# A charge of -2 C is at the origin. How much energy would be applied to or released from a  4 C charge if it is moved from  (-7 ,1 )  to (4 ,-6 ) ?

Nov 19, 2017

$197$ $M J$ of energy is released.

#### Explanation:

Electrostatic Potential Energy: The electrostatic potential energy of a charge pair $Q$ and $q$ separated by a distance $r$ is :

U(r) = -k(Qq)/r; \qquad k \equiv 1/(4\pi\epsilon_0) = 8.99\times10^9 (N.m^2)/C^2

Assuming that the units of the coordinates are in meters.
${r}_{i} = \setminus \sqrt{{\left(- 7 m\right)}^{2} + {\left(1 m\right)}^{2}} = \setminus \sqrt{50}$ $m = 7.071$ $m$
${r}_{f} = \setminus \sqrt{{\left(4 m\right)}^{2} + {\left(- 6 m\right)}^{2}} = 7.211$ $m$

$\setminus \Delta U = {U}_{f} - {U}_{i} = - k Q q \left(\frac{1}{r} _ f - \frac{1}{r} _ i\right)$
$= - \left(8.99 \setminus \times {10}^{9} \frac{N {m}^{2}}{C} ^ 2\right) \left(- 2 C\right) \left(4 C\right) \left(\frac{1}{7.211 m} - \frac{1}{7.071 m}\right)$

$= - 197$ $M J$

Change in potential energy is negative. This means the potential energy has decreased. So the energy is released.