## Filters

Q&A - Ask Doubts and Get Answers
Q

# 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 :

for  r< R

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

• Option 1)

• Option 2)

• Option 3)

• Option 4)

Answers (2)
129 Views
N

As we discussed in comcept

If P lies inside -

$\dpi{100} E_{in}=\frac{1}{4\pi \epsilon _{0}}\frac{Qr}{R^{3}}$   $\dpi{100} V_{in}=\frac{Q}{4\pi \epsilon _{0}}\frac{3R^{2}-r^{2}}{2R^{3}}$

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

-

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

where $(x

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

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

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

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

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

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

Option 1)

Option 2)

Option 3)

Option 4)

Exams
Articles
Questions