Question

Question 5: An unknown charge particle is placed near to an ion of Fe2+. The unknown charge particle was then displaced to 0.1 mm away from the Fe ion. Data from the velocity of displacement shows that the repelling force was 0.05 N. Determine the charge of the unknown particle.

Answer 1

To solve the problem we will use Coulomb's law:

$F=k_{e}\frac{q_{1}q_{2}}{r^{2}}$

where $q_{1}$ will be the charge of the ion $Fe^{2+}$  and $q_{2}$ will be the unknown charge.

$F=0.05N$

$r=0.1mm=1x10^{-4}m$.

Now let's determine the charge of $Fe^{2+}$ ion ($q_{1}$).

The atom Fe is neutral, that is, it has 26 electrons and 26 protons. In this case we have an $Fe^{2+}$ ion, this means that we have 2 less electrons (2 more protons). The $Fe^{2+}$ is not neutral and has an excess of positive charge given by the two protons.

$q_{1}=(2)(+1.6021917x10^{-19}C)$

$q_{1}=+3.20x10^{-19}C$

Now that we have the charge of the ion we are going to determine the unknown charge.

$F=k_{e}\frac{q_{1}q_{2}}{r^{2}}$

Solving the previous equation for $q_{2}$, we obtain:

$q_{2}=\frac{F.r^{2}}{k_{e}.q_{1}}$

where ke is the Coulomb constant given by

$k_{e}=8.9875x10^{9}\frac{N.m^{2}}{C^{2}}$

$q_{2}=\frac{F.r^{2}}{k_{e}.q_{1}}=\frac{(0.05N)(1x10^{-4}m)^{2}}{(8.9875x10^{9}N.m^{2}/C^{2})(+3.20x10^{-19}C)}$

$q_{2}=\frac{(0.05N)(1x10^{-8}m^{2})}{(8.9875x10^{9}N.m^{2}/C^{2})(+3.20x10^{-19}C)}$

$q_{2}=\frac{5x10^{-10}N.m^{2}}{2.876x10^{-9}N.m^{2}/C}$

$q_{2}=+0.17C$