What is the new force of repulsion between A and B?

3.802 \times 10^{- 3} N

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8.336 \times 10^{- 3} N

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15.21 \times 10^{- 3} N

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5.704 \times 10^{- 3} N

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Solution

The correct option is D $5.704 \times 10^{- 3} N$
Therefore, new force of repulsion between the spheres A and B,

$F_{a i r} = \frac{1}{4 π ϵ_{0}} \cdot \frac{q_{1}^{'} q_{2}^{'}}{d^{2}}$

$= 9 \times 10^{9} \times \frac{3.25 \times 10^{- 7} \times 4.875 \times 10^{- 7}}{(0.05)^{2}}$
$= 5.704 \times 10^{- 3} N$

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What is the new force of repulsion between A and B?

The correct option is D 5.704×10−3NTherefore, new force of repulsion between the spheres A and B,Fair=14πϵ0⋅q′1q′2d2=9×109×3.25×10−7×4.875×10−7(0.05)2=5.704×10−3N