Answer:
Part a)
F = 0.056 N
Part b)
a = 28.13 m/s/s
Explanation:
Part a)
Electrostatic force between two charged balls is given as
[tex]F = \frac{kq_1q_2}{r^2}[/tex]
now we will have
[tex]q_1 = q_2 = 50 nC[/tex]
r = 2 cm
now we will have
[tex]F = \frac{(9\times 10^9)(50 \times 10^{-9})(50 \times 10^{-9})}{0.02^2}[/tex]
so here we have
[tex]F = 0.056 N[/tex]
Part b)
Now due to above force the acceleration of one ball which is released is given as
[tex]a = \frac{F}{m}[/tex]
[tex]a = \frac{0.056}{2 \times 10^{-3}}[/tex]
[tex]a = 28.13 m/s^2[/tex]