A Charge Oscillating Above A Charged Square Plate.

Consider a square loop of thickness $dx$ at a distance $x$ from center, such that the side of square is $2x$ . Due to each of 4 sides, carrying charge $\sigma dx$ each electric field's magnitude is $\displaystyle \frac{\sigma x dx}{2 \pi \epsilon_{0} \sqrt{x^2+h^2} \sqrt{2x^2+h^2}}$

Since the net field is vertical, we take only vertical component , multiplying above expression by $\displaystyle \frac{h}{\sqrt{x^2+h^2}}$ and then multiply by $4$ , as there are $4$ sides.

Hence, the electric field due to square loop is :

$dE = \displaystyle \frac{2 \sigma h x dx}{\pi \epsilon_{0} (x^2+h^2) \sqrt{2x^2+h^2}}$

$E = \displaystyle \frac{2 \sigma h}{ \pi \epsilon_{0}} \int_{0}^{\frac{a}{2}} \frac{x dx}{(x^2+h^2) \sqrt{2x^2+h^2}}$

Let $2x^2+h^2 = t^2$ , and obtain:

$E = \displaystyle \frac{2 \sigma h}{\pi \epsilon_{0}} \int_{h}^{\sqrt{\frac{a^2}{2} + h^2}} \frac{dt}{h^2+t^2}$

= $\frac{2 \sigma}{\pi \epsilon_{0}} \bigg(\tan^{-1} \bigg(\frac{\sqrt{\frac{a^2}{2}+h^2}}{h}\bigg) - \frac{\pi}{4}\bigg)$

Note that you get $E = \frac{\sigma}{2 \epsilon_{0}}$ for $a \to \infty$ , and also for $h \to 0$ . This verifies the result.

Now, in our case, when $a = 2h$ , $E = \frac{\sigma}{6 \epsilon_{0}}$ as in another problem posted by kartik kannan.

Now, $\displaystyle \frac{q \sigma}{6 \epsilon_{0}} = mg$

Let us change $h$ by a small distance $\delta x << 1$ . Then by taking derivative of the force,

$F_{net} = qE - mg$

$\Rightarrow \displaystyle \frac{\delta F}{\delta x} = q \frac{\delta E}{\delta x} \approx -\frac{q \sigma}{\sqrt{3} \pi \epsilon_{0} h}$

$\Rightarrow \displaystyle F_{new} = \delta F = - \frac{2 \sqrt{3} mg}{\pi h} \delta x$

Hence, $\displaystyle \boxed{ T = 2 \pi \sqrt{\frac{\pi h}{2 \sqrt{3} g}}}$

Put values to get the answer as $308$