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

 

[HalfLostAndSearching]

$V=-\frac{Q}{2L}(\ln(b)-\ln(a)+\ln(R))$
$V=\frac{Q}{2L}\ln\left(\frac{Rb}{a}\right)$

I would like to first clarify that the solution provided is correct and follows the appropriate mathematical equations. Additionally, I would like to point out that the surface charge density at each point is dependent on the charge and dimensions of the cylinder, as well as the distance from the center. This means that the values of $\sigma$ may change if any of these variables are altered.

Furthermore, it is important to note that the potential at the center of the cylinder is determined by the reference point chosen, in this case being $r=b$. If a different reference point were chosen, the potential may have a different value.

Overall, this problem highlights the relationship between charge, surface charge density, and potential in a metal cylinder surrounded by a coaxial shell. It is a good exercise in applying mathematical equations and understanding the concept of electric potential.