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

Answer:

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

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.

$E_T = E_p = \frac{k q_1 q_2}{r}$

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

$r = \sqrt{(x_2-x_1)^2 + (y_2- y_1)^2} = \sqrt{(1.25)^2 + (0.57)^2} = 1.374\ m$, then if

$k =8.987\times 10^9 N m^2/C^2$ and   $q_1 = q_2 =3.2\times 10^{-6} C$,

$E_T = E_p = \frac{k q_1 q_2}{r} = \frac{8.987\times 10^9 * 3.2\times 10^{-6}*3.2\times 10^{-6}}{1.374} = 6.698 \times 10^{-2} Nm$.  

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

$E_T = E_k = \frac{1}{2}mv_f^2$, so

$v_f = \sqrt{2E_T/m} =\sqrt{2 *6.698 \times 10^{-2} /0.00263} = 7.137 m/s$ (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

$E_k = \frac{1}{2}mv^2 = 1.674 \times 10^{-2} Nm$

and potential energy is

$E_p = E_T - E_k = 6.698 \times 10^{-2} - 1.674 \times 10^{-2} = 5.024 \times 10^{-2} Nm$

so the distance is

$r = \frac{k q_1 q_2}{E_p} = 1.832\ m$