Question

Two similar balls, each of mass $m$ and charge $q$, are hung from a common point from two silk threads, each of length $l$. Prove that separation between the balls is $x = {\left[ {\dfrac{{{q^2}l}}{{2\pi {\varepsilon _0}mg}}} \right]^{\dfrac{1}{3}}}$, if $\theta $ is small. Find the rate $dq/dt$ with which the charge should leak off each sphere if the velocity of approach varies as $v = \dfrac{a}{{\sqrt x }}$, where $a$ is a constant.

Answer 1

Hint: Balance all the forces on the balls to find the distance between the balls. Use the fact that the angle between the strings is very small,secondly, use the relation of $v$ given in the question to find the rate of leakage of charge.

Complete answer:
First, we make the free body diagram of the 2 balls and the forces which are acting on the ball.
The free body diagram is given below
From the figure, we can easily balance the equations,
$T\sin \theta = \dfrac{{k{q^2}}}{{{x^2}}}$
We know that sin is the ratio of perpendicular to hypotenuse of the triangle.
So,
$\sin \theta = \dfrac{x}{{2l}}$
Putting the value of sin in above equation
$\dfrac{{Tx}}{{2l}} = \dfrac{{k{q^2}}}{{{x^2}}}$
This is our equation 1
Now,
Again, from the figure,
$T\cos \theta = mg$
It is given in the question that the angle is small.
And,
$\mathop {\lim }\limits_{x \to 0} \cos \theta = 1$
So, the equation becomes,
$T = mg$
Putting the value of $T$ in equation 1
$\dfrac{{\left( {mg} \right)x}}{{2l}} = \dfrac{{k{q^2}}}{{{x^2}}}$
$x = {\left[ {\dfrac{{{q^2}l}}{{2\pi {\varepsilon _0}mg}}} \right]^{\dfrac{1}{3}}}$
From the above expression, we can write as:
$q = {\left[ {\dfrac{{2\pi {\varepsilon _0}mg{x^3}}}{l}} \right]^{\dfrac{1}{2}}}$

$q = {\left[ {\dfrac{{2\pi {\varepsilon _0}mg}}{l}} \right]^{\dfrac{1}{2}}}{x^{\dfrac{3}{2}}}$
Differentiating with respect to time, we get,
$\dfrac{{dq}}{{dt}} = \dfrac{3}{2}{\left[ {\dfrac{{2\pi {\varepsilon _0}mg}}{l}} \right]^{\dfrac{1}{2}}}{x^{\dfrac{1}{2}}}\dfrac{{dx}}{{dt}}$
We know that,$\dfrac{{dx}}{{dt}} = v$
$\dfrac{{dq}}{{dt}} = \dfrac{3}{2}{\left[ {\dfrac{{2\pi {\varepsilon _0}mg}}{l}} \right]^{\dfrac{1}{2}}}{x^{\dfrac{1}{2}}}v$
But, $v = \dfrac{a}{{\sqrt x }}$
$\dfrac{{dq}}{{dt}} = \dfrac{3}{2}{\left[ {\dfrac{{2\pi {\varepsilon _0}mg}}{l}} \right]^{\dfrac{1}{2}}}{x^{\dfrac{1}{2}}}\left( {\dfrac{a}{{\sqrt x }}} \right)$

$\dfrac{{dq}}{{dt}} = \dfrac{{3a}}{2}{\left[ {\dfrac{{2\pi {\varepsilon _0}mg}}{l}} \right]^{\dfrac{1}{2}}}$
This is the rate of leakage of charge from the ball.

Additional information:
Coulomb's law states that the magnitude of the electrostatic force of attraction or repulsion between two point charges is directly proportional to the product of the magnitudes of charges and inversely proportional to the square of the distance between them. The force is along the straight line joining them.

Note: It is given in the question the angle formed by the strings is very small. That’s why we are able to take the value of cosine of the angle between the strings equal to 1.