# The final velocity of a positive test charge after being repelled to infinity by another positive charge

Let's say that we have a positive charge $$Q$$ and another positive test charge $$q$$ in the electric field of $$Q$$ placed at a distance of $$R$$ from it. A force is being applied on the charge equal to $$\dfrac{K_eQq}{R^2}$$ in the direction opposite to that of $$F_e$$ (the electrostatic force exerted on $$q$$ by $$Q$$). Now, we suddenly stop applying the force and hence, the electron flies away to infinity. Now, I was thinking that the farther it goes, the lesser force it experiences and hence, the lesser it accelerates and as the distance approaches infinity, the force's magnitude approaches zero. So, at infinity, the electron would drift at a constant velocity (almost) across space. Now, I was wondering if there might be a way to evaluate what this final velocity will be, given the values of $$Q$$, $$q$$ and $$R$$.

Now, at every point, some electrostatic force is being exerted on $$q$$ by $$Q$$ and at hence, it's position changes and it goes farther from $$Q$$ and hence, the electrostatic force on it changes as well due to a change in position. I see what would happen for a very small displacement $$ds$$ assuming that the force at all points between, say, $$R$$ and$$R+ds$$ will be the same. Now, the initial velocity $$(u)$$ at a distance of $$R$$ from $$Q$$ will be $$0 \text{ m/s}$$. Using the third equation of motion, $$v^2 = u^2+2as$$, we get the value of $$v_1$$, which is the final velocity at $$R+ds$$. So, $$v_1 = \sqrt{u^2+2a\ ds} = \sqrt{\dfrac{2F_1}{m}ds}$$ where $$F_1$$ is the force exerted by $$Q$$ at $$R$$ i.e. $$F_1 = \dfrac{K_eQq}{R^2}$$ and $$m$$ is the mass of $$q$$.

Now, at a distance of $$R+ds$$, the initial velocity of the charge $$q$$ becomes $$v_1$$ and the final velocity at $$R+2ds$$ $$(v_2)$$ will be $$\sqrt{v_1^2+\dfrac{2F_2}{m}\ ds}$$ $$F_2$$ is the force exerted by $$Q$$ on $$q$$ at a distance of $$R+ds$$ i.e. $$F_2 = \dfrac{K_eQq}{(R+ds)^2}$$ Now, $$v_2 = \sqrt{\dfrac{2F_1}{m}ds + \dfrac{2F_2}{m}}ds$$

As you can see, a pattern emerges and now, the total velocity at a distance of $$n.ds$$ from $$Q$$ will be equal to :

$$\sqrt{\dfrac{2F_1}{m}\ ds + \dfrac{2F_2}{m}\ ds + \dfrac{2F_3}{m}\ ds+...+\dfrac{2F_{n+1}}{m}\ ds}$$

This can be simplified as : $$\sqrt{\dfrac{2\ ds}{m}} \sqrt{F_1+F_2+F_3+...+F_{n+1}}$$

Now, $$\sqrt{F_1+F_2+F_3+...+F_{n+1}}$$ can be simplified as $$\sqrt{\dfrac{Qq}{4\pi\varepsilon_0R^2}+\dfrac{Qq}{4\pi\varepsilon_0(R+dr)^2}+...+\dfrac{Qq}{4\pi\varepsilon_0(R+n\ dr)^2}}=\sqrt{\dfrac{Qq}{4\pi\varepsilon_0}}\sqrt{\dfrac{1}{R^2}+\dfrac{1}{(R+dr)^2}+...+\dfrac{1}{(R+n\ dr)^2}}$$

Now, how do I continue? Were these simplifications even necessary? I took the assistance of a friend of mine and we thought that this can be solved using integration if we can just figure out how to express it in the form of an integral, but we weren't able to do it since we're both absolute beginners to calculus.

Thank You!

• Why don't you use conservation of energy? – Dirichlet Jul 10 at 7:11
• @Ohw So, the value of $U_e$ at $\infty$ will be $0$ and the value of $KE$ will be the sum of $KE$ and $U_e$ at any point. How do I evaluate that? – Rajdeep Sindhu Jul 10 at 7:15
• "Now when we suddenly stop" there KE =0 and U= kQq/r. Now equate it to 1/2mv^2. – Dirichlet Jul 10 at 7:21
• Thanks, that clarifies it, but, is there a way to continue it the way I did it? – Rajdeep Sindhu Jul 10 at 7:23
• I don't know, i told u the way i would do this problem. Your method is very complicated. – Dirichlet Jul 10 at 7:32

Now, the initial velocity $$(u)$$ at a distance of $$R$$ from $$Q$$ will be $$0 \text{ m/s}$$. Using the third equation of motion, $$v^2 = u^2+2as$$, we get the value of $$v_1$$, which is the final velocity at $$R+ds$$.
Whenever we take infinitesimal displacement $$ds$$ then we assume velocity to be constant in that infinitesimal period of time $$dt$$.
In fact this is how we define instantaneous velocity for a point object in space i.e. $$\vec{v}=\frac{d\vec{s}}{dt}$$.
• I'm sorry, I don't quite get what you mean. I have assumed an infinitely small displacement, $ds$ so that it can be treated exactly like the regular $s$ used in equations of motion. A constant (assumed) force is acting on the charge when it's position is between $R$ and $R+ds$. Hence, the acceleration is constant and is equal to $\dfrac{F_1}{m}$. The final velocity $(v_1)$ is equal to $\sqrt{u^2+2\dfrac{F_1}{m}.ds}$. Can you please point out what's wrong in this? I couldn't quite grasp much in your answer. Thanks! Sorry for the inconvenience. – Rajdeep Sindhu Jul 11 at 5:59
• We have to always be careful in dealing with infinitesimals because they aren't treated like usual notations.Imagine a position time curve in which take small elements $dt$ and $dx$. I meant to say that $\frac{dx}{dt}$ doesn't changes in the interval of $dt$.In this case $v_{1}$ is nothing but $u+du$. – Shreyansh Pathak Jul 11 at 14:44
If we start with the equation you used: $$v^2 = u^2 + 2as$$ keeping in mind that is only accurate if a is constant, we can multiply by m and write it as: $$(1/2)mv^2 – (1/2)mu^2 = K_2 – K_1 = mas = Fs$$ For a very short distance a is nearly constant and we get dK = Fdr. Put in your function for F and integrate from R to infinity.