Question

two point charges q1 and q2 are separated by a distance r12 are kept in external electric field. derive an expression for the potential energy of the system of two charges in the field​

Answer 1

Answer:

(iii) work done on q

2

to move it against the force of q

1

=

r

12

Kq

1

q

2

∴ The potential energy of the system =q

1

V(r

1

)+q

2

(V)(r

2

)+

r

12

Kq

1

q

2

(b) work done in moving charges = Potential of the initial position - Potential of the final position.

Now, Potential of the initial system =V

12

+V

23

+V

31

=2KQ

2

+(−3KQ

2

)+(−6KQ

2

)=−7KQ

2

J

potential energy after moving to position, A

1

,B

1

&C

1

=

1/2

K(2Q)(Q)

+

1/2

K(Q)(−3Q)

+

1/2

K(−3Q)(2Q)

=4KQ

2

−6KQ

2

−12KQ

2

=−14KQ

2

J

∴ work done =−7Q

2

K−(−14KQ

2

)=7KQ

2

J