Electrostatics

Kindly go throiugh the solution 

 $\lambda =\frac{c}{v}=\frac{Q^3}{W^2}=\frac{I^3{T}^3}{M^2{L}^4{T}^{-4}}\Rightarrow \left[\lambda \right]=\left[M^{-2}{L}^{-4}{T}^7{l}^3\right]$

Let the charge of on each drop be q and radius of each drop be r.

  $\frac{kq}{r}=2$

When all drops are joined, radius,

$r\text{'}={\left(512\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}r=8r$

Potential of the new drop,

$V=\frac{k.512q}{8r}=64\frac{kq}{r}=128V$

According to relation between field and potential, we can write

  ^{$\overrightarrow{E}=-\frac{dV}{dx}\left(\widehat{i}\right)=-\frac{d\left(3x^2\right)}{dx}\left(\widehat{i}\right)=-6x\left(\widehat{i}\right)$}

^{${\overrightarrow{E}}_{\left(1,0,3\right)}=-6\left(\widehat{i}\right)N/C$}

According to Law of discharging of capacitor, we can write

$Q=Q_0{e}^{-\frac{t}{\tau }}\Rightarrow \frac{Q_2}{8}=Q_0{e}^{-\frac{t_2}{\tau }}\Rightarrow t_{2}3\tau \mathcal{l}n\left(2\right),and$

^{$U=\frac{Q^2}{2C}=\frac{Q_0^2}{2C}{e}^{-\frac{2t}{\tau }}\Rightarrow \frac{1}{2}\left(\frac{Q_0^2}{2C}\right)=\frac{Q_0^2}{2C}{e}^{-\frac{2t}{\tau }}\Rightarrow t_1=\frac{\tau }{2}\mathcal{l}n\left(2\right)$}

^{$\Rightarrow \frac{t_1}{t_2}=\frac{1}{6}$}

According to relation between field and potential, we can write

$\overrightarrow{E}=-\frac{dV}{dx}\left(\widehat{i}\right)=-\frac{d\left(3x^2\right)}{dx}\left(\widehat{i}\right)=-6x\left(\widehat{i}\right)$

${\overrightarrow{E}}_{\left(1,0,3\right)}=-6\left(\widehat{i}\right)N/C$

According to Law of discharging of capacitor, we can write

$Q=Q_0{e}^{-\frac{t}{\tau }}\Rightarrow \frac{Q_2}{8}=Q_0{e}^{-\frac{t_2}{\tau }}\Rightarrow t_{2}3\tau \mathcal{l}n\left(2\right),and$

$U=\frac{Q^2}{2C}=\frac{Q_0^2}{2C}{e}^{-\frac{2t}{\tau }}\Rightarrow \frac{1}{2}\left(\frac{Q_0^2}{2C}\right)=\frac{Q_0^2}{2C}{e}^{-\frac{2t}{\tau }}\Rightarrow t_1=\frac{\tau }{2}\mathcal{l}n\left(2\right)$

$\Rightarrow \frac{t_1}{t_2}=\frac{1}{6}$

$\frac{kQe}{0.12}+\frac{1}{2}m{u}^2=\frac{kQe}{0.15}+\frac{1}{2}m{u}^2$

$\frac{kQe*0.03}{0.12*0.15}=\frac{1}{2}m*\left(v^2-u^2\right)$

$Q=\frac{9.1*99*10^{-11}}{12}$

$=\frac{75.075*10^{-12}}{4\pi *{\left(0.1\right)}^2}=5.97nC/m^2$

$WD=\Delta QV$

$\Rightarrow \Delta Q=4\mu C$

$Q_i=\frac{2*4}{2+4}*6=8\mu C$

$Q_f=12\mu C=C_{eq}*6$

$C_{eq}=2\mu F\Rightarrow C_1=C_2=4\mu F$

$k=\frac{C_1}{C_0}=2$

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

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

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