At equilibrium, the pendulum bob experiences three forces:
The electric field ($E$) produced by a vertical sheet of charge with surface charge density $\sigma$ at a distance from the sheet is given by:
$E = \frac{\sigma}{2 \varepsilon_0}$
This electric field exerts a horizontal electrostatic force ($F_e$) on the charged bob ($q$):
$F_e = qE = q \left( \frac{\sigma}{2 \varepsilon_0} \right) = \frac{\sigma q}{2 \varepsilon_0}$
When the string makes an angle $\theta$ with the vertical, the forces are balanced. Resolving the tension force into horizontal and vertical components:
Dividing the horizontal component equation by the vertical component equation:
$\frac{T \sin\theta}{T \cos\theta} = \frac{\frac{\sigma q}{2 \varepsilon_0}}{mg}$
$\tan\theta = \frac{\sigma q}{2 \varepsilon_0 m g}$
The equilibrium condition is met when $\tan\theta = \frac{\sigma q}{2 \varepsilon_0 m g}$.
A uniform time-varying magnetic field exists in a circular region of radius $R$, directed perpendicular into the plane of the paper, increasing at a constant rate $\alpha$. A straight conducting rod of length $2R$ is placed exactly along the diameter of the circular region (passing through the centre). Find the induced emf across the rod.

There is a ring of radius $r$ having linear charge density $\lambda$ and rotating with a uniform angular velocity $\omega$. The magnitude of the magnetic field produced by this ring at its own centre would be
($\mu_0 = \text{permeability of air}$)