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=(kqQ)/(r^2)`

`ma=(kq^2)/(r^2)`

Put the value into the formula

`33*10^(-3)* a=(9*10^(9)*(16*10^(-9))^2)/((27*10^(-2))^2)`

`a=(9*10^(9)*(16*10^(-9))^2)/((27*10^(-2))^2*33*10^(-3))`

`a=0.0009577\ m/s^2`

`a=9.577*10^(-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*9.577*10^(-4)*16*10^(-2)`

`v=\sqrt{0+2*9.577*10^(-4)*16*10^(-2)}`

`v=0.0175\ m/s`

Hence, The speed of the charge is 0.0175 m