Suppose two point objects charged with opposite charges $q_1$ and $q_2$ at a distance $r$ in a vaccum.
So, the net electrostatic force on both objects $= F_c = \frac {q_1q_2}{4π\epsilon_0r²}$ [$\epsilon_0$ is vaccum permittivity]
There should be also gravitational force working on those objects. Suppose, the masses of two objects is $m_1$ and $m_2$
Then, the gravitational force $= F_g = \frac {Gm_1m_2} {r²}$
So, the net force working on the objects $= F_{net} = \frac {4π\epsilon_0Gm_1m_2 + q_1q_2} {4π\epsilon_0r²}$
I tried to calculate the time taken by the two objects to collide with each other with the net force but failed. I want to find out the equation. So can anyone help me to find out the period of collision in such a situation mentioned above?
Best Answer
In this video by Flammable Maths, the solution to a similar problem is given.
The only difference is that we just need to include the electrostatic force, besides that the process is exactly the same.
Let's say we have two objects $1$ and $2$ with mass $m_1,m_2$ and charge $q_1,q_2$ respectivey separated by distance $R$ then-
$$\textstyle\displaystyle{F=F_C+F_G=\frac{Gm_1m_2+kq_1q_2}{R^2}}$$
Where $G$ is the Newtonian constant of gravitation and $$\textstyle\displaystyle{k=\frac{1}{4\pi\epsilon_0}}$$
By newton's third law we have $F_{12}=-F_{21}$ so
$$\textstyle\displaystyle{F_{12}=\frac{Gm_1m_2+kq_1q_2}{(r_2-r_1)^2}=m_1\frac{d^2r_1}{dt^2}}$$
$$\textstyle\displaystyle{F_{21}=-\frac{Gm_1m_2+kq_1q_2}{(r_2-r_1)^2}=m_2\frac{d^2r_2}{dt^2}}$$
Where $R=r_2-r_1$
$$\therefore\textstyle\displaystyle{\frac{d^2r_2}{dt^2}-\frac{d^2r_1}{dt^2}}$$
$$\textstyle\displaystyle{=-\frac{Gm_1m_2+kq_1q_2}{(r_2-r_1)^2}\bigg(\frac{1}{m_1}+\frac{1}{m_2}\bigg)}$$
$$\implies\textstyle\displaystyle{\frac{d^2R}{dt^2}=-\frac{\kappa}{R^2}}$$
Now we just need to solve this differential equation-
$$\textstyle\displaystyle{\frac{dv}{dt}=-\frac{\kappa}{R^2}=\frac{dv}{dR}\frac{dR}{dt}}$$
$$\implies\textstyle\displaystyle{-\frac{\kappa}{R^2}=v\frac{dv}{dR}}$$
$$\implies\textstyle\displaystyle{-\kappa\int\frac{1}{R^2}dR=\int vdv}$$
At $t=0$, $R(0)=R_i$ [The initial radius] $v(0)=0$ [velocity at the beginning]
$$\therefore\textstyle\displaystyle{\int_{0}^{v(t)}vdv=-\kappa\int_{R_i}^{R(t)}R^{-2}dR}$$
$$\implies\textstyle\displaystyle{\frac{v^2}{2}=\kappa\bigg(\frac{1}{R}-\frac{1}{R_i}\bigg)}$$
$$\implies\textstyle\displaystyle{v=\frac{dR}{dt}=\pm\sqrt{2\kappa\bigg(\frac{R_i-R}{R_iR}\bigg)}}$$
$$\implies\textstyle\displaystyle{\int_{0}^{T_c}dt=\pm\int_{R_i}^{0}\frac{1}{\sqrt{2\kappa\bigg(\frac{R_i-R}{R_iR}\bigg)}}dR}$$
$$\implies\textstyle\displaystyle{T_c=\pm\sqrt{\frac{R_i}{2\kappa}}\int_{R_i}^{0}\sqrt{\frac{R}{R_i-R}}dR}$$
Solving the integral is simple, if you would like to see the steps then see here. Noting that time can't be negative, we have-
$$\textstyle\displaystyle{T_c=\frac{\pi}{2}\sqrt{\frac{R^3}{2\kappa}}}$$
Now simply substituting the value for $\kappa$ and $k$ gives us less cleaner formula-
$$\textstyle\displaystyle{T_c=\sqrt{\frac{\pi^3\epsilon_0m_1m_2R^3}{2(m_1+m_2)(4\pi\epsilon_0Gm_1m_2+q_1q_2)}}}$$