Question

A charge of `-2 C` is at the origin. How much energy would be applied to or released from a `4 C` charge if it is moved from `(-7 ,5 )` to `(9 ,-2 )`?

Answer 1

The energy released is `=0.56*10^9J`

Explanation:

The potential energy is

`U=k(q_1q_2)/r`

The charge `q_1=-2C`

The charge `q_2=4C`

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

The distance

`r_1=sqrt((-7)^2+(5)^2)=sqrt74m`

The distance

`r_2=sqrt((9)^2+(-2)^2)=sqrt(85)`

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*((-2)*(4))(1/sqrt74-1/sqrt(85))`

`=-0.56*10^9J`

The energy released is `=0.56*10^9J`