Physics Electrostatic Potential and Capacitance

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

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.

V = 10 $\left(1-e^{-t/RC}\right)$  

2 = 20 $\left(1-e^{-t/RC}\right)$  

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

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

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

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

$\Delta v={\displaystyle \int \overrightarrow{E}.d\overrightarrow{x}}$

$\Delta v=\frac{\sigma }{\in o}\frac{d}{2}$

$C_{New}=\frac{Q}{\Delta v}=\frac{\sigma A2\in o}{\sigma d}=\frac{2\in oA}{d}$

$C_{New}=2{\begin{array}{l}C\\ original\\ \end{array}}_{}$

$\frac{C_{New}}{C_{Original}}=2:1$

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

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

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

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

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

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

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

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

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

$\Delta v={\displaystyle \int \overrightarrow{E}.d\overrightarrow{x}}$  

$\Delta v=\frac{\sigma }{\in o}\frac{d}{2}$

$C_{New}=\frac{Q}{\Delta v}=\frac{\sigma A2\in o}{\sigma d}=\frac{2\in oA}{d}$

$C_{New}=2{\begin{array}{l}C\\ original\\ \end{array}}_{}$

$\frac{C_{New}}{C_{Original}}=2:1$

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

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

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

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

$C_1=4\pi R_{1}and$

$C_2=\frac{4\pi R_2{R}_1}{R_2-R_1}=\frac{R_2{C}_1}{R_2-R_1}$

$\Rightarrow \frac{C_2}{C_1}=\frac{4\pi R_2{R}_1}{R_2-R_1}=\frac{R_2}{R_2-R_1}=n$

$\Rightarrow \frac{R_2}{R_1}=\frac{n}{n-1}$