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, -1 )` to `(-9 ,-2 )`?

Answer 1

Energy is released, `E ~~ 1.7619134*10^10J`

Explanation:

Let charge at origin be called `q_1` and other is `q_2`. The initial position vector of charge `q_2` is `vec(r_1)` and its final position vector is `vec(r_2)`.

So,

`|vec(r_1)|=sqrt(6^2+(-1)^2)=sqrt(37)`

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

Since,

`DeltaW=q_2*DeltaV=E`

`=>E=q_2*(V_({-9, -2}) -V_({6, -1}))=q_2*((k_eq_1)/|vec(r_2)|-(k_eq_1)/|vec(r_1)|)`

`=>E=(k_eq_1q_2)/(|vec(r_1)|*|vec(r_2)|)*(|vec(r_1)|-|vec(r_2)|)`

`=>E=(9*10^9*7*5)/(sqrt(37*85))*(sqrt(37)-sqrt(85))`

`=>E~~-1.7619134*10^10J`

Negative sign indicates that energy is released.