Question

Ans.
8. Two identical metallic balls A and B have charges 40
and - 10 uC respectively. The distance between them
is 2.0 m. What is the magnitude and type of force
acting between them? They are touched to each other
and again separated by a distance of 2.0 m from each 14
other. Calculate the new force between them. What
will be the force if one ball is connected to earth?
Ans. 0.90 N (attraction), 0.51 N (repulsion), zero.​

Answer 1

Answer:

The magnitude and type of the force will be 0.5 N and repulsive.

Explanation :

When both balls are touched to each other, then charge between them will be distributed equally,

New charge

Q = (+40 - 10)/2 = 15 μC = 15 x 10⁻⁶ C

distance r = 2 m

Hence force between them,

F = 1/4πε₀ x QQ/r²

= 9 x 10⁹ x 15 x 10⁻⁶ x 15 x 10⁻⁶/4

= 506 x 10⁻³ N

= 0.5 N

So, the magnitude of the force will be 0.5 N

The nature of the force will be repulsive as these balls have same charges.

When one of the ball will be touched to the ground, then it completely becomes neutral i.e. 0 charge, hence the new force between them will be zero.

Hope This Helps