Question

A charge of `7 C` is at the origin. How much energy would be applied to or released from a `9 C` charge if it is moved from `(-5, 1 )` to `(2 ,-1 )`?

Answer 1

The energy applied is `=142.4*10^9J`

Explanation:

The potential energy is

`U=k(q_1q_2)/r`

The charge `q_1=7C`

The charge `q_2=9C`

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

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_1-1/r_2)`

`=9*10^9*((7)*(9))(1/sqrt5-1/(sqrt26))`

`=142.4*10^9J`

The energy released is `=142.4*10^9J`