Total Charge on the Cylinder (Gauss' Law)

[Alouette]

Homework Statement

The cylindrical drum of a copy machine is .38 m long with .05 m radius. If charge is uniformly distributed over the cylindrical surface (the top and bottom flat circular ends of the drum have no charge) and the electric field very close to this surface midway along the length of the drum is 1.0x10^5 N/C, what is the total charge on the cylinder?

Homework Equations

Electric flux $\Phi$ = Q / $\epsilon$o

Electric flux $\Phi$ = E*2*$\pi$*r*L

The Attempt at a Solution

I thought I could use these two equations to solve for Q, just by dividing:

E2πrL*Eo

...but that doesn't give the right answer. I'm unsure if the top and bottom of cylinder having no charge has something to do with it not being this simple. Is this hunch right?

I also have the equation for surface charge density which is E = σ/2εo ,yet I'm unsure if this has much use here...so lost..

 

[SammyS]

Homework Statement

The cylindrical drum of a copy machine is .38 m long with .05 m radius. If charge is uniformly distributed over the cylindrical surface (the top and bottom flat circular ends of the drum have no charge) and the electric field very close to this surface midway along the length of the drum is 1.0x10^5 N/C, what is the total charge on the cylinder?

Homework Equations

Electric flux $\Phi$ = Q / $\epsilon$o

Electric flux $\Phi$ = E*2*$\pi$*r*L

The Attempt at a Solution

I thought I could use these two equations to solve for Q, just by dividing:

E2πrL/Eo

...but that doesn't give the right answer. I'm unsure if the top and bottom of cylinder having no charge has something to do with it not being this simple. Is this hunch right?

I also have the equation for surface charge density which is E = σ/2εo ,yet I'm unsure if this has much use here...so lost..

Check your algebra.

Solving Q/ε₀ = 2π r L E for Q does not give: Q = E2πrL/ε₀ .

 

[Alouette]

Sorry meant E*Eo*2*π*r*L

 

[SammyS]

Sorry meant E*Eo*2*π*r*L

That should give the correct answer.

 

[Alouette]

Duh..thanks a lot! Actually answered, I appreciate it! :D