Question

A charge of `-6 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 ,-8 )` to `(2 ,-4 )`?

Answer 1

The energy applied is `=57.2*10^9J`

Explanation:

The potential energy is

`U=k(q_1q_2)/r`

The charge `q_1=-6C`

The charge `q_2=9C`

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

The distance

`r_1=sqrt((5)^2+(-8)^2)=sqrt89m`

The distance

`r_2=sqrt((2)^2+(-4)^2)=sqrt(20)`

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*((-6)*(9))(1/sqrt20-1/sqrt(89))`

`=-57.2*10^9J`

The energy applied is `=57.2*10^9J`