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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 5.4×10−8j. when the second charge is moved to point b, the electric force on the charge does −1.9×10−8j of work.



Answer :

The electric potential energy of the pair of charges when the second charge is at point b is U at b = +7.3×10⁻⁸J.

Electric potential is the amount of work done in moving point charge from infinity to aspecific point.

Calculation of the electric potential energy:

Here the particle should be moved from the point at the time when the potential energy should be U a

The U b should be the change in the potential energy

W is work done.

(W at a )-(W at b)  = U a - U b

U b = U a - (W at a )+(W at b)

So,

= 5.4 × 10⁻⁸J - ( -1.9×10⁻⁸ J)

= +7.3×10⁻⁸J.

Learn more about electric potential here:

https://brainly.com/question/24284560

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