# A charge of -1 C is at the origin. How much energy would be applied to or released from a  6 C charge if it is moved from  (-5, 1 )  to (2 ,4 ) ?

Mar 1, 2016

$\Delta W = 6 k \left(- \frac{1}{\sqrt{20}} + \frac{1}{\sqrt{26}}\right)$

#### Explanation:

$\text{you can solve with two different ways}$
$\text{1)}$
$\text{find the potential energy of system for both situation:}$
${E}_{p} = \frac{k \cdot {q}_{1} \cdot {q}_{2}}{d}$
${E}_{p 1} = \frac{k \cdot \left(- 1\right) \cdot 6}{\sqrt{{5}^{2} + {1}^{2}}} \text{ } {E}_{p 1} = - \frac{6 k}{\sqrt{26}}$
${E}_{p 2} = \frac{k \cdot 6}{\sqrt{{2}^{2} + {4}^{2}}} \text{ } {E}_{p 2} = \frac{6 k}{\sqrt{20}}$
$\text{Find work doing:}$
$W = \Delta {E}_{p} = {E}_{p 2} + {E}_{p 1}$
$W = \Delta {E}_{p} = 6 k \left(- \frac{1}{\sqrt{20}} + \frac{1}{\sqrt{26}}\right)$
$\text{2)}$
$\text{find the potential of both points created by a charge of -1C:}$
$V = k \cdot \frac{q}{d}$
${v}_{1} = \frac{- 1 \cdot k}{\sqrt{26}} \text{ } {v}_{2} = \frac{- 1 \cdot k}{\sqrt{20}}$
$\text{now,find the work :}$
$\Delta W = q \left({V}_{2} - {V}_{1}\right)$
$\Delta W = 6 k \left(- \frac{1}{\sqrt{20}} + \frac{1}{\sqrt{26}}\right)$