Physics Electrostatic Potential and Capacitance

In JEE Main Physics, electric potential and capacitance chapter has a weightage of 3% to 6%. You can expect around 1 or 2 questions from this chapter which carries 4 to 8 marks.

In NEET Physics exam, the chapter electric potential and capacitance carries a weightage of 2% to 5%, which means you can expect one question out of 45 questions.

The ability to do work on a charge is called the electric potential and the ability to store charge is termed as the capacitance.

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

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

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 bigger drop =   $U_2=\frac{Q^2}{8\pi {\epsilon }_{0}R}=\frac{27*27*q^2}{8\pi {\epsilon }_0\left(3r\right)}=243\left(\frac{q^2}{8\pi {\epsilon }_{0}r}\right)=243U_1$

$\Rightarrow \frac{U_2}{U_1}=243$           

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

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

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

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