Question

JEE ADVANCED PHYSICS QUESTION SEPTEMBER 2020

Answer 1

Answer:

6.2

Explanation:

*Given :-

$\sigma(r) = \sigma \circ(1 - \frac{r}{R} )$

*We have to find ratio of Electric flux

$\frac{ \phi \circ}{ \phi}$from spherical surface of radius R and radius R/4

Consider a ring element of radius ,thickness in disc charge of elemet=dQ = $(2 \pi \: rdr) \sigma \:$.

dQ = $2 \pi \:r \: dr \sigma \circ(1 - \frac{1}{r} )$

Charge of disc upto r = $\int \: charge \: of \: elementring$

$Q = \int \: 2 \pi \: r \sigma \circ(1 - \frac{r}{R} )dr$

$Q= 2 \pi \sigma \circ( \frac{r {}^{2} }{2} - \frac{r {}^{3} }{3R} )$

$\phi \circ \: = \frac{charge \: upto \: r}{ \epsilon \circ}$ _____(1)

Put the value of Q ,which have been already get .

$\phi \: = \frac{charge \: upto \: \frac{r}{4} }{ \epsilon \circ} = 2 \pi r \sigma \circ( \frac{r {}^{2} }{16 \times 2} - \frac{r {}^{3} }{\ 64 \times 3R} )$ /ۥ____(2)

Dividing (1) and (2) equation

Since,

$\pink{\frac{ \phi \circ}{ \phi} = 6.2}$

Answer 2

Consider an elemental ring of radius $\sf{r}$ and width $\sf{dr}$ from the circular disc.

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Area of this elemental ring,

$\sf{\longrightarrow dA=\pi[(r+dr)^2-r^2]}$

$\sf{\longrightarrow dA=2\pi r\,dr\quad[(dr)^2\approx0]}$

Charge on this elemental ring,

$\sf{\longrightarrow dq=\sigma(r)\cdot dA}$

$\sf{\longrightarrow dq=\sigma_0\left(1-\dfrac{r}{R}\right)\cdot2\pi r\,dr}$

$\sf{\longrightarrow dq=2\pi\sigma_0r\left(1-\dfrac{r}{R}\right)\,dr}$

So, total charge on a circular disc of radius $\sf{r}$ will be given by,

$\displaystyle\sf{\longrightarrow q=2\pi\sigma_0\int\limits_0^rr\left(1-\dfrac{r}{R}\right)\,dr}$

$\displaystyle\sf{\longrightarrow q=2\pi\sigma_0\int\limits_0^r\left(r-\dfrac{r^2}{R}\right)\,dr}$

$\displaystyle\sf{\longrightarrow q=2\pi\sigma_0\left(\dfrac{r^2}{2}-\dfrac{r^3}{3R}\right)}$

Then, electric flux through this disc,

$\sf{\longrightarrow \Phi_E=\dfrac{q}{\epsilon_0}}$

$\displaystyle\sf{\longrightarrow \Phi_E=\dfrac{2\pi\sigma_0}{\epsilon_0}\left(\dfrac{r^2}{2}-\dfrac{r^3}{3R}\right)}$

This implies,

$\displaystyle\sf{\longrightarrow \Phi_E\propto\dfrac{r^2}{2}-\dfrac{r^3}{3R}}$

$\displaystyle\sf{\longrightarrow \Phi_E\propto r^2\left(\dfrac{1}{2}-\dfrac{r}{3R}\right)}$

Therefore,

$\displaystyle\sf{\longrightarrow\dfrac{\Phi_{E_1}}{\Phi_{E_2}}=\dfrac{r_1\left(\dfrac{1}{2}-\dfrac{r_1}{3R}\right)}{r_2\left(\dfrac{1}{2}-\dfrac{r_2}{3R}\right)}}$

$\displaystyle\sf{\longrightarrow\dfrac{\Phi_{E_1}}{\Phi_{E_2}}=\dfrac{r_1(3R-2r_1)}{r_2(3R-2r_2)}}$

In the question,

• $\sf{\Phi_{E_1}=\Phi_0}$

• $\sf{\Phi_{E_2}=\Phi}$

• $\sf{r_1=R}$

• $\sf{r_2=\dfrac{R}{4}}$

Then,

$\displaystyle\sf{\longrightarrow\dfrac{\Phi_0}{\Phi}=\dfrac{R(3R-2R)}{\dfrac{R}{4}\left(3R-\dfrac{2R}{4}\right)}}$

$\displaystyle\sf{\longrightarrow\underline{\underline{\dfrac{\Phi_0}{\Phi}=1.6}}}$