Question

Consider two point particles that have charge +e, are at rest, and are separated by 2.7 x 10-15 m.

Answer 1

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

$U=\frac{Kq_{1}q_{2}}{d}$

By substituting the values

$U=\frac{9\times 10^{9}\times 1.6\times 10^{-19}\times 1.6\times 10^{-19}}{2.7\times 10^{-15}}$

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

$U=\frac{Kq_{1}q_{2}}{d}$

By substituting the values

$U=\frac{9\times 10^{9}\times 1.6\times 10^{-19}\times 1.6\times 10^{-19}}{5.4\times 10^{-15}}$

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.

$2 \times \frac{1}{2}mv^{2}=8.53\times 10^-14$

$2 \times \frac{1}{2}\times 1.67\times 10^{-27}\times v^{2}=8.53\times 10^{-14}$

v = 7.14 x 10^6 m/s