Question

A particle with a charge of -60.0 nC is placed at the center of a onconducting spherical shell of inner radius 20.0 cn and outer radius 36.0 crn. The spherical shell carries charge with a uniform density of-3.55 HC/m3. A proton moves in a circular orbit just outside the spherical shell. Calculate the speed of the proton. m/s

Answer 1

Charge at the center of spherical shell $q_c=-60.0\,nC$

Charge density on spherical shell $\rho=-3.55\,\mu C/m^3$

Inner radius of shell $a=20.0\,cm$

Outer radius of shell $b=36.0\,cm$

Consider a thin spherical shell of radius $r$ where $a<r<b$ and thickness $dr$ as the element.

Volume of element $dV=4\pi r^2dr$

Charge on element $dQ=\rho dV=4\pi\rho r^2dr$

Total charge on shell $Q=\int dQ=\int \rho dV=\int_a^b 4\pi\rho r^2dr$

$Q=\int_a^b 4\pi\rho r^2dr=4\pi\rho\left.\frac{r^3}{3}\right|_a^b=\frac{4\pi\rho}{3}\left(b^3-a^3 \right )$

Radius of proton's orbit is $b$.

Apply Gauss's law to find the electric field at the location of proton.

$\oint_S\bold{E}\cdot d\bold{S}=\frac{Q_{enc}}{\epsilon_0}$

Consider a Gaussian surface $S$ of radius $r$ where $r >b$. By symmetry magnitude of electric field is uniform on the surface $S$.Also direction of $\bold{E}$ is parallel to $d\bold{S}$. Both are radially outward.

Hence $\oint_S\bold{E}\cdot d\bold{S}=\oint_S E d{S}=E\oint_S d{S}=E(4\pi r^2)$

Charge enclosed $Q_{enc}=q_c&plus;Q$

$E=\frac{Q_{enc}}{4\pi\epsilon_0 r^2}=\frac{q_c&plus;Q}{4\pi\epsilon_0 r^2}$

When proton is just outside the shell $r=b$ and $E=\frac{q_c&plus;Q}{4\pi\epsilon_0 b^2}$

Since proton is moving in a circular orbit, $\frac{mv^2}{b}=e|E|$

$\frac{mv^2}{b}=\frac{e(|q_c|&plus;|Q|)}{4\pi \epsilon_0 b^2}$

$v=\sqrt{\frac{e(|q_c|&plus;|Q|)}{4\pi \epsilon_0 bm}}$

$v=\sqrt{\frac{(1.602*10^{-19})(60.0*10^{-9}&plus;\frac{4\pi}{3}(3.55*10^{-6})\left(36.0^3-20.0^3 \right )(10^{-6}))}{4\pi (8.854*10^{-12}) (36.0*10^{-2})(1.6726*10^{-27})}}$

$v=1.232*10^{6}\,m/s$