Question

# Two identical small conducting spheres, having charges of opposite sign, attract each other with a force of $$0.108N$$ when separated by $$0.5m$$. The spheres are connected by a conducting wire, which is then removed, and thereafter, they repel each other with a force of $$0.036N$$. The initial charges of the spheres are :

A
±5×106Cand15×106C
B
±1.0×106Cand3.0×106C
C
±2.0×106Cand6.0×106C
D
±0.5×106Cand1.5×106C

Solution

## The correct option is B $$\pm 1.0\times { 10 }^{ -6 }C\quad and\quad \mp 3.0\times { 10 }^{ -6 }C$$Let initial charge be Q1 and Q2.Initial force $$= F = \displaystyle\frac{K \times Q1 \times Q2}{r^2} = -0.108$$    $$F = \displaystyle\frac{9 \times 10^9 \times Q1 \times Q2}{0.25} = -0.108$$$$Q1 \cdot Q2 = - 3 \times 10^{-12}$$Now on connecting them, since they are identical so they will have equal charges.Final charge $$= \displaystyle\frac{Q1 + Q2}{2}$$Final force $$= F = \displaystyle\frac{9 \times 10^9 \times (Q1+Q2)^2 }{4 \times 0.25} = 0.036$$$$Q1 + Q2 = \pm 2 \times 10^{-6}$$$$Q2 = \pm 1.0\times { 10 }^{ -6 }C\quad , Q1 = \mp 3.0\times { 10 }^{ -6 }C$$Physics

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