A simple pendulum of length $L$, mass $m$ and electric charge $q$ on its bob is oscillating with a time period $T$ under uniform gravity which is in the $-\hat{z}$ direction. Upon applying a uniform electric field $|E|\hat{n}$ (where $\hat{n}$ is a unit vector in the plane of oscillation), the time period of the pendulum decreases. Which of the following statements is NOT correct?

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

For the time period to decrease, the effective acceleration due to gravity $g_{eff}$ must be greater than $g$.

The effective acceleration is given by $\vec{g}_{eff} = \vec{g} + \dfrac{\vec{F}_e}{m} = -g\hat{z} + \dfrac{q|E|\hat{n}}{m}$.

If $q$ is positive and $\hat{n} = \hat{z}$, the electric force is in the $+\hat{z}$ direction. The effective gravity is $\vec{g}_{eff} = -(g - \dfrac{q|E|}{m})\hat{z}$. Assuming the electric force is smaller than the weight, $g_{eff} < g$, which would cause the time period to increase. Thus, this statement is incorrect.

If $q$ is positive and $\hat{n} = -\hat{z}$, the electric force is in the $-\hat{z}$ direction. The effective gravity is $\vec{g}_{eff} = -(g + \dfrac{q|E|}{m})\hat{z}$. Here, $g_{eff} > g$, so the time period decreases.

If $q$ is negative and $\hat{n} = \hat{z}$, the electric force is in the $-\hat{z}$ direction. The effective gravity is $\vec{g}_{eff} = -(g + \dfrac{|q||E|}{m})\hat{z}$. Here, $g_{eff} > g$, so the time period decreases.

If $q$ is positive and $\hat{n} \cdot \hat{z} = -\dfrac{1}{\sqrt{2}}$, the magnitude of effective gravity is $|\vec{g}_{eff}| = \sqrt{g^2 + (\dfrac{q|E|}{m})^2 - 2g(\dfrac{q|E|}{m})(\hat{n} \cdot \hat{z})} = \sqrt{g^2 + (\dfrac{q|E|}{m})^2 + \sqrt{2}g(\dfrac{q|E|}{m})}$. Since this is strictly greater than $g$, the time period decreases.

Therefore, the statement that $q$ is positive and $\hat{n} = \hat{z}$ is not correct.

Answer: $q$ is positive and $\hat{n} = \hat{z}$