# A charge of -1 C is at the origin. How much energy would be applied to or released from a  3 C charge if it is moved from  (-5 ,1 )  to (2 ,-6 ) ?

Feb 13, 2016

$W = 27 \cdot {10}^{9} \left(\frac{\sqrt{40} - \sqrt{26}}{\sqrt{40 \cdot 26}}\right) J$

#### Explanation:

$\text{potential of any point :} V = k \cdot \frac{q}{d}$
$\text{The potential at the point(-5,1) by the charge of(-1C)}$
${V}_{A} = k \cdot \frac{- 1}{\sqrt{{\left(- 5\right)}^{2} + 1 {'}^{2}}}$
${V}_{A} = \frac{- k}{\sqrt{26}}$
$\text{The potential at the point(2,-6) by the charge of(-1C)}$
${V}_{B} = \frac{k \cdot \left(- 1\right)}{\sqrt{{2}^{2} + \left(- {6}^{2}\right)}}$
${V}_{B} = \frac{- k}{\sqrt{40}}$
$W = q \left({V}_{B} - {V}_{A}\right)$
$W = 3 \left(- \frac{k}{\sqrt{40}} - \left(- \frac{k}{\sqrt{26}}\right)\right)$
$W = 3 \left(- \frac{k}{\sqrt{40}} + \frac{k}{\sqrt{26}}\right)$
$W =$
$W = 3 \cdot 9 \cdot {19}^{9} \left(\right)$
$W = 27 \cdot {10}^{9} \left(\frac{\sqrt{40} - \sqrt{26}}{\sqrt{40 \cdot 26}}\right) J$