Question

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 `( 6 , 4 )` to `(-3 ,2 )`?

Answer 1

The key to this problem is to find the change in the distance between the two charges. The change in energy will be given by `U=(kq_1q_2)/r`. The energy lost to the environment is `1.5xx10^10` `J`.

Explanation:

Initial separation:

`r_1=sqrt((6-0)^2+(4-0)^2)=sqrt52~~7.2` `m` (we have to assume the units are `m`)

Final separation:

`r_2=sqrt((-3-0)^2+(2-0)^2)=sqrt13~~3.6` `m`

Difference: `7.2-3.6 = 3.6` `m`

The change in the energy will be:

`DeltaU = (kq_1q_2)/(Delta r) = (9xx10^9xx2xx-3)/3.6 = -1.5xx10^10` `J`.

Since the charges are opposite and they have become closer, the energy of the charge system has decreased, so energy will have been lost to the environment.