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 `(-2 ,3 )` to `(2 ,-4 )`?

Answer 1

The equation for Electrostatic potential energy between two charges is:

`E = k_e(Q_1Q_2)/r`

where `k_e ~~ 8.99xx10^9" JmC"^-2`

The change in energy by moving a charge is:

`DeltaE = k_eQ_1Q_2(1/r_2-1/r_1)`

Compute `r_1` using the point `(-2,3)`

`r_1 = sqrt((-2)^2+ 3^2)`

`r_1 = sqrt(13)"m"`

Compute `r_2` using the point `(2,-4)`

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

`r_2 = sqrt(20)"m"`

We are given `Q_1 = -6"C" and Q_2=9"C"`

`DeltaE = (8.99xx10^9" JmC"^-2)(-6"C")(9"C")(1/(sqrt(20)"m")-1/(sqrt(13)"m"))`

`DeltaE = 2.6xx10^10J`