The distance between charges 5×10−11C and −2.7×10−11C is 0.2 m. The distance at which a third charge should be placed in order that it will not experience any force along the line joining the two charges is

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0.44 m

0.65 m

0.556 m

0.350 m

answer is C.

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Detailed Solution

If two opposite charges are separated by a certain distance, then for it’s equilibrium a third charge should be kept outside and near the charge which is smaller in magnitude. Here, suppose third charge q is placed at a distance x from -2.7×10−11C then for it’s equilibrium |F1|=F2⇒kQ1q(x+0.2)2=kQ2qx2⇒x=0.556m Here k=14πε0 and Q1=5×10−11C,Q2=−2.7×10−11C

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