Two small equal point charges of magnitude $q$ are suspended from a common point on the ceiling by insulating mass less strings of equal lengths. They come to equilibrium with each string making angle $\theta$ from the vertical. If the mass of each charge is $m$, then the electrostatic potential at the centre of line joining them will be $\left(\frac{1}{4\pi\,\in_{0}}=k\right).$
To solve this problem, we need to determine the electrostatic potential at the center of the line joining two point charges, each of magnitude \(q\), which are suspended in equilibrium. The main concept involves using the equilibrium condition and electrostatic potential formula.
The charges are suspended with strings making an angle \(\theta\) from the vertical. This suggests a formation where the forces due to tension, gravity, and electrostatic repulsion are balanced.
The angle \(\theta\) provides a geometric basis to evaluate the distance between the charges. In equilibrium, the vertical component of the tension balances the weight of the charges. Hence:
The equilibrium condition can be expressed as:
\[T \cos \theta = mg\]
where \(T\) is the tension in the string, and \(m\) is the mass of each charge.
For the horizontal component, which balances the electrostatic force, we have:
\[T \sin \theta = \frac{kq^2}{r^2}\]
where \(r\) is the separation distance between the two charges.
Dividing these two equations gives:
\[\tan \theta = \frac{k q^2}{mg \cdot r^2}\]
Solving for \(r^2\), we get:
\[r^2 = \frac{k q^2}{mg \cdot \tan \theta}\]
The electrostatic potential at the midpoint between the charges, where each charge contributes equally, is given by:
