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.

Difficulty level: Level 2 – Upper secondary level
Qualitative task
Tasks focused on analysis
Original source: Kohout, J. (2010). Studijní materiály ke cvičením z Elektřiny a
magnetismu. Interní materiál, Plzeň.
×Original source: Kohout, J. (2010). Studijní materiály ke cvičením z Elektřiny a magnetismu. Interní materiál, Plzeň.
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