A charge Q is placed at each of the two opposite corners of a square. A charge q is placed at each of the other two comers. If the resultant force on each charge q is zero, then :
A
q=√2Q
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B
q=−√2Q
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C
q=2√2Q
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D
q=−2√2Q
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Solution
The correct option is Dq=−2√2Q The net force on q at one corner is zero if →F1+→F2+→F3=0 or F1cos45o^i−F1sin45o^j−F2^j+F3^i=0 so, F1cos45o=−F3....(1) and F1sin45o=−F2....(2) using (1) , kq2(√2a)2×1√2=−kqQa2 or q=−2√2Q