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JEE Main 202229 JulyMorning Shift
Exam
JEE Main
Year
2022
Date
29 July
Shift
Morning Shift
Subject
Physics
Chapter
Electrostatics

JEE Main 2022 · 29 July · Morning Shift · Physics · Electrostatics

A spherically symmetric charge distribution is considered with...

A spherically symmetric charge distribution is considered with charge density varying as

ρρ\left(r\right) = \left\{{ρ0(34−rR)forr≤RZeroforr>R\begin{cases}{ρ}_{0}\left(\dfrac{3}{4} - \dfrac{r}{R}\right) & \text{for}r \le R \\ Zero & & & forr > R\end{cases}\right)

Where, r(r<R)r\left(r < R\right) is the distance from the centre OO (as shown in figure). The electric field at point PP will be :

A spherically symmetric charge distribution is considered with charge density varying as ρ (r) = \ cases ρ _ 0 ((3)/(4)

Options

  1. Aρ0r4ε0(34−rR)\dfrac{{ρ}_{0}r}{4{ε}_{0}}\left(\dfrac{3}{4} - \dfrac{r}{R}\right)
  2. Bρ0r3ε0(34−rR)\dfrac{{ρ}_{0}r}{3{ε}_{0}}\left(\dfrac{3}{4} - \dfrac{r}{R}\right)
  3. Cρ0r4ε0(1−rR)\dfrac{{ρ}_{0}r}{4{ε}_{0}}\left(1 - \dfrac{r}{R}\right)
  4. Dρ0r5ε0(1−rR)\dfrac{{ρ}_{0}r}{5{ε}_{0}}\left(1 - \dfrac{r}{R}\right)

Correct answer

C. ρ0r4ε0(1−rR)\dfrac{{ρ}_{0}r}{4{ε}_{0}}\left(1 - \dfrac{r}{R}\right)

Step-by-step solution

A spherically symmetric charge distribution is considered with charge density varying as ρ (r) = \ cases ρ _ 0 ((3)/(4)

According to Gauss's law, the electric flux through a closed surface area is equal to the charge inside the surface divided by ε0{ε}_{0}.

⇒∮E⋅→ds→=Qinε0⇒ ∮ \overset{\to}{E \cdot}d\overset{\to}{s} = \dfrac{{Q}_{in}}{{ε}_{0}}

⇒E4πr2=∫0rρ0(34−rR)4πr2drε0⇒ E4\pi{r}^{2} = \dfrac{{∫}_{0}^{r}{ρ}_{0}\left(\dfrac{3}{4} - \dfrac{r}{R}\right)4\pi{r}^{2}dr}{{ε}_{0}}

⇒E4πr2=ρo4πεo(34r33−r44R)⇒ E4\pi{r}^{2} = \dfrac{{ρ}_{o}4\pi}{{ε}_{o}}\left(\dfrac{3}{4}\dfrac{{r}^{3}}{3} - \dfrac{{r}^{4}}{4R}\right)

⇒E=ρor4εo(1−rR)⇒ E = \dfrac{{ρ}_{o}r}{4{ε}_{o}}\left(1 - \dfrac{r}{R}\right)

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