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Electrostatics question

2022 · 29 Jul · Shift 1 · Q57
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  5. /2022 · 29 Jul · Shift 1 · Q57

Electrostatics question

2022 · 29 Jul · Shift 1 · Q57

JEE MainPhysicsElectrostaticsMCQ+4 / −1
A spherically symmetric charge distribution is considered with charge density varying as ρ(r)={ρ0(34−rR) for r≤R zero  for r>R\rho(r)= \begin{cases}\rho_{0}\left(\frac{3}{4}-\frac{r}{R}\right) & \text { for } r \leq R \\ \text { zero } & \text { for } r>R\end{cases}ρ(r)={ρ0​(43​−Rr​) zero ​ for r≤R for r>R​ Where, r(r<R)r(r \lt R)r(r<R) is the distance from the centre O (as shown in figure). The electric field at point P will be: JEE Main 2022 (Online) 29th July Morning Shift Physics - Electrostatics Question 106 English
  1. A
    ρ0r4ε0(34−rR)\frac{\rho_{0} \mathrm{r}}{4 \varepsilon_{0}}\left(\frac{3}{4}-\frac{r}{R}\right)4ε0​ρ0​r​(43​−Rr​)
  2. B
    ρ0r3ε0(34−rR)\frac{\rho_{0} r}{3 \varepsilon_{0}}\left(\frac{3}{4}-\frac{r}{R}\right)3ε0​ρ0​r​(43​−Rr​)
  3. C
    ρ0r4ε0(1−rR)\frac{\rho_{0} r}{4 \varepsilon_{0}}\left(1-\frac{r}{R}\right)4ε0​ρ0​r​(1−Rr​)
  4. D
    ρ0r5ε0(1−rR)\frac{\rho_{0} r}{5 \varepsilon_{0}}\left(1-\frac{r}{R}\right)5ε0​ρ0​r​(1−Rr​)
View written solutionFree

Correct answer: C

  1. Use Gauss's law for spherical symmetry

For a spherically symmetric charge distribution, the electric field at a distance r<Rr<Rr<R from the centre is radial and depends only on the charge enclosed within radius rrr.

By Gauss's law,

E(4πr2)=Qencε0E(4\pi r^2)=\frac{Q_{\text{enc}}}{\varepsilon_0}E(4πr2)=ε0​Qenc​​

so

E=Qenc4πε0r2.E=\frac{Q_{\text{enc}}}{4\pi \varepsilon_0 r^2}.E=4πε0​r2Qenc​​.
  1. Compute enclosed charge QencQ_{\text{enc}}Qenc​

Given

ρ(r)=ρ0(34−rR),r≤R\rho(r)=\rho_0\left(\frac34-\frac{r}{R}\right), \qquad r\le Rρ(r)=ρ0​(43​−Rr​),r≤R

For a thin spherical shell of radius r′r'r′ and thickness dr′dr'dr′, volume element is

dV=4πr′2 dr′.dV=4\pi r'^2\,dr'.dV=4πr′2dr′.

Hence,

Qenc=∫0rρ(r′) dV=4πρ0∫0r(34−r′R)r′2 dr′.Q_{\text{enc}}=\int_0^r \rho(r')\, dV =4\pi \rho_0 \int_0^r \left(\frac34-\frac{r'}{R}\right) r'^2 \, dr'.Qenc​=∫0r​ρ(r′)dV=4πρ0​∫0r​(43​−Rr′​)r′2dr′.
  1. Evaluate the integral
Qenc=4πρ0[34∫0rr′2dr′−1R∫0rr′3dr′].Q_{\text{enc}}=4\pi \rho_0 \left[ \frac34 \int_0^r r'^2 dr' - \frac{1}{R}\int_0^r r'^3 dr' \right].Qenc​=4πρ0​[43​∫0r​r′2dr′−R1​∫0r​r′3dr′].

Now,

∫0rr′2dr′=r33,∫0rr′3dr′=r44.\int_0^r r'^2 dr' = \frac{r^3}{3}, \qquad \int_0^r r'^3 dr' = \frac{r^4}{4}.∫0r​r′2dr′=3r3​,∫0r​r′3dr′=4r4​.

Therefore,

Qenc=4πρ0[34⋅r33−1R⋅r44]=4πρ0(r34−r44R).Q_{\text{enc}}=4\pi \rho_0 \left[ \frac34\cdot \frac{r^3}{3} - \frac{1}{R}\cdot \frac{r^4}{4} \right] =4\pi \rho_0 \left(\frac{r^3}{4}-\frac{r^4}{4R}\right).Qenc​=4πρ0​[43​⋅3r3​−R1​⋅4r4​]=4πρ0​(4r3​−4Rr4​).

So,

Qenc=πρ0(r3−r4R)=πρ0r3(1−rR).Q_{\text{enc}}=\pi \rho_0 \left(r^3-\frac{r^4}{R}\right) =\pi \rho_0 r^3\left(1-\frac{r}{R}\right).Qenc​=πρ0​(r3−Rr4​)=πρ0​r3(1−Rr​).
  1. Substitute into Gauss's law
E=Qenc4πε0r2=πρ0r3(1−rR)4πε0r2.E=\frac{Q_{\text{enc}}}{4\pi \varepsilon_0 r^2} =\frac{\pi \rho_0 r^3\left(1-\frac{r}{R}\right)}{4\pi \varepsilon_0 r^2}.E=4πε0​r2Qenc​​=4πε0​r2πρ0​r3(1−Rr​)​.

Simplifying,

E=ρ0r4ε0(1−rR).E=\frac{\rho_0 r}{4\varepsilon_0}\left(1-\frac{r}{R}\right).E=4ε0​ρ0​r​(1−Rr​).
  1. Match with the options

This matches:

C: ρ0r4ε0(1−rR)\boxed{\text{C: } \frac{\rho_0 r}{4\varepsilon_0}\left(1-\frac{r}{R}\right)}C: 4ε0​ρ0​r​(1−Rr​)​
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