Chapter 17, Problem 1OQ

To determine

The impact on potential energy, if two positively charged small spheres are pushed towards each other.

Answer to Problem 1OQ

Solution:

(c) It increases

Explanation of Solution

Given data:

Given options

1. It remains unchanged
2. It decreases
3. It increases
4. There is no potential energy in this solution

Formula used:

The potential energy of two charges (point) separated by distance $r$ is :

  $U=\frac{1}{4\pi {\in }_0}\frac{q_1{q}_2}{r^2}$

  $U=\frac{1}{4\pi {\in }_0}\frac{q_1{q}_2}{r^2}$

  $\begin{array}{l}{q}_2.q_2>0\left({\text{For the given q}}_1>0,q_2>0\right)\\ U\alpha \frac{1}{r^2}\end{array}$

From above, it becomes clear that potential energy is inversely proportional to the square of the distance between charges. As distance decreases, potential energy should increase. Hence, the correct option is $\left(c\right)$ .

Conclusion:

By decreasing the distance between two positive point charges, potential energy is increased.