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

Feb 15, 2016

${W}_{12} = 8.99 \times {10}^{1} \frac{2}{\sqrt{20} - \sqrt{5}} = 8.04 \times {10}^{9} J$

#### Explanation:

Given two charges:
${q}_{1} = 2 C , \text{ located at } {P}_{0} \left(0 , 0\right)$
${q}_{2} = 1 C , \text{ located at } {P}_{1} \left(- 2 , 1\right)$
${q}_{2} \text{ is moved to } {P}_{1} \left(- 2 , 1\right) \implies {P}_{2} \left(- 4 , 2\right)$
Find the energy required to move from ${P}_{1} \left(- 2 , 1\right) \implies {P}_{2} \left(- 4 , 2\right)$

Solution:
First let's list the key definition and principles we need to solve this problem:
1) $\textcolor{g r e y}{\text{Electric Force between two charged particles}}$:
F_e = K(q_1q_2)/(Deltar)^2; q_(i=1,2) = "charge"; K = 8.99xx10^9 Nm^2/C^2
$\Delta r = \text{distance between charges 1 and 2}$

2) $\text{Work done to move } {q}_{2} : {P}_{1} \implies {P}_{2}$
color(red)(W_(12) = K(q_1q_2)/(Deltar_(12))
$\Delta {r}_{12} = | {r}_{01} - {r}_{02} | = r$ change in distance of separation
3) $\textcolor{red}{\text{Distance formula}}$ :
${r}_{12} = \sqrt{\left({x}_{0} - {x}_{2}\right) + \left({y}_{0} - {y}_{2}\right)} - \sqrt{\left({x}_{0} - {x}_{1}\right) + \left({y}_{0} - {y}_{1}\right)} =$
${r}_{12} = \sqrt{20} - \sqrt{5}$
${W}_{12} = 8.99 \times {10}^{1} \frac{2}{\sqrt{20} - \sqrt{5}} = 8.04 \times {10}^{9} J$

That is a lot of energy, the reason is that a Coulomb is a huge charge. Typically you would have smaller charges and shorter distances. BTW I made an assumption that all distances are meters your question is silent on that...

The relevant principles are 20 Work formula and distance formula
Technically what we need is:
${W}_{12} = {\int}_{{r}_{1}}^{{r}_{2}} {F}_{e} \left(r\right) \cdot \mathrm{dr} = K {q}_{1} {q}_{2} {\int}_{{r}_{1}}^{{r}_{2}} \frac{1}{r} ^ 2 \mathrm{dr}$
but by inspection all you need the linear separation
$\Delta {r}_{02} - \Delta {r}_{01} = \sqrt{20} - \sqrt{5}$