Question

A metal ball has a net charge of 4.5x10-7 C

Answer 1

a) the number of protons is $2.81\cdot 10^{12}$ more than the electrons

b) $4.69\cdot 10^{-15} kg$

Explanation:

The net electric charge on the ball is

$Q=+4.5\cdot 10^{-7}C$

This electric charge is given by the algebraic sum of the charge of the protons and of the charge of the electrons.

The charge of one proton is:

$q_p =+e= +1.6\cdot 10^{-19}C$

While the charge of one electron is

$q_e = -e=-1.6\cdot 10^{-19}C$

So the net charge on the metal ball will be given by

$Q=N_p q_p + N_e q_e = (N_p -N_e)e$

where

$N_p$ is the number of protons

$N_e$ is the number of electrons

So we find:

$N_p-N_e=\frac{Q}{e}=\frac{4.5\cdot 10^{-7}}{1.6\cdot 10^{-19}}=2.81\cdot 10^{12}$

This means that the number of protons is $2.81\cdot 10^{12}$ more than the electrons.

b)

In this case, we want to make the ball neautral, so we have to remove a net charge of Q' such that the new charge is zero:

$Q-Q'=0$

This implies that the charge that we must remove is

$Q'=Q=4.5\cdot 10^{-7}C$

To do that (and to make the ball losing mass at the same time), we have to remove protons, since they have positive charge.

The number of protons that must be removed is:

$N_p = \frac{Q'}{q_p}=\frac{4.5\cdot 10^{-7}}{1.6\cdot 10^{-19}}=2.81\cdot 10^{12}$

The mass of one proton is

$m_p = 1.67\cdot 10^{-27}kg$

Therefore, the total mass that must be removed from the ball is

$M=m_p N_p = (1.67\cdot 10^{-27})(2.81\cdot 10^{12})=4.69\cdot 10^{-15} kg$