Question

The external work required to reduce the
distance between two charge q, = 4 HC and
9, = 20 pc from 2 m to 1 m will be​

Answer 1

Answer: 3.6 J

Explanation:

Since, there are none any non - conservative forces act.

So, work required will be given as change in potential energy of the system of charges.

$q_1=40\mu C\\ \\q_2=20 \mu C\\ \\r_i=2m \\ \\ r_f=1m$

We have,

$W_{req}=V_f-V_i\\ \\=\frac{kq_1q_2}{r_f}-\frac{kq_1q_2}{r_i}\\ \\=\frac{kq_1q_2}{1}-\frac{kq_1q_2}{2}\\ \\=\frac{kq_1q_2}{2}\\ \\=\frac{9*10^9*40*10^{-6}*20*10^{-6}}{2}\\ \\=3.6J$

Hence, work required will of 3.6J .

Answer 2

Answer:

work required will of 3.6J