Question

An alpha particle has a mass of 6.6 × 10−27 kg and charge of Q=+2e=3.2×10-19 compare the magnitude of the electron repulsion between two apha particles with that of the gravitaiontal field between them.

Answer 1

Solution. Find the gravitational attraction force of two alpha particles using the formula

$$F_1=\frac {G*m_1*m_2} {r^2}$$

$$G=6.67*10^{-11} N*m^2/kg^2$$

where F1 is attraction force; m1, m2 are mass of the alpha particles; r is distance between alpha particles.

The electric repulsion force can be found using the formula

$$F_2=\frac {k*q_1*q_2} {r^2}$$ $$k=9*10^9 N*m^2/C^2$$

where F2 is repulsion force; q1 and q2 are alpha particle charges; r is distance between alpha particles.

Find the ratio of the two forces

$$n=\frac {F_2} {F_1}=\frac {k*q_1*q_2} {r^2} * \frac {r^2}{G*m_1*m_2}$$

$$n=\frac {k*q_1*q_2} {G*m_1*m_2}$$

According to the condition of the problem

$$q_1=q_2=3.2*10^{-19} C$$

$$m_1=m_2=6.6*10^{-27} kg$$

Therefore

$$n=\frac {9*10^9*3.2*10^{-19}*3.2*10^{-19}} {6.67*10^{-11}*6.6*10^{-27}*6.6*10^{-27}}$$

$$n=3.17*10^{35}$$

The repulsive force is greater than the force of gravitational attraction

$$3.17*10^{35}$$

times

Answer. The repulsive force is greater than the force of gravitational attraction

$$3.17*10^{35}$$

times