Question

A 20 C positive charge is held a distance x away from a stationary 10 C negative charge. The positive charge is released and accelerates toward the negative charge, attaining a velocity of 50 m/s when it reaches a distance of 0.5x from the 10 C charge. How far apart were the two particles initially?

Answer 1

Answer:

Distance between the two charges is given as

$x = \frac{7.2 \times 10^8}{m}$

where m = mass of the charge

Explanation:

As we know that the charge is released from rest

Here it will gain kinetic energy when it reaches near to other stationary charge.

So we can say that change in electrostatic potential energy = gain in kinetic energy of the charge

So we have

$\frac{kq_1q_2}{r_2} - \frac{kq_1q_2}{r_1} = \frac{1}{2}mv^2$

$kq_1q_2(\frac{1}{0.5x} - \frac{1}{x}) = \frac{1}{2}m(50^2)$

$\frac{(9\times 10^9)(20)(10)}{x} = 2500 m$

so we have

$x = \frac{7.2 \times 10^8}{m}$