A simple pendulum having length l and mass 'm' is suspended between...

The forces acting on the bob of the pendulum are the gravitational force $mg$ acting downwards and the electrostatic force $qE$ acting downwards (since the electric field $E$ is directed downwards and the charge $q$ is assumed positive).

The net downward force on the bob is $F_{net} = mg + qE$.

The effective acceleration due to gravity is $g_{eff} = \dfrac{F_{net}}{m} = g + \dfrac{qE}{m}$.

The time period of a simple pendulum is given by $T = 2\pi\sqrt{\dfrac{l}{g_{eff}}}$.

Substituting the value of $g_{eff}$, we get $T = 2\pi\sqrt{\dfrac{l}{g + \dfrac{qE}{m}}}$.

Answer: $2\pi\sqrt{\dfrac{l}{g + \dfrac{Eq}{m}}}$