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Question

Physics Question on Electric charges and fields

Let there be a spherically symmetric charge distribution with charge density varying as ρ(r)=ρ0(54rR)\rho\left(r\right) = \rho_{0} \left(\frac{5}{4}-\frac{r}{R}\right) upto r=Rr = R, and ρ(r)=0\rho \left(r\right) = 0 for r>Rr > R, where r is the distance from the origin. The electric field at a distance r(r<R)r\left(r < R\right) from the origin is given by

A

4πρ0r3ε0(54rR)\frac{4\pi\rho_{0}r}{3\varepsilon_{0}} \left(\frac{5}{4}-\frac{r}{R}\right)

B

ρ0r4ε0(54rR)\frac{\rho_{0}r}{4\varepsilon_{0}} \left(\frac{5}{4}-\frac{r}{R}\right)

C

4ρ0r3ε0(54rR)\frac{4\rho_{0}r}{3\varepsilon_{0}} \left(\frac{5}{4}-\frac{r}{R}\right)

D

ρ0r3ε0(54rR)\frac{\rho_{0}r}{3\varepsilon_{0}} \left(\frac{5}{4}-\frac{r}{R}\right)

Answer

ρ0r4ε0(54rR)\frac{\rho_{0}r}{4\varepsilon_{0}} \left(\frac{5}{4}-\frac{r}{R}\right)

Explanation

Solution

Apply shell theorem the total charge upto distance r can be calculated as followed dq=4πr2.dr.ρdq = 4\pi r^{2}.dr.\rho =4πr2.dr.ρ0[54rR]= 4\pi r^{2}.dr.\rho_{0} \left[\frac{5}{4}-\frac{r}{R}\right] =4πρ0[54r2drr3Rdr]= 4\pi\rho_{0}\left[\frac{5}{4}r^{2}dr-\frac{r^{3}}{R}dr\right] dq=q=4πρ00r(54r2drr3Rdr)\int dq = q = 4\pi \rho _{0}\int\limits^{r}_{0}\left(\frac{5}{4}r^{2}dr-\frac{r^{3}}{R}dr\right) 4πρ0=[54r331Rr44]4\pi \rho _{0} = \left[\frac{5}{4} \frac{r^{3}}{3}-\frac{1}{R} \frac{r^{4}}{4}\right] E=kqr2E = \frac{kq}{r^{2}} =14πε01r2.[54(r33)r44R]= \frac{1}{4\pi\varepsilon_{0}} \frac{1}{r^{2}}. \left[\frac{5}{4} \left(\frac{r^{3}}{3}\right)- \frac{r^{4}}{4R}\right] E=ρ0r4ε0[54rR]E = \frac{\rho_{0}r}{4\varepsilon_{0}} \left[\frac{5}{4}-\frac{r}{R}\right]