# A charge of -5 C is at the origin. How much energy would be applied to or released from a  4 C charge if it is moved from  (-5, 3 )  to (2 ,-7 ) ?

Mar 30, 2017

#### Answer:

$W = 6.12 \cdot {10}^{9} \text{ Joules}$
$\text{since The Potential of the point B is greater than the potential}$
$\text{of the point of A, work is done by an external electrical force. }$

#### Explanation:

$\text{Since The electrical force is conservative,it is not important}$
$\text{that the charge have gone from A to B by which path.(f or c).}$ $\text{Work doing both situation is the same.}$

$\text{Since a charge of 4C moves in the electric field of the charge -5C,}$
$\text{we should find the potentials of the points A and B.}$

${V}_{a} = - k \cdot \frac{5}{{r}_{a}} \text{ , } {V}_{b} = - k \frac{5}{{r}_{b}}$

${r}_{a} = \sqrt{{\left(- 5\right)}^{2} + {3}^{2}} = \sqrt{25 + 9} = \sqrt{34} = 5.83$

${r}_{b} = \sqrt{{2}^{2} + {\left(- 7\right)}^{2}} = \sqrt{4 + 49} = \sqrt{53} = 7.28$

$W = Q \left({V}_{b} - {V}_{a}\right)$

$W = 4 \left(- k \cdot \frac{5}{r} _ b + k \cdot \frac{5}{r} _ a\right)$

$W = 4 \cdot 5 k \left(\frac{1}{r} _ a - \frac{1}{r} _ b\right)$

$W = 20 k \left(\frac{{r}_{b} - {r}_{a}}{{r}_{a} \cdot {r}_{b}}\right)$

$W = 20 k \left(\frac{7.28 - 5.83}{7.28 \cdot 5.83}\right)$

$k = 9 \cdot {10}^{9}$

$W = 20 k \left(0.034\right)$

$W = 0.68 \cdot 9 \cdot {10}^{9}$

$W = 6.12 \cdot {10}^{9} \text{ Joules}$