A point charge q is fixed at each of the two opposite corners of a square. A third point charge Q is placed at each of the other two corners. If one of the charge Q experiences no force,then
A
Q=√2q
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B
Q=−2√2q
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C
Q=√2q2
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D
Q=−2√2q2
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Solution
The correct option is BQ=−2√2q If the charge at B experiences zero net force, then FABcos45+FCBcos45+FDB=0 14πϵ0[qQa21√2+qQa21√2+QQ(a√2)2]=0 √2qQa2=−QQ2a2⇒Q=−2√2q