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 `(-5, 1 )` to `(2 ,-6 )`?

Answer 1

The energy released is `=11.97*10^9J`

Explanation:

The potential energy is

`U=k(q_1q_2)/r`

The charge `q_1=7C`

The charge `q_2=5C`

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

The distance

`r_1=sqrt((-5)^2+(1)^2)=sqrt26m`

The distance

`r_2=sqrt((2)^2+(-6)^2)=sqrt(40)`

Therefore,

`U_1=k(q_1q_2)/r_1`

`U_2=k(q_1q_2)/r_2`

`DeltaU=U_2-U_1=k(q_1q_2)/r_2-k(q_1q_2)/r_1`

`=k(q_1q_2)(1/r_2-1/r_1)`

`=9*10^9*((7)*(5))(1/sqrt40-1/(sqrt26))`

`=-11.97*10^9J`

The energy released is `=11.97*10^9J`