# A charge of 2 C is at the origin. How much energy would be applied to or released from a  3 C charge if it is moved from  ( -4, 0 )  to ( -6 ,-3 ) ?

Feb 15, 2016

change in U = dU = $k Q q \left(\frac{1}{{r}_{\textrm{\in i t i a l}} - {r}_{\textrm{f \in a l}}}\right)$
$\mathrm{dU} \cong - 1.99 \times {10}^{10} J$
Therefore around 19.9 GJ has been released by moving the charge.

#### Explanation:

Electrical potential energy = $U = \frac{k Q q}{r}$
k = Coulomb's constant $\cong 8.9876 \times {10}^{9} J m {C}^{-} 2$
${r}_{\textrm{\in i t i a l}} = \sqrt{{\left(- 4\right)}^{2} + {0}^{2}} m = \sqrt{16} m = 4 m$
${r}_{\textrm{f \in a l}} = \sqrt{{\left(- 6\right)}^{2} + {\left(- 3\right)}^{2}} m = \sqrt{45} m = 3 \sqrt{5} m$

change in U = dU = $k Q q \left(\frac{1}{{r}_{\textrm{\in i t i a l}} - {r}_{\textrm{f \in a l}}}\right)$

$\mathrm{dU} = \left(8.9876 \times {10}^{9} J m {C}^{-} 2\right) \left(2 C\right) \left(3 C\right) \left(\frac{1}{4 m - 3 \sqrt{5} m}\right)$

$\mathrm{dU} \cong \left(5.3923 \times {10}^{9} J m\right) \left(\frac{1}{4 m - 3 \sqrt{5} m}\right)$

$\mathrm{dU} \cong - 1.99 \times {10}^{10} J$

Therefore around 19.9 GJ has been released by moving the charge.

Feb 15, 2016

proceed as follows

#### Explanation:

Let the 2C charge is at origin O (0,0)
The initial position of 3C is at A (-4,0)
The final position of 3C is at B (--6,-3)
find distance OA,OB
find initial Pot Energy of the system when 3C is at A
then when at B using formula (q_1*q_2)/(4piepsilonr
calculate the difference ----Final - initial