Electrostatic Potential and Capacitance

$C_{eq}=C_1+C_2+C_3+C_4$

= 1 + 2 + 4 + 3

$=10\mu F$

Q = C_eq v

= (10 * 20) $\mu C$

Q = 200 $\mu C$

F = qE Þ Electric force acting on the body

$a=\frac{F}{m}=\frac{q}{m}E=\frac{8*10^{-6}}{10^{-3}}*100=0.80m/s^2$ ->Acceleration of body

$T=2\sqrt[]{\frac{2\mathcal{l}}{a}}=2\sqrt[]{\frac{2*0.1}{0.8}}=1s\Rightarrow$ Time period of Motion

V = 10 $\left(1-e^{-t/RC}\right)$  

2 = 20 $\left(1-e^{-t/RC}\right)$  

$\frac{1}{10}=1-e^{-t/RC}$

$e^{t/RC}=\frac{10}{9}$

$\frac{t}{RC}=ln\left(\frac{10}{9}\right)=0.105$

$C=\frac{t}{R*0.105}=\frac{10^{-6}}{10*0.105}=0.95\mu F$              

Battery is connected while dielectric is inserted so potential difference will be  remains same.

$U_i=\frac{1}{2}c{V}^2$

$U_f=\frac{1}{2}Kc{V}^2$

$\Rightarrow \Delta U=\frac{1}{2}\left(K-1\right)c{V}^2=\frac{1}{2}*1*200*10^{-9}*200^2=4$

From observation, we can say that right of $-5\mu c$ charge, net electric force (or electric field) can be zero.

$So,\frac{k\left(5\mu c\right)q}{x^2}=\frac{k\left(20\mu c\right)q}{{\left(5+x\right)}^2}$

$\Rightarrow \frac{1}{x^2}=\frac{4}{{\left(5+x\right)}^2}$

$\Rightarrow \frac{5+x}{x}=2$

x = 5 cm

$V_{AB}=\frac{5}{5}*4=4V$

$\Rightarrow Q=CeqV=2x=8\mu c$

$C_{eq}=C+\frac{C_1.C_2}{C_1+C_2}$              

$=\frac{{\epsilon }_0\raisebox{1ex}{$A$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}{d}+\frac{\frac{k_1{\epsilon }_0\raisebox{1ex}{$A$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}{\raisebox{1ex}{$d$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}.\frac{k_2{\epsilon }_0\raisebox{1ex}{$A$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}{\raisebox{1ex}{$d$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}{\frac{k_1{\epsilon }_0\raisebox{1ex}{$A$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}{\raisebox{1ex}{$d$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}+\frac{k_2{\epsilon }_0\raisebox{1ex}{$A$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}{\raisebox{1ex}{$d$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}$              

= $\frac{{\epsilon }_{0}A}{d}\left(\frac{1}{2}+\frac{k_1{k}_2}{k_1+k_2}\right)$  

$V_A=\frac{1}{4\pi {\epsilon }_0}\left[\frac{1}{a}-\frac{1}{\sqrt[]{a^2+s^2}}\right]$

$V_B=\frac{-1}{4\pi {\epsilon }_0}\left[\frac{1}{a}-\frac{1}{\sqrt[]{a^2+s^2}}\right]$

$V_A-V_B=\frac{1}{2\pi {\epsilon }_0}\left[\frac{1}{a}-\frac{1}{\sqrt[]{a^2+s^2}}\right]$

for discharging RC – circuit

$q=q_0{e}^{\raisebox{1ex}{$-t$}\!\left/ \!\raisebox{-1ex}{$\tau $}\right.}$         

$l=\frac{dq}{dt}=\frac{q_0}{\tau }{e}^{-t/\tau }$       

$\tau =CR=\left(\frac{k{\epsilon }_{0}A}{d}\right)\left(\rho \frac{d}{A}\right)\left[?\mathcal{l}=d\right]$       

$l_{max}=\frac{q_0}{\rho k{\epsilon }_0}=\frac{2*10^{-12}*40}{200*50*8.85*10^{-12}}$

= 0.9 mA

Assuming shell to be conducting

For  $\frac{0\le r<R}{q_{in}=0},$

E = 0

$\frac{R\le r<a}{q_{in}=Q}$       

$E=\frac{kQ}{r^2}$            

$\frac{a\le r<b}{q_{in}=Q-Q=0}$          

E = 0

$\frac{r\ge b}{q_{in}=Q}$

$E=\frac{kQ}{r^2}$