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 `(6, -4 )` to `(-7 ,-2 )`?

Answer 1

The energy released is `=0.414*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((6)^2+(-4)^2)=sqrt52m`

The distance

`r_2=sqrt((-7)^2+(-2)^2)=sqrt(53)`

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)*(5))(1/sqrt53-1/(sqrt52))`

`=-0.414*10^9J`

The energy released is `=0.414*10^9J`