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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).

sammy gerbil
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user27657
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4 Answers4

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In your equation

$$\frac{1}{4 \pi \epsilon_0} \frac{q^2}{x^2} - \frac{mgx}{2l} = m \ddot{x}$$

$ \ddot{x}$ can be written as $$\frac{dv}{dt} = \frac{dv}{dx}\frac{dx}{dt}= v\frac{dv}{dx}$$

so the first equation can be written in terms of $x$ for all the terms (as you are given $v$ in terms of $x$ in the question).

After doing that and multiplying through by $x^2$, you get

$$\frac{q^2}{4 \pi \epsilon_0} - \frac{mgx^3}{2l} = -\frac{a^2}{2}$$

i.e. $x$ is constant

This means that we must imagine that $x$ can only vary really slowly, that's the justification - from the things given in the question, for deciding we can put $\ddot{x}=0$

so you get $q^2 \propto x^3$

$$2q\frac{dq}{dt} = 3x^2\frac{dx}{dt}$$

putting $q$ and $\frac{dx}{dt} (= v)$ in terms of $x$ gives the result that $\frac{dq}{dt}$ is constant.

John Hunter
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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.

sammy gerbil
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Yes, your reasoning is right. This is a wrong question. This question has been rectified in the new edition of the book.

S Das
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-1

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.

Billiehuman
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