# A charge of -1 C is at the origin. How much energy would be applied to or released from a  -1 C charge if it is moved from  ( -1, 3 )  to (-2 ,7 ) ?

Jul 3, 2016

$\approx 1.61 \times {10}^{9} J$

#### Explanation:

Given two point charges
$\text{At origin"Q=-1C" and another } q = - 1 C$
Let the point of origin be O(0,0) where -1C charge is placed.The initial or first position of another -1C charge is A (-1,3)and its final or displaced position is B (-2.7)

The initial distance of the charge from origin O is $O A = {r}_{i}$

${r}_{i} = \sqrt{{\left(- 1 - 0\right)}^{2} + {\left(3 - 0\right)}^{2}} = \sqrt{10}$

The final distance of the charge from origin O is $O B = {r}_{f}$

${r}_{f} = \sqrt{{\left(- 2 - 0\right)}^{2} + {\left(7 - 0\right)}^{2}} = \sqrt{53}$

Formula for calculation PE

$P E = \left(U\right) = \frac{k Q q}{r}$,
$\text{where }$
$k = \text{Coulomb'sconstant} = 9 \times {10}^{9} \frac{N {m}^{2}}{C} ^ 2$

The change in PE for movement of -1C charge from A to B is
$\Delta U = {U}_{f} - {U}_{i} = \left(k Q q\right) \left(\frac{1}{r} _ f - \frac{1}{r} _ i\right)$
$= 9 \times {10}^{9} \times \left(- 1\right) \left(- 1\right) \left(\frac{1}{\sqrt{53}} - \frac{1}{\sqrt{10}}\right) J \approx - 1.61 \times {10}^{9} J$

As the change in PE is negative the energy will be released for the movement of the charge.