Question #211534

 A hollow spherical shell carries charge density ρ = k/r2 in the region a < r < b (see Figure). Find the electric field in the three regions : (I) r < a (II) a < r < b, (III) r > b. Plot the magnitude of E~ as a function of r.


1
Expert's answer
2021-07-06T11:50:47-0400

Gives ρ=kr2\rho=\frac{k}{r^2}


r<a

Qr=0(1)Q_r=0\rightarrow(1)

r>b

Qr=QQ_r=Q

Qr=srρdτQ_r=\int _{s_r}\rho d\tau

Qr=4πarρ(s)s2dsQ_r=4 \pi\int_{a}^{r}\rho (s)s^2 ds

Qr=4πarks2s2dsQ_r=4 \pi\int_{a}^{r}\frac{k}{s^2}s^2 ds

Qr=4πk(ra)(2)Q_r=4\pi k(r-a)\rightarrow(2)

a<r<ba<r<b

Similarly

Qr=4πabks2s2dsQ_r=4 \pi\int_{a}^{b}\frac{k}{s^2}s^2 ds

Qr=4πk(ba)(3)Q_r=4\pi k(b-a)\rightarrow(3)

Gauss law


E.da=E.da=4πr2(Er)=Qrϵ0\oint E.da=\oint E.da=4\pi r^2( E_r)=\frac{Q_r}{\epsilon_0}

Er=Qr4πϵ0r2(4)E_r=\frac{Q_r}{4\pi\epsilon_0 r^2}\rightarrow(4)

Put QrQ_r Value

equation (1) and equation (4)

r<ar<a

Er=04πϵ0r2=0E_r=\frac{0}{4\pi\epsilon_0 r^2}=0

equation (2) and (4)


a<r<ba<r<b

Er=4πk(ra)4πϵ0r2E_r=\frac{4\pi k(r-a)}{4\pi\epsilon_0 r^2}

Er=kϵ0(rar2),E_r=\frac{k}{\epsilon_0}(\frac{r-a}{r^2}),

Equation (3) and (4)

r>br>b

Er=Qr4πϵ0r2E_r=\frac{Q_r}{4\pi\epsilon_0 r^2}

Er=4πk(ba)4πϵ0r2E_r=\frac{4\pi k(b-a)}{4\pi\epsilon_0 r^2}

Er=kϵ0(bar2)E_r=\frac{k}{\epsilon_0}(\frac{b-a}{r^2})


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