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Can someone help me with this, - Electrostatics - JEE Main-9

 A spherically symmetric charge  distribution is characterised by a charge density having the following variation :

 

\rho (r)=\rho _{0}(1-\frac{r}{R}) for  r< R

\rho (r)=0\; \; \; \; \; for\; r\geq R

Where r is the distance from the centre of the charge distribution and \rho _{0} is a constant. The electric field at an internal point (r < R) is :

  • Option 1)

    \frac{\rho _{0}}{4\epsilon _{0}}\left ( \frac{r}{3} -\frac{r^{2}}{4R}\right )

  • Option 2)

    \frac{\rho _{0}}{\epsilon _{0}}\left ( \frac{r}{3} -\frac{r^{2}}{4R}\right )

  • Option 3)

    \frac{\rho _{0}}{3\epsilon _{0}}\left ( \frac{r}{3} -\frac{r^{2}}{4R}\right )

  • Option 4)

    \frac{\rho _{0}}{12\epsilon _{0}}\left ( \frac{r}{3} -\frac{r^{2}}{4R}\right )

 
Answers (2)
129 Views
N neha

As we discussed in comcept

If P lies inside -

E_{in}=\frac{1}{4\pi \epsilon _{0}}\frac{Qr}{R^{3}}   V_{in}=\frac{Q}{4\pi \epsilon _{0}}\frac{3R^{2}-r^{2}}{2R^{3}}

\dpi{100} E_{in}=\frac{\rho r}{3 \epsilon _{0}}        V_{in}=\frac{\rho \left ( 3R^{2}-r^{2} \right )}{6 \epsilon _{0}}

-

 

 dq=\int 4 \pi x^{2} dx=\rho _{0} (1- \frac{x}{R})4 \pi x^{2} dx

where (x<R)

q=\int dq= 4\pi \rho _{0}\int_{0}^{r}(1-\frac{x}{R})x^{2}dx

=4\pi\rho _{0}[\frac{x^{2}}{3}-\frac{x^{4}}{4R}]^{r}

=4\pi\rho _{0}[\frac{r^{2}}{3}-\frac{r^{4}}{4R}]

q= 4\pi\rho _{0}r^{3}[\frac{1}{3}-\frac{r}{4R}]

E=\frac{1}{4 \pi \varepsilon _{0}}\frac{q}{r^{2}}= \frac{1}{4 \pi \varepsilon _{0}}\frac{4 \pi\ \varepsilon _{0}r^{3}}{r^{2}}[\frac{1}{3}-\frac{r}{4R}]

E=\frac{\rho _{0}}{\varepsilon ^{0}}[\frac{r}{3}-\frac{r^{2}}{4R}]


Option 1)

\frac{\rho _{0}}{4\epsilon _{0}}\left ( \frac{r}{3} -\frac{r^{2}}{4R}\right )

Option 2)

\frac{\rho _{0}}{\epsilon _{0}}\left ( \frac{r}{3} -\frac{r^{2}}{4R}\right )

Option 3)

\frac{\rho _{0}}{3\epsilon _{0}}\left ( \frac{r}{3} -\frac{r^{2}}{4R}\right )

Option 4)

\frac{\rho _{0}}{12\epsilon _{0}}\left ( \frac{r}{3} -\frac{r^{2}}{4R}\right )

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