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The question is-

Two point charges each having mass m are suspended from same point by silk thread of length l. The distance between them is x (x << l). Find the rate ${dq\over dt}$ with which the charge leaks off each sphere if their velocity of approach varies as v = $\frac{a}{\sqrt x}$ where a is positive constant.

I tried this problem by assuming equilibrium at the instant. From here I got-

$$\frac{mgx^3}{2l} = \frac{q^2}{4\pi\epsilon_o}$$

Differentiating this equation and substituting the value of $v = {dx\over dt}$ = $\frac{a}{\sqrt x}$ and obtaining an expression for ${dq\over dt}$, I substituted the value of q from the above written expression. On solving I got my answer as-

$$\frac{dq}{dt} = \frac{3}{2}a\sqrt{\frac{2\pi\epsilon_o mg}{l}}$$

Which is the answer given in the answer key. But I do not understand that why are the pith balls at equilibrium when they have given an expression for velocity and we can obtain some value of acceleration which is not negligible.

Arnav Mahajan
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1 Answers1

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The idea is that they are in fact not really at equilibrium, as otherwise they won't move at all; however the problem implicitly assumes that the difference of force is so small that there is almost no acceleration(quasi static). This is quite the same as assuming $\tan x = x$ as $x$ is far less than $l$; however the assumption that the process is gradual is implicit but $x<<l$ is explicit, which is the issue of the problem.

It is possible to include the acceleration factor in your equation; the one invented the problem just assumed that it is way too complex and you will give up :-( Maybe show them you won't.

user12986714
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