A charged sphere of radius $R$ carries a positive charge whose volume charge density depend only on the distancer from the ball's centre as $ρ = ρ_{0} (1 - \frac{r}{R})$ , where $ρ_{0}$ is a constant. Assume the permittivity of space $\in .$
The magnitude of electric field as a function of the distance $r$ inside the sphere is given by

see full answer

Start JEE / NEET / Foundation preparation at rupees 99/day !!

21% of IItians & 23% of AIIMS delhi doctors are from Sri Chaitanya institute !!

An Intiative by Sri Chaitanya

$E = \frac{ρ_{0}}{ε} [\frac{r}{3} - \frac{r^{2}}{4 R}]$

$E = \frac{ρ_{0}}{ε} [\frac{r}{4} - \frac{r^{2}}{3 R}]$

$E = \frac{ρ_{0}}{ε} [\frac{r}{3} + \frac{r^{2}}{4 R}]$

$E = \frac{ρ_{0}}{ε} [\frac{r}{4} + \frac{r^{2}}{3 R}]$

answer is A.

(Unlock A.I Detailed Solution for FREE)

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

Detailed Solution

According to Gauss's theorem,

$E = (\frac{1}{4 π ε_{0}}) (\frac{q_{in}}{r^{2}})$

For $r \leq R$

$q_{i n} = \int_{0}^{r} (4 π r^{2}) \cdot d r \cdot ρ = \int_{0}^{r} (4 π r^{2}) (ρ_{0}) (1 - \frac{r}{R}) d r = 4 π ρ_{0} (\frac{r^{3}}{3} - \frac{r^{4}}{4 R})$

Substituting in Eq. (i), we get

$E = \frac{ρ_{0}}{ε} [\frac{r}{3} - \frac{r^{2}}{4 R}]$

Watch 3-min video & get full concept clarity

Result Producing online coaching for JEE/NEET of south India

Largest All India Test Series for JEE/NEET

Best Books for IIT JEE

Best Books for NEET

How to Join IIT after 10th

🔥 Start Your JEE/NEET Prep at Just ₹1999 / month - Limited Offer! Check Now!

Get Expert Academic Guidance – Connect with a Counselor Today!

Result Producing online coaching for JEE/NEET of south India

Largest All India Test Series for JEE/NEET

Best Books for IIT JEE

Best Books for NEET

How to Join IIT after 10th