Answer:
W = 5.76 J
Explanation:
Given that,
Charge 1, [tex]q_1=12\times 10^{-6}\ C[/tex]
Charge 2, [tex]q_2=8\times 10^{-6}\ C[/tex]
The distance between charges is 10 cm.
Initial distance = 10 cm = 0.1 m
Final distance = 10 cm - 4 cm = 6 cm = 0.06 m
We need to find the work done in bringing the 4 cm together.
We know that,
Work done = change in kinetic energy
[tex]W=kq_1q_2(\dfrac{1}{r_2}-\dfrac{1}{r_1})\\\\\text{Put all the values}\\\\W=9\times 10^9\times 12\times 10^{-6}\times 8\times 10^{-6}(\dfrac{1}{0.06}-\dfrac{1}{0.1})\\\\W=5.76\ J[/tex]
So, the required work done is 5.76 J.
