15×10−2=27×104225=1.2×103 N/CEγAC=KQr2=9×109×3×10−915×10−2×15×10−2=27×104225=1.2×103 N/CWe can observe from the figure that EγAB⊥EγACUsing parallelogram law of vector addition,Eγr1=√(EAB)2+(EAC)2=1.7×103 N/C direction along the diagonal.For, EγAC=KQ(r′)2=9×109×3×10−9(21×10−2)2=6.1×10−210−4=6.1×102=0.61×103 N/CDirected along the diagonal.Therefore EγAC∥Eγr1, resultant Eγr will be given by,Eγ=2.31×103 N/C=2130 V/m directed along the diagonal of the square.

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