# A charge of 2 C is at the origin. How much energy would be applied to or released from a  3 C charge if it is moved from  ( -4, 6 )  to ( 7 , -4 ) ?

Jun 5, 2018

$= - 0.7895$ $G J$

#### Explanation:

The potential energy can be given by:

$V \left(r\right) = \frac{{Q}_{1} {Q}_{2}}{4 \pi {\epsilon}_{0} r}$

Where $Q$ are the magnitudes of the charges and $r$ is the distance between the charges.

We already know that the charges (which are constant) are:

${Q}_{1} = 2 C$ and ${Q}_{2} = 3 C$ so we can plug these in to the top (ignoring units for now) to get:

$V \left(r\right) = \frac{6}{4 \pi {\epsilon}_{0} r} = \frac{3}{2 \pi {\epsilon}_{0} r}$.

We wish to see the energy difference in moving the $3 C$ charge from $\left(- 4 , 6\right)$ to $\left(7 , - 4\right)$.

So we need to find $r$ for both these cases (the scalar distance from the origin) which can be done by simple Pythagoras.

${r}_{1} = \sqrt{{\left(- 4\right)}^{2} + {6}^{2}} \approx 7.2111 m$
${r}_{2} = \sqrt{{7}^{2} + {\left(- 4\right)}^{2}} \approx 8.0623 m$

assuming the coordinates given are in metres.

As it is a repulsive force (same sign charges) and we are moving further away then energy is being released.

To calculate how much we simply need:

$\Delta V = V \left({r}_{2}\right) - V \left({r}_{1}\right)$
$= \frac{3}{2 \pi {\epsilon}_{0} \times 8.0623} - \frac{3}{2 \pi {\epsilon}_{0} \times 7.2111}$

$= 6.6887 \times {10}^{9} - 7.4783 \times {10}^{9}$

$= - 7.895 \times {10}^{8} J$

$= - 0.7895$ $G J$