Finding surface charge densities and potential (metal cylinder)

[ghostfolk]

Homework Statement

A solid metal cylinder of radius $R$ and length $L$, carrying a charge $Q$, is surrounded by a thick coaxial metal shell of inner radius $a$ and outer radius $b$. The shell carries no net charge.
a) Find the surface charge desnities $\sigma$ at $R$, at $a$, and at $b$.
b) Find the potential at the center using $r=b$ as the reference point.

Homework Equations

$\sigma=q/A$, $A_{cyl}=2\pi rh$, $V(r)=-\int_{r_1}^{r_2} \frac{\sigma}{r} da$

The Attempt at a Solution

a) $\sigma_R=\frac{Q}{2\pi RL}$ $\sigma_a=\frac{Q}{2\pi aL}$ $\sigma_b=\frac{Q}{2\pi bL}$

b) $V=-\int_b^a \frac{\sigma_b}{r}da-\int_a^R \frac{\sigma_a}{r}da-\int_R^0 \frac{\sigma_R}{r}da$
$da=2\pi RLdr$

 

[ehild]

Homework Statement

A solid metal cylinder of radius $R$ and length $L$, carrying a charge $Q$, is surrounded by a thick coaxial metal shell of inner radius $a$ and outer radius $b$. The shell carries no net charge.
a) Find the surface charge desnities $\sigma$ at $R$, at $a$, and at $b$.
b) Find the potential at the center using $r=b$ as the reference point.

Homework Equations

$\sigma=q/A$, $A_{cyl}=2\pi rh$, $V(r)=-\int_{r_1}^{r_2} \frac{\sigma}{r} da$

The last equation is not correct. How are the potential function V(r) and the electric field E(r) related ?

You should integrate with respect to the variable the integrand depends on. If you integrate a function depending on r , write it as $\int f(r)dr$ instead of $\int f(r) da$. If you use fixed boundaries in the integral for r the result can not depend on r .
Moreover, a means the inner radius of the shell, do not use it as an integrating variable.

The Attempt at a Solution

a) $\sigma_R=\frac{Q}{2\pi RL}$ $\sigma_a=\frac{Q}{2\pi aL}$ $\sigma_b=\frac{Q}{2\pi bL}$

b) $V=-\int_b^a \frac{\sigma_b}{r}da-\int_a^R \frac{\sigma_a}{r}da-\int_R^0 \frac{\sigma_R}{r}da$
$da=2\pi RLdr$

The surface charge densities are all right, but the sign of the surface charge density at a is not correct.
I do not understand what you tried to say with line b). Does the potential change inside a conductor?

ehild

 

[ghostfolk]

The last equation is not correct. How are the potential function V(r) and the electric field E(r) related ?
You should integrate with respect to the variable the integrand depends on. If you integrate a function depending on r , write it as $\int f(r)dr$ instead of $\int f(r) da$. If you use fixed boundaries in the integral for r the result can not depend on r .
Moreover, a means the inner radius of the shell, do not use it as an integrating variable.

I should've made it made it more clear that $da$ is the normal vector of the surface area of the cylinder so that $d\vec{a}=2 \pi RLdr\hat{r}$

The surface charge densities are all right, but the sign of the surface charge density at a is not correct.
I do not understand what you tried to say with line b). Does the potential change inside a conductor?

Yeah I forgot the minus sign. Also the potential does not change in the conductor so I should remove the integral that goes from 0 to $R$

 

[ehild]

I should've made it made it more clear that $da$ is the normal vector of the surface area of the cylinder so that $d\vec{a}=2 \pi RLdr\hat{r}$

Note that the potential is line integral of the negative electric field instead of a surface integral.

Yeah I forgot the minus sign. Also the potential does not change in the conductor so I should remove the integral that goes from 0 to $R$

That is right, but it is zero also somewhere else.

ehild