# Two identical insulated charged copper spheres A and B have their centres separated by a distance of 50 cm. What is the mutual force of electrostatic repulsion if the charge on each is 6.5 × 10-7 C? The radii of A and B are negligible compared to the distance of separation. (b) What is the force of repulsion if each sphere is charged double the above amount, and the distance between them is halved? (c). A third sphere of the same size but uncharged is brought in contact with the first, then brought in contact with the second, and finally removed from both. What is the new force of repulsion between A and B?

Dear Student,

The solution to first 2 parts has been provided below:

(a)Charge on sphere A, *q*_{A} = Charge on sphere B, *q*_{B} = 6.5 × 10^{−7} C

Distance between the spheres, *r* = 50 cm = 0.5 m

Force of repulsion between the two spheres,

Where,

∈_{0} = Free space permittivity

= 9 × 10^{9} N m^{2} C^{−2}

∴

= 1.52 × 10^{−2} N

Therefore, the force between the two spheres is 1.52 × 10^{−2} N.

**(b) **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.

∴

Force of repulsion between the two spheres,

= 16 × 1.52 × 10^{−2}

= 0.243 N

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

For remaining part, please post them in a separate thread.

Regards,

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