Question

Consider a situation where two point charges of charge Q = 16 nC and mass m = 33 g are 27 cm apart. Then, one of these point charges are let go (NOTE: ignore gravity).

After that charge has moved 16 cm, what will be its speed?

Answer 1

Given that,

Charge = 16 nC

Mass = 33 g

Distance = 27 cm

We need to calculate the acceleration

Using formula of electrostatic force

$F=\dfrac{kqQ}{r^2}$

$ma=\dfrac{kq^2}{r^2}$

Put the value into the formula

$33\times10^{-3}\times a=\dfrac{9\times10^{9}\times(16\times10^{-9})^2}{(27\times10^{-2})^2}$

$a=\dfrac{9\times10^{9}\times(16\times10^{-9})^2}{(27\times10^{-2})^2\times33\times10^{-3}}$

$a=0.0009577\ m/s^2$

$a=9.577\times10^{-4}\ m/s^2$

We need to calculate the speed of charge

Using equation of motion

$v^2=u^2+2as$

Where, v= speed

u = initial speed

a = acceleration

s = distance

$v^2=0+2\times9.577\times10^{-4}\times16\times10^{-2}$

$v=\sqrt{0+2\times9.577\times10^{-4}\times16\times10^{-2}}$

$v=0.0175\ m/s$

Hence, The speed of the charge is 0.0175 m