Let there be a spherically symmetric charge distribution with charge density varying as (r)=₀≤ft(5/4-r/R) upto r=R, and (r)=0 for r>R, where r is the distance from the erigin. The electric field at a distance r(r [AIEEE 2010]

Correct Answer is : ρ0r4ε0(53−rR)₀ r/4 ε₀≤ft(5/3-r/R)4ε0ρ0r(35−Rr)

Correct Answer is : ρ0r4ε0(53−rR)₀ r/4 ε₀≤ft(5/3-r/R)4ε0ρ0r(35−Rr)

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Electrostatics Part 1

Section: Physics

Question : 76 of 100

Marks: +1, -0

\rho(r)=\rho_0\left(\frac{5}{4}-\frac{r}{R}\right)

r=R

\rho(r)=0

r>R

r

r(r < R)

Solution:

Let us consider a spherical shell of radius

x

and thickness

dx

Charge on this shell

dq=\rho\cdot 4\pi^2 dx=\rho_0\left(\frac{5}{4}-\frac{x}{R}\right)\cdot 4\pi x^2 dx

\therefore

Total charge in the spherical region from center to

r(r < R)

q=\int dq=4\pi\rho_0\int\limits_{0}^{r}\left(\frac{5}{4}-\frac{x}{R}\right)x^2 dx

=4\pi\rho_0\left[\frac{5}{4}\cdot\frac{r^3}{3}-\frac{1}{R}\cdot\frac{r^4}{4}\right]

=\pi\rho_0 r^3\left(\frac{5}{3}-\frac{r}{R}\right)

\therefore

Electric field at

r, E=\frac{1}{4\pi E_0}\cdot\frac{q}{r^2}

=\frac{1}{4\pi E_0}\cdot\frac{\pi\rho_0 r^3}{r^2}\left(\frac{5}{3}-\frac{r}{R}\right)

=\frac{\rho_0 r}{4 E_0}\left(\frac{5}{3}-\frac{r}{R}\right)

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