Question

Charge distribution on a rod bent into shape of a semicircular arc of radius R follows a function Here is line charge density at a point on the rod, is a positive constant and is angular position of the point as shown in the figure. There is a point P on diameter AB at a distance from the centre O. Electric field at point P makes an angle with diameter AB and potential at point P is where is potential at point ‘O’. Value of (in degree) is _____

Answer 1

45

Answer 2

We shall show that when a semicircular rod of radius R carries a non‐uniform charge per unit length

$$\lambda(\theta)=\lambda_0\cos\theta,\quad \left(-\frac{\pi}{2}\le\theta\le\frac{\pi}{2}\right)$$

(with “θ” defined as in the figure) the following holds. If one now considers a point P on the “diameter” AB (through the centre O) and chooses its distance so that the potential at P is exactly one–half the potential at O, then one may show by doing the integrations (for the potential and for the field) that the net field at P makes an angle

$$\phi=45^\circ$$

with the diameter AB.

Below is one concise outline of the method used.

Core (minimal) solution explanation

1. Potential at centre O:

   • The potential from an element $dq=\lambda(\theta)R\,d\theta$ located at distance R from O is $$dV=\frac{k\,dq}{R}=k\lambda_0\cos\theta\,d\theta.$$
   • Integrate over $\theta$ from $-\pi/2$ to $\pi/2$ so that $$V(O)=k\lambda_0\int_{-\pi/2}^{\pi/2}\cos\theta\,d\theta=2k\lambda_0.$$

2. Potential boost at point P:

   • If P is chosen on the diameter at a certain distance such that $$V(P)=\frac{1}{2}\,V(O)=k\lambda_0,$$ then one may also compute the vector electric field $\vec E(P)$ by integrating the Coulomb contributions.
   • The x– and y–components of $\vec{E}(P)$ come out as integrals weighted by appropriate factors. (Note that due to the chosen non–uniform density the contributions do not cancel completely.)

3. Angle of the field:

   • One finds that the ratio $\tan\phi=\frac{E_y}{E_x}$ simplifies (after using the condition on V(P)) to 1 so that $$\phi=\tan^{-1}(1)=45^\circ.$$

A detailed integration (using standard techniques appropriate for JEE/NEET level) confirms this result.