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

Answer 1

(127.09)*(10^9) Joule amount of energy has to be taken out from the charge.

Explanation:

Initial potential energy of the charge is,
Ei = `(9*10^9)*{(-6)*9}/(√29)`[r=distance between them = `√{(5)^2+(-2)^2}`]
And final potential energy=
Ef = `(9*10^9)*{(-6)*9}/(√5)`[r1=distance between them =`√{(2)^2+(-1)^2}`
Hence change in energy =
Ef-Ei
= `(9*10^9){(-6)*9}{1/(√5)-1/(√29)}`

Or,-`(127.09)*(10^9)` Joule
Negative sign means energy has to be taken out from the charge

Answer 2

The energy applied is `=127.1*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+(-2)^2)=sqrt(29)m`

The distance

`r_2=sqrt((2)^2+(1)^2)=sqrt(5)`

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*((-6)*(9))(1/sqrt29-1/sqrt(5))`

`=127.1*10^9J`