Question

They are then fixed at positions that are 4.30 x 10-11 m apart. What is EPEfinal - EPEinitial, which is the change in the electric potential energy?

Answer 1

This is an incomplete question, here is a complete question.

Two particles with charges +6 e⁻ and -10 e⁻ are initially very far apart (effectively an infinite distance apart). They are then fixed at positions that are 4.30 × 10⁻¹¹ m apart. What is EPE(final) - EPE(initial), which is the change in the electric potential energy?

Answer : The change in the electric potential energy is,
`-2.92* 10^(-6)J`

Explanation : Given,

Formula used for electric potential energy of the two charges when they are separated is:

`EPE=(1)/(4\pi \epsilon_0)* {(q_1* q_2)/(r^2)`

`EPE=(k* q_1* q_2)/(r^2)`

where,

EPE = electric potential energy

k =
`(1)/(4\pi \epsilon_0)=8.99* 10^9`

`q_1` = charge on 1st particle = +6 e⁻ =
`6* 10^(-19)C`

`q_2` = charge on 2nd particle = -10 e⁻ =
`-10* 10^(-19)C`

r = distance between two charges =
`4.30* 10^(-11)m`

Now put all the given values in the above formula, we get:

`EPE=((8.99* 10^9)* (6* 10^(-19))* (-10* 10^(-19)))/((4.30* 10^(-11))^2)`

`EPE=-2.92* 10^(-6)J`

Initially EPE = 0 J

Thus,
`EPE_(final)-EPE_(initial)=-2.92* 10^(-6)J`

The positive sign indicate the attractive force and negative sign indicate the repulsive force.

Thus, the change in the electric potential energy is,
`-2.92* 10^(-6)J`

Answer 2

The change in the electric potential energy is
`-3.2*10^(-16)\ J`

Step-by-step explanation:

Given that,

Distance
`d=4.30*10^(-11)\ m`

suppose, Two particles with charges +6 e and -10 e are initially very far apart

We need to calculate the change in the electric potential energy

Using formula of energy

`\text{electric potential energy}=\text{final electric potential energy-initial electric potential energy}`

`EPE=EPE_(f)-EPE_(i)`

Here, initial electric potential energy= 0

final electric potential energy
`EPE_(f)=(kq_(1)q_(2))/(r_(2)^2)`

Put the value into the formula

`EPE=(kq_(1)q_(2))/(r_(2)^2)+0`

Put the value into the formula

`EPE=(9*10^(9)*6*1.6*10^(-19)*(-10*1.6*10^(-19)))/((4.30*10^(-11)))`

`EPE=-3.2*10^(-16)\ J`

Hence, The change in the electric potential energy is
`-3.2*10^(-16)\ J`