1
$\begingroup$

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?

$\endgroup$
8
  • 1
    $\begingroup$ Use Newton's second law and solve the resulting differential equation. You'll need two initial conditions. $\endgroup$
    – kricheli
    May 22 at 9:48
  • $\begingroup$ @kricheli How do I use F = ma for Coulomb's law? $\endgroup$ May 22 at 14:46
  • $\begingroup$ This is called a two-body problem. And you'll be hard-pressed not to find an answer to this... physics.stackexchange.com/a/156748/297348 $\endgroup$
    – kricheli
    May 22 at 15:07
  • $\begingroup$ Also note physics.stackexchange.com/a/580174/297348 $\endgroup$
    – kricheli
    May 22 at 15:07
  • $\begingroup$ That is a two body problem with vanishing angular momentum... $\endgroup$
    – kricheli
    May 22 at 15:26

2 Answers 2

1
$\begingroup$

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-

$$\textstyle\displaystyle{T_c=\sqrt{\frac{\pi^3\epsilon_0m_1m_2R^3}{2(m_1+m_2)(4\pi\epsilon_0Gm_1m_2+q_1q_2)}}}$$

$\endgroup$
0
$\begingroup$

Start by writing the problem in the form $$ F = -\frac{ C }{ r} $$ for the force between the two particles separated by the distance $r$ where $ C $ is easily identified (note that generally, the gravitational force is much weaker than the Coulomb force and is usually ignored, as well, I have included a $-ve$ sign because you are considering opposite charges and the particles attract each other).

Suppose that you hold only particle and let the second particle be released from rest at an initial distance $ r_0 $ from the first. Because the force is central this is really a one-dimensional problem.

Rather than solve the system by applying the equation $ F = m_2 \frac{d^2 x}{d t^2} $ (where for convenience $ x $ is the separation of the second particle from the first), it is best to consider the conservation of energy. Then the equation you have to consider is $$ -\frac{ C }{ x } + \frac{1}{2} m_2 \left( \frac{ d x }{ d t } \right)^2 = \mbox{ constant}$$ where you can determine the constant given that the second particle is initially at rest and separated by the distance $ x = r_0 $.

From the above you can easily find $ d x / d t $ as a (rather) complicated function of $ x $ that may possibly be solved analytically for $ x $ (see the link in Sarka's answer).

So, what you should find is that when you release the second particle it will move towards the first, increasing in speed, getting closer to the fixed particle, the force increases, ensuring its speed increases further. In fact, the speed will increase without limit (in practice, rather than dealing with point particles you would need to take into account the structure of the "particles").

However there is one important issue that has been ignored, namely that as charges accelerate you will need to take electromagnetic radiation into account (as charges accelerate electromagnetic radiation is emitted by the charge), that is a non-trivial problem to include.

$\endgroup$
4
  • $\begingroup$ What if the masses of those two point objects become larger as sun? $\endgroup$ May 22 at 18:39
  • $\begingroup$ @DebanjanBiswas if the masses are big enough then, yes, you need to include them. $\endgroup$
    – jim
    May 22 at 18:47
  • $\begingroup$ @DebanjanBiswas whether you include gravity or not isn't too important, the potential is still of the form $ - C/r $. $\endgroup$
    – jim
    May 23 at 10:08
  • $\begingroup$ What is C in this equation? $\endgroup$ May 23 at 11:47

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.