Question

A small metal sphere has a mass of 0.19 g and a charge of -26.0 nC . It is 10 cm directly above an identical sphere with the same charge. This lower sphere is fixed and cannot move. What is the magnitude of the force between the spheres? If the upper sphere is released, it will begin to fall. What is the magnitude of its initial acceleration?

Answer 1

Electrostatic force between two balls is given as

`F = 6.08 * 10^(-4) N`initial acceleration of the ball when it falls down is given as

`a = 6.61 m/s^2`

Step-by-step explanation:

As we know that the electrostatic force between the two charges is given as

`F = (kq_1q_2)/(r^2)`

so we will have

`F = ((9 * 10^9)(26 * 10^(-9))^2)/((0.10)^2)`

`F = 6.08 * 10^(-4) N`

Now when the sphere is released from rest

Net force on it is given as

`F = F_g - F_(electrostatic)`

`F = mg - (6.08 * 10^(-4))`

`F = (0.19 * 10^(-3) * 9.81) - (6.08 * 10^(-4))`

`F = 1.26 * 10^(-3) N`

Now the acceleration of the ball is given as

`a = (F)/(m)`

`a = (1.26 * 10^(-3))/(0.19 * 10^(-3))`

`a = 6.61 m/s^2`