Finding surface charge densities and potential (metal cylinder)

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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##
 

Answers and Replies

  • #2
ehild
Homework Helper
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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
 
  • #3
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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##
 
Last edited:
  • #4
ehild
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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
 

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