Question

two identical metallic balls A and B have charges + 40 and - 10 microcoulomb respectively in the distance between them is 2 metre what is the magnitude and type of force acting between them​

Answer 1

F ={ (9×10^-9) (40×10^-6) (10×10^-6)}/2^2

F = {(9× 10^-9)(400×10^-12)}/4

F =( 9×10^-9)(100×10^-12)

F = 9×10^-19 N

F = 0.9×10^-18 N

Nature of force will be attractive as both charges are of opposite nature

Answer 2

Answer:

The force is 0.9 N and this force is attractive.

Explanation:

Given that,

Charge of ball A $q_{1} =40\times10^{-6}\ C$

Charge of ball B$q_{2} = -10\times10^{-6}\ C$

Distance d = 2 m

The force acting between the charges is

$F=\dfrac{kq_{1}q_{2}}{d^2}$

$F=\dfrac{9\times10^{9}\times40\times10^{-6}\times(-10)\times10^{-6}}{2^2}$

$F=-0.9\ N$

The negative sign represents the attractive force.

Hence, The force is 0.9 N and this force is attractive.