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 Q1, Q2 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 m1, m2 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 Q, we can simplify the equation:
\[F_e\,=\,\frac{1}{4 \pi \varepsilon_0}\,\frac{Q^2}{r^2}\,.\]We will calculate the value of the gravitational force using Newton’s law. Since both particles have the same mass m, we can simplify the equation once again:
\[F_g\,=\,\kappa \,\frac{m_1 m_2}{r^2}\,=\,\kappa \,\frac{m^2}{r^2}\,.\]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
Fe = ? (N)
Fg = ? (N)
From tables:
ε0 = 8.85·10−12 C2N−1m−2
κ = 6.67·10−11 m3kg−1s−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
α-particles repel with force of
\[F_e\,=\,\frac{1}{4 \pi \varepsilon_0}\,\frac{Q^2}{r^2}\,=\,9.2 {\cdot} 10^{-2}\,\mathrm{N}.\]The 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.
