Question

A charge of 3.20 μC is held fixed at the origin. A second charge of 3.20 μC is released from rest at the position (1.25 m, 0.570 m).

a) If the mass of the second charge is 2.63 g , what is its speed when it moves infinitely far from the origin? (m/s)

b) At what distance from the origin does the second charge attain half the speed it will have at infinity?

Answer 1

a)
`v = 7.137 m/s` and b)
`r = 1.832\ m`

Step-by-step explanation:

We have to analyze this problem from the point of view of energy conservation. In this case there are two kind of energy, electric potential energy and kinetic energy. First, at
`t = t_1` there isn't relative movement between the two charges, so kinetic energy is zero and the total energy (
`E_T = E_p + E_k`) is just potential.

`E_T = E_p = (k q_1 q_2)/(r)`

where
`k` is the Coulomb constant,
`q_1` and
`q_2` are the two interacting charges, and
`r` is the distance between them.

Considering the fixed charge at (x1,y1) = (0,0) and the second one at (x2, y2) = (1.25, 0.57), the initial distance is

`r = √((x_2-x_1)^2 + (y_2- y_1)^2) = √((1.25)^2 + (0.57)^2) = 1.374\ m`, then if

`k =8.987* 10^9 N m^2/C^2` and
`q_1 = q_2 =3.2* 10^(-6) C`,

`E_T = E_p = (k q_1 q_2)/(r) = (8.987* 10^9 * 3.2* 10^(-6)*3.2* 10^(-6))/(1.374) = 6.698 * 10^(-2) Nm`.

Now, at
`t = t_2`,
`r \rightarrow \infty` and
`E_p \rightarrow 0`. This means all the energy is kinetic

`E_T = E_k = (1)/(2)mv_f^2`, so

`v_f = √(2E_T/m) =\sqrt{2 *6.698 * 10^(-2) /0.00263} = 7.137 m/s` (mass in Kg).

That would be the velocity when the second charge moves infinitely far from the origin.

For the second part we have that
`v = v_f/2`, so kinetic energy is

`E_k = (1)/(2)mv^2 = 1.674 * 10^(-2) Nm`

and potential energy is

`E_p = E_T - E_k = 6.698 * 10^(-2) - 1.674 * 10^(-2) = 5.024 * 10^(-2) Nm`

so the distance is

`r = (k q_1 q_2)/(E_p) = 1.832\ m`