# A charge of 7 C is at the origin. How much energy would be applied to or released from a  5  C charge if it is moved from  (-5, 1 )  to (2 ,-1 ) ?

Jun 24, 2017

The potential energy difference between the initial and final positions of the charges will be $8.2 \times {10}^{10}$ $J$.

#### Explanation:

We care only about the distance between the two charges. We can use Pythagoras' theorem to calculate the distances and the change:

${r}_{\text{initial}} = \sqrt{\left(- {5}^{2}\right) + {1}^{2}} = \sqrt{25 + 1} = \sqrt{26} = 5.1$ $m$ (we have to assume the units are $m$, although we were not told)

${r}_{\text{final}} = \sqrt{{2}^{2} + {\left(- 1\right)}^{2}} = \sqrt{4 + 1} = \sqrt{5} = 2.2$ $m$

Since both charges are positive, and they are moved closer together, electric potential energy will be added to the system.

${E}_{\text{initial}} = \frac{k {q}_{1} {q}_{2}}{r} = \frac{9 \times {10}^{9} \times 7 \times 5}{5.1} = 6.1 \times {10}^{10}$ $J$

${E}_{\text{final}} = \frac{k {q}_{1} {q}_{2}}{r} = \frac{9 \times {10}^{9} \times 7 \times 5}{2.2} = 1.2 \times {10}^{11}$ $J$

${E}_{\text{change"=E_"final"-E_"initial}}$
$= 1.4 \times {10}^{11} - 6.1 \times {10}^{10} = 8.2 \times {10}^{10}$ $J$