I am struggling with the following problem (Irodov 3.3):

Two small equally charged spheres, each of mass $m$, are suspended from the same point by silk threads of length $l$. The distance between the spheres $x \ll l$. Find the rate $\frac{dq}{dt}$ with which the charge leaks off each sphere if their approach velocity varies as $v = \frac{a}{\sqrt{x}}$, where $a$ is a constant.

This is embarrassingly simple; we make an approximation for $x \ll l$ and get $$ \frac{1}{4 \pi \epsilon_0} \frac{q^2}{x^2} - \frac{mgx}{2l} = m \ddot{x}. $$ We can get $\ddot{x}$ from our relation for $v$, so we can solve for $q$ and then find $\frac{dq}{dt}$.

However, in general, $\frac{dq}{dt}$ will depend on $x$ and hence on $t$. The answer in the back of the book and other solutions around the web have $\frac{dq}{dt}$ a constant.

You can get this by assuming that at each moment the spheres are in equilibrium, so that you have $\ddot{x} = 0$ in the equation of motion above.

Does the problem tacitly imply we should assume equilibrium and hence $\frac{dq}{dt}$ is constant, or am I missing something entirely? I.e. why is the assumption of equilibrium justified? I understand reasoning like "the process happens very gradually, so the acceleration is small compared to other quantities in the problem," but I don't understand how that is justified by the problem itself, where we are simply given that the spheres are small (so we can represent them as points) and $x \ll l$ (which we have used to approximate the gravity term in the equation of motion).

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    $\begingroup$ As far as I can remember, the assumption of equilibrium was given explicitly in the problem, but if it doesn't say anything about it, you certainly do have acceleration: $\dfrac{dv}{dt}\neq 0$. $\endgroup$
    – Mostafa
    Jul 31, 2013 at 10:30
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    $\begingroup$ I don't see it stated explicitly (I copied the exact problem above). So yes, it definitely looks like $\dot{v} \neq 0$, but all the solutions I've seen solve it as if there is none. $\endgroup$
    – user27657
    Jul 31, 2013 at 14:08
  • $\begingroup$ Binominal expansion shows that result (dq/dt) must contain power of (3/2) of x. Even (dv/dt)=(a^2/x^2) which tends to infinity, so it is not logical to ignore acceleration in any case. $\endgroup$ Apr 25, 2017 at 17:01
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    $\begingroup$ Possible duplicate of Charge leakage from two suspended charged spheres $\endgroup$ Apr 25, 2017 at 22:26
  • 4
    $\begingroup$ @sammy gerbil This question asks about the justification of spheres in equilibrium while that question asks about the meaning of charge leakage. $\endgroup$ Apr 26, 2017 at 6:09

2 Answers 2


If we continue with the suggestion you made, of obtaining $\ddot x$ from the equation of motion $v=a/\sqrt{x}$ which was provided, and substituting this into the equation $F=m\ddot x$, then we do indeed find that $\dot q$ is not constant. It is only by ignoring the $m\ddot x$ term - by assuming that $v\approx 0$ - that we can reach the result which Irodov intended.

But there is nothing in the question statement which justifies the assumption that $v \approx 0$. No values are given which would enable us to conclude that $\dot v=-(a/2x\sqrt{x})v$ can be neglected so that there is a quasi-static equilibrium.

The conclusion must be that Irodov made an error. He deliberately imposed an unrealistic but fairly simple equation of motion $v=a/\sqrt{x}$ in order to derive an equally unrealistic but simple result (that $\dot q$ is constant). While doing so he failed to state the assumptions which were necessary to obtain this result.

Even the most respected authors and textbooks are fallible.


I think the answer is sonething that you have overlooked, a (. ) AKA FULL STOP.

You state that the web results say the answe is dt/dq a this is a constant because (a) is A constant.

The question you ask is " Does the problem tacitly imply we should assume equilibrium and hence dt/dq is constant, or am I missing something entirely?"

I reckon you've overlooked the fact its (a) NOT (dt/dq) thats the constant.

Have I answered your question?

Sorry for the technical lomotations of my keyboard.


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