Question

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 `(-2 ,-2 )`?

Answer 1

The energy to be applied is `=73*10^9J`

Explanation:

Calculate the potential energy at each point.

`E_1=k(q_1q_2)/d_1`

`E_2=k(q_1q_2)/d_2`

The distance `d_1=sqrt((-3)^2+(7)^2)=sqrt58`

The distance `d_2=sqrt((-2)^2+(-2)^2)=sqrt8`

The charges are `q_1=7C` and `q_2=5C`

Coulomb's constant is `k=9*10^9Nm^2C^-2`

`E_1=9*10^9*(7)(5)/sqrt58=41.4*10^9J`

`E_2=9*10^9*(7)(5)/sqrt8=111.4*10^9J`

The work done is

`=E_2-E_1`

`=(114.4-41.4)*10^9J`

`=73*10^9J`