## Gravitational and electric force acting on particles

### Task number: 2324

Calculate the force by which two α-particles repel each other when their distance is 10^{−13} m and compare this force with the gravitational force. The charge of an α-particle is 3.2·10^{−19} C and its mass is 6.68·10^{−27} kg.

#### Hint: gravitational and electric force

This relation applies for the value of the electric force:

\[F_e\,=\,\frac{1}{4 \pi \varepsilon_0}\,\frac{Q_1 Q_2}{r^2}\,,\]where

*Q*_{1},*Q*_{2}are the values of charges and*r*is the distance of the charges that act upon eachother.This applies for the gravitational force:

\[F_g\,=\,\kappa \,\frac{m_1 m_2}{r^2}\,,\]where

*m*_{1},*m*_{2}are masses of charges that act upon eachother.#### Solution

We will calculate the value of electric force that the particles act on each other with using Coulomb law:

\[F_e\,=\,\frac{1}{4 \pi \varepsilon_0}\,\frac{Q_1 Q_2}{r^2}\,.\]Since both particles are charged with the same charge

\[F_e\,=\,\frac{1}{4 \pi \varepsilon_0}\,\frac{Q^2}{r^2}\,.\]*Q*, we can simplify the equation:We will calculate the value of the gravitational force using Newton’s law. Since both particles have the same mass

\[F_g\,=\,\kappa \,\frac{m_1 m_2}{r^2}\,=\,\kappa \,\frac{m^2}{r^2}\,.\]*m*, we can simplify the equation once again:Next, we will find the ratio of both forces to see how many times is one force bigger than the other.

\[\frac{F_e}{F_g}\,=\, \frac{\frac{1}{4 \pi \varepsilon_0}\,\frac{Q^2}{r^2}}{ \kappa\,\frac{m^2}{r^2}}\]We adjust the fraction:

\[\frac{F_e}{F_g}\,=\, \frac{1}{4 \pi \varepsilon_0\kappa}\,\frac{Q^2r^2}{m^2r^2}\,=\, \frac{1}{4 \pi \varepsilon_0\kappa}\,\frac{Q^2}{m^2}\]We can see that the ratio of the electric and the gravitational force is independent of particle distance.

#### List of known information and numerical calculation

*r*= 10^{−13}m*Q*= 3.2·10^{−19}C*m*= 6.68·10^{−27}kg*F*_{e}= ? (N)*F*_{g}= ? (N)**From tables:***ε*_{0}= 8.85·10^{−12}C^{2}N^{−1}m^{−2}*κ*= 6.67·10^{−11}m^{3}kg^{−1}s^{−2}

\[\begin{eqnarray} F_e\,&=&\,\frac{1}{4 \pi \varepsilon_0}\,\frac{Q^2}{r^2}\,=\,\frac{1}{4 \pi \cdot 8.85{\cdot}10^{-12}}\cdot\frac{\left(3.2 {\cdot} 10^{-19}\right)^2}{\left(10^{-13}\right)^2}\,=\,9.2{\cdot}10^{-2}\,\mathrm{N}.\\ F_g\,&=&\,\kappa \,\frac{m^2}{r^2}\,=\,6.67{\cdot}10^{-11} \cdot\frac{\left(6.68 {\cdot} 10^{-27}\right)^2}{\left(10^{-13}\right)^2}\,=\, 3.0 {\cdot} 10^{-37}\,\mathrm{N}.\\ \frac{F_e}{F_g} \,&=&\, \frac{1}{4 \pi \varepsilon_0\kappa}\,\frac{Q^2}{m^2}\,=\,\frac{1}{4 \pi \cdot 8.85{\cdot}10^{-12}\cdot6.67{\cdot}10^{-11}}\cdot\frac{\left(3.2 {\cdot} 10^{-19}\right)^2}{\left(6.68{\cdot}10^{-27}\right)^2}\\ \frac{F_e}{F_g}\,&=&\,3{\cdot}10^{35} \end{eqnarray}\]The electric force is much larger than the gravitational force in these conditions.

#### Answer

\[F_e\,=\,\frac{1}{4 \pi \varepsilon_0}\,\frac{Q^2}{r^2}\,=\,9.2 {\cdot} 10^{-2}\,\mathrm{N}.\]*α*-particles repel with force ofThe value of the gravitational force is

\[F_g\,=\,\kappa \,\frac{m^2}{r^2}\,=\,3.0 {\cdot} 10^{-37}\,\mathrm{N}.\]The ratio of the electric and the gravitational force is

\[\frac{F_e}{F_g} \,=\, \frac{1}{4 \pi \varepsilon_0\,\kappa}\,\frac{Q^2}{m^2}=\,3{\cdot} 10^{35}.\]The electric force is thus much larger than the gravitational force.