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