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Numerical

What is the force of repulsion if each sphere is charged double the above amount, and the distance between them is halved?

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#### Solution

After doubling the charge, charge on sphere A, q_{A} = Charge on sphere B, q_{B} = 2 × 6.5 × 10^{−7} C = 1.3 × 10^{−6} C

The distance between the spheres is halved.

∴ `"r" = 0.5/2 = 0.25` m

Force of repulsion between the two spheres,

`"F" = ("q"_"A""q"_"B")/(4piin_0r^2)`

= `(9 xx 10^9 xx 1.3 xx 10^-6 xx 1.3 xx 10^-6)/((0.25)^2)`

= 16 × 1.52 × 10^{−2}

= 0.243 N

Therefore, the force between the two spheres is 0.243 N.

Concept: Charging by Induction

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Chapter 1: Electric Charges and Fields - Exercise [Page 47]