Question

Two identical charged spheres suspended from a common point by two massless strings of length $$l$$ are initially a distance $d(d<<l)$ apart because of their mutual repulsion. The charge begins to leak from both the spheres at a constant rate. As a result the charges approach each other with a velocity $v$ . Then as a function of distance $x$ between them:
A) $v\propto x^{-1/2}$
B) $v\propto x^{-1}$
C) $v\propto x^{1/2}$
D) $v\propto x$

Answer 1

Hint: In order to solve this you have to draw a diagram first which shows the equilibrium positions of the two identical charged spheres that are suspended from a common point. Also indicate all the forces acting on two spheres and then apply the equilibrium conditions and write the equations.

Formula used:
The formula for coulombic force is given by
$F={\displaystyle \frac{K{Q}_1{Q}_2}{r^2}}$
Where, $K$ is the proportionality constant and $K={\displaystyle \frac{1}{4\pi {\epsilon }_0}}$
$Q_1,Q_2$ are the charges
$r$ is the distance between the two charges

Complete step by step solution:

Here, in the above diagram there are two spheres A and C with identical charges $q$ and connected to a common point O at an angle of $\theta$ with a massless string of length $l$ and both the spheres are at a distance of $d$. The mass of both the spheres is given by $m$ due to which a gravitational force $mg$ acts downward on both the spheres and due to having identical charge a coulomb force $F$ is acted on both the spheres equal and opposite to each other. Let us assume the tension in the string is $T$.
Firstly break the components of tension force and then
By applying the equilibrium condition for sphere A, we get
$T\cos \theta =mg$ ………….(i)
And, $T\sin \theta =F$ ………….(ii)
Now, on dividing both the equations (i) and (ii), we get
$\Rightarrow$ $\tan \theta ={\displaystyle \frac{F}{mg}}$ …………(iii)
As we know that the coulomb force is given by,
$\Rightarrow$ $F={\displaystyle \frac{K{Q}_1{Q}_2}{r^2}}$
Here given that the charge is same on both the spheres, that is $q$
On putting all the values, we get
$\Rightarrow$ $F={\displaystyle \frac{K{q}^2}{d^2}}$
Now, put the above value in equation (iii), we get
$\therefore \tan \theta ={\displaystyle \frac{K{q}^2}{d^{2}mg}}$ ……….(iv)
It is given in the question that the charge begins to leak from both the spheres at a constant rate with a function of distance $x$ between them. Then the above equation becomes
$\Rightarrow \tan \theta ={\displaystyle \frac{K{q}^2}{x^{2}mg}}$
Now, from $△AOB$, we have
$\Rightarrow$ $\tan \theta ={\displaystyle \frac{{\displaystyle \frac{x}{2}}}{\sqrt{l^2-{\displaystyle \frac{x^2}{4}}}}}$ ……….(v)
As $l>>x\Rightarrow l^2>>{\displaystyle \frac{x^2}{4}}$
So, neglect the term ${\displaystyle \frac{x^2}{4}}$.
Hence, $l^2-{\displaystyle \frac{x^2}{4}}\approx l^2$
Now, the equation (v) becomes,
$\Rightarrow$ $\tan \theta ={\displaystyle \frac{x}{2l}}$ …………(vi)
Now, from equation (iv) and (vi), we have
$\Rightarrow$ ${\displaystyle \frac{x}{2l}}={\displaystyle \frac{K{q}^2}{x^{2}mg}}$
Now write the above equation in terms of x, we have
$\Rightarrow$ $x^3={\displaystyle \frac{2K{q}^{2}l}{mg}}$
Here, all the terms are constant except $x$ and $q$
So, it is clear that,
$\Rightarrow$ $x^3\propto q^2$
On differentiating both sides with respect to time $t$, we get
$\Rightarrow$ ${\displaystyle \frac{3}{2}}{x}^{1/2}{\displaystyle \frac{dx}{dt}}\propto {\displaystyle \frac{dq}{dt}}$
Here, ${\displaystyle \frac{dx}{dt}}$ is the rate of change of distance, which is known as velocity. So, the charges approach each other with a velocity $v$.
Hence, ${\displaystyle \frac{dx}{dt}}=v$ and ${\displaystyle \frac{dq}{dt}}=$ constant
So, our above equation becomes
$\Rightarrow$ $v\propto x^{-1/2}$

Therefore, the correct option is (A).

Note: Remember that when two charges separated by a distance $r_0$ in a vacuum and the force between them is same as the force between the same charges separated by a distance $r$ in a medium, then from Coulomb’s Law; $K{r}^2={r_0}^2$. Also remember that the coulomb’s law is only applicable for the point charges at rest.