Question

A point charge q1 is held stationary at the origin. A second charge q2 is placed at point a, and the electric potential energy of the pair of charges is 5.4 x 10^-8J. When the second charge is moved to point b, the electric force on the charge does -1.9x10^-8J of work. What is the electric potential energy of the pair of charges when the second charge is at point b?

Answer 1

Answer:

$U_{b}=+7.3*10^{-8}J$

Explanation:

When a particle moves from a point where the potential energy is Ua to point where it is Ub,the change in potential energy is is equal to work done

So

$W_{a-b}=U_{a}-U_{b}\\ U_{b}=U_{a}-W_{a-b}$

Where Wa-b here is negative this means Ub is greater Ua. Therefore potential energy increases

So

$U_{b}=(+5.4*10^{-8}J )-(-1.9*10^{-8}J)\\U_{b}=+7.3*10^{-8}J$  

Answer 2

The electric potential energy of the pair of charges when the second charge is at point b is U_b = +7.3*10^-8J.

Calculation of the electric potential energy:

here the particle should be moved from the point at the time when the potential energy should be Ua

The Ub should be the change in the potential energy

W_a - b = U_a - U_b

U_b = U_a - W_a - b

So,

= 5.4 x 10^-8J - ( -1.9x10^-8J)

= +7.3*10^-8J.

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