Electrostatics

$F=\frac{q\left(Q-q\right)}{4\pi {\epsilon }_0{r}^2}\Rightarrow \frac{dF}{dq}=\frac{1}{4\pi {\epsilon }_0{r}^2}\left(Q-2q\right)\Rightarrow \frac{dF}{dq}=\frac{1}{2\pi {\epsilon }_0{r}^2}$

Here r is fixed

For maxima or minima of force, its first derivative should be zero.

$\frac{dF}{dq}=\frac{1}{4\pi {\epsilon }_0{r}^2}\left(Q-2q\right)=0\Rightarrow q=\frac{Q}{2}$                

Since second derivative is always negative so maxima will occur at this value of q.

$\lambda =\frac{-Q}{R*2\pi /3}$

$\overrightarrow{E}=\frac{2k\lambda }{R}sin\left(\frac{\theta }{2}\right)\left(-\widehat{i}\right)$

$\Rightarrow \overrightarrow{E}=\frac{2k}{R}*\frac{-3Q}{2\pi R}sin60°\left(-\widehat{i}\right)$

$\Rightarrow \overrightarrow{E}=\frac{3\sqrt[]{3}Q}{8{\pi }^2{\epsilon }_0{R}^2}\left(\widehat{i}\right)$

$Q_1=C_{1}V=2V\mu C$

$Q_2=Q_3=\frac{C_2{C}_3}{C_2+C_3}V=\frac{6*12}{6+12}V=4V\mu C$

$Q_1:Q_2:Q_3=1:2:2$

Volume of 27 identical drops = volume of a bigger drop

$\Rightarrow 27*\frac{4}{3}\pi r^3=\frac{4}{3}\pi R^3$              

R³ = 27r³

R = 3r

Given potential of a small drop = 22v

$V_{bigger}=\frac{k\left(27q\right)}{R}=\frac{27kq}{3r}=9\frac{kq}{r}=9*22=198volt$

$F_R=\left\{\frac{kQ{q}_0}{{\left(a-x\right)}^2}-\frac{kQ{q}_0}{{\left(a+x\right)}^2}\right\}$

$=-kQ{q}_0\left\{\frac{a^2+x^2+2ax-a^2-x^2+2ax}{\left(a^2-x^2\right)}\right\}$

$F_R=-\frac{kQ{q}_0\left(4ax\right)}{{\left(a^2-x^2\right)}^2}$

$T=2\pi \sqrt[]{\frac{a^{3}m}{4kQ{q}_0}=\sqrt[]{\frac{4{\pi }^2{a}^{3}m*4\pi {\epsilon }_0}{4Q{q}_0}}}=\sqrt[]{\frac{4{\pi }^3{\epsilon }_{0}m{a}^3}{Q{q}_0}}$

$\mu mg=\frac{k{q}_1{q}_2}{L^2}$

$\Rightarrow 0.25*0.01*10=\frac{9*10^9*2*10^{-7}*2*10^{-7}}{L^2}$

$=\frac{9*2^2*10^{-5+4}}{25*10}$

L = 12 cm

A = 30p cm² = 30p * 10^-4 m²

d = 1mm            = 10^-3 m

E = (dielectric strength for breakdown)

= 3.6 * 10⁷ v/m

Q = 7 * 10^-6 C

$\frac{\sigma }{{\in }_0}=E$

        $\Rightarrow \frac{Q}{kA{\in }_0}=E$

$k=\frac{Q}{A{\in }_{0}E}$        

$=\frac{7*10^{-6}*4\pi *9*10^9}{30\pi *10^{-4}*1*3.6*10^7}$

$=\frac{7*4\pi *9}{30\pi *3.6}$

$=\frac{7*4*9*10}{30*36}$

$=\frac{70}{30}$

$k=\frac{7}{3}=2.33$

qE = mg

$\Rightarrow q=\frac{mg}{E}=\frac{0.1}{1000}*\frac{9.8}{4.9*10^5}$         

q = 2 * 10^-9 C

$\Rightarrow$ Electric field due to infinite sheet is uniform.

 $C_0=\frac{{\epsilon }_{0}A}{d}$

$C_1=\frac{4{\epsilon }_{0}A}{d}=4C_0$

$C_Z=\frac{k{\epsilon }_{0}A}{\frac{3}{4}d}=\frac{4k{c}_0}{3}$

Series

$C_a=\frac{C_1{C}_2}{C_1+C_2}$

$\frac{16k}{3d}\frac{C_0^2}{4C_0\left(1+\frac{k}{3}\right)}$

$=\frac{4k{C}_0}{k+3}$