Question

Two protons are aimed directly toward each other by a cyclotron accelerator with speeds of 1200 km/s , measured relative to the earth. Find the maximum electrical force that these protons will exert on each other?

Answer 1

Answer:

The maximum electrical force is $2.512\times10^{-2}\ N$.

Explanation:

Given that,

Speed of cyclotron = 1200 km/s

Initially the two protons are having kinetic energy given by

$\dfrac{1}{2}mv^2=\dfrac{1}{2}mv^2$

When they come to the closest distance the total kinetic energy is converts into potential energy given by

Using conservation of energy

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

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

Put the value into the formula

$r=\dfrac{8.99\times10^{9}\times(1.6\times10^{-19})^2}{1.67\times10^{-27}\times(1200\times10^{3})^2}$

$r=9.57\times10^{-14}\ m$

We need to calculate the maximum electrical force

Using formula of force

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

$F=\dfrac{8.99\times10^{9}\times(1.6\times10^{-19})^2}{(9.57\times10^{-14})^2}$

$F=2.512\times10^{-2}\ N$

Hence, The maximum electrical force is $2.512\times10^{-2}\ N$.