Electrostatics

For a region of constant electric potential

(1) The electric field may be uniform

(2) The electric field may be zero

(3) There can be no charge inside the region

 Kindly go through the solution

C = 3A? ₀/d

With the help of conservation of volume, we can write

  $27\times \frac{4}{3}\pi r^3=\frac{4}{3}\pi R^3\Rightarrow R=3r.......\left(1\right)$                            

With the help of conservation of charge, we can write

Q = 27 q.(2)

Potential energy of single drop = U₁ =   $\frac{q^2}{8\pi {\epsilon }_{0}r}$

Potential energy of bigg

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

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

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

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

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

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

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

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