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

Jan 7, 2016

$0.565 \text{k}$ units

#### Explanation:

I'll work out the electric potential at points P and Q.

Then I will use this to work out the potential difference between the 2 points.

This is the work done by moving a unit charge between the 2 points.

The work done in moving a 5C charge between P and Q can therefore be found by multiplying the potential difference by 5.

We need to find the length of PR and PQ which can be done using Pythagoras:

$P {R}^{2} = {2}^{2} + {7}^{2}$

$\therefore P R = \sqrt{4 + 49}$

$P R = 7.8$

and:

$Q {R}^{2} = {3}^{2} + {7}^{2}$

$\therefore Q R = \sqrt{9 + 49} = 6.93$

The electric potential due to a charge $q$ at a distance $r$ is given by:

$V = k . \frac{q}{r}$

$k$ is a constant.

So the potential at point $P$ is given by:

${V}_{P} = k \times \frac{7}{7.8} = 0.897 k$

The potential at $Q$ is given by:

${V}_{Q} = k \times \frac{7}{6.93} = 1.01 k$

So the potential difference is given by:

${V}_{P} - {V}_{Q} = k \left(0.897 - 1.01\right) = - 0.113 k$

So the work done in moving a 5C charge between these 2 points is given by:

$W = - 0.113 k \times 5 = - 0.565 k \text{units}$

This is the work done on the charge.

There are no units of distance given. If this was in meters then $k = 9 \times {10}^{9} \text{m/F}$ and the answer would be in Joules.