Answer:
Explanation:
q1 = q2 = e = 1.6 x 10^-19 C
d = 2.7 x 10^-15 m
1. Work done is equal to the potential energy stored between the two charges.
The formula for the potential energy is given by
[tex]U=\frac{Kq_{1}q_{2}}{d}[/tex]
By substituting the values
[tex]U=\frac{9\times 10^{9}\times 1.6\times 10^{-19}\times 1.6\times 10^{-19}}{2.7\times 10^{-15}}[/tex]
U = 8.53 x 10^-14 J
So, the work done is 8.53 x 10^-14 J.
(2) d = 2 x 2.7 x 10^-15 m = 5.4 x 10^-15 m
So, the potential energy is
[tex]U=\frac{Kq_{1}q_{2}}{d}[/tex]
By substituting the values
[tex]U=\frac{9\times 10^{9}\times 1.6\times 10^{-19}\times 1.6\times 10^{-19}}{5.4\times 10^{-15}}[/tex]
U = 4.27 x 10^-14 J
(3) mass, m = 1 u = 1.67 x 10^-27 kg
So, the kinetic energy is equal to the potential energy.
Let v be the velocity.
[tex]2 \times \frac{1}{2}mv^{2}=8.53\times 10^-14[/tex]
[tex]2 \times \frac{1}{2}\times 1.67\times 10^{-27}\times v^{2}=8.53\times 10^{-14}[/tex]
v = 7.14 x 10^6 m/s
