Answer:
Electrostatic force between two balls is given as
[tex]F = 6.08 \times 10^{-4} N[/tex]initial acceleration of the ball when it falls down is given as
[tex]a = 6.61 m/s^2[/tex]
Explanation:
As we know that the electrostatic force between the two charges is given as
[tex]F = \frac{kq_1q_2}{r^2}[/tex]
so we will have
[tex]F = \frac{(9 \times 10^9)(26 \times 10^{-9})^2}{(0.10)^2}[/tex]
[tex]F = 6.08 \times 10^{-4} N[/tex]
Now when the sphere is released from rest
Net force on it is given as
[tex]F = F_g - F_{electrostatic}[/tex]
[tex]F = mg - (6.08 \times 10^{-4})[/tex]
[tex]F = (0.19 \times 10^{-3} \times 9.81) - (6.08 \times 10^{-4})[/tex]
[tex]F = 1.26 \times 10^{-3} N[/tex]
Now the acceleration of the ball is given as
[tex]a = \frac{F}{m}[/tex]
[tex]a = \frac{1.26 \times 10^{-3}}{0.19 \times 10^{-3}}[/tex]
[tex]a = 6.61 m/s^2[/tex]
