# Two protons, each of charge +1.60 * 10-19 C, are 2.00 * 10-5 m apart. What is the change in the electric potential energy of this pair of charges if they are brought 1.00 * 10-5 m closer together?

## Answer is A, why? A) 1.15 10-23J B) 3.20 10-19J C) 3.20 10-16J D) 1.60 10-19J E) 1.60 * 10-14J

Initial electrostatic potential energy was $P . E = \frac{9 \cdot {10}^{9} \cdot {\left(1.6 \cdot {10}^{-} 19\right)}^{2}}{2 \cdot {10}^{-} 5} J$ (using the formula, $P . E = \frac{1}{4 \pi \epsilon} \frac{{q}_{1} {q}_{2}}{x}$)
When they were brought closer by $1 \cdot {10}^{-} 5 m$,new distance between them became, $\left(2 - 1\right) \cdot {10}^{-} 5 = 1 \cdot {10}^{-} 5 m$
So,new potential energy is $P . E ' = \frac{9 \cdot {10}^{9} \cdot {\left(1.6 \cdot {10}^{-} 19\right)}^{2}}{1 \cdot {10}^{-} 5} J$
So,change in potential energy =$P . E ' - P . E = \frac{9 \cdot {10}^{9} \cdot {\left(1.6 \cdot {10}^{-} 19\right)}^{2}}{{10}^{-} 5} \cdot \left(1 - \frac{1}{2}\right) = 11.52 \cdot {10}^{-} 24 = 1.15 \cdot {10}^{-} 23 J$