Potential of Concentric Cylindrical Insulator and Conducting Shell

[dancingmonkey]

An infinitely long solid insulating cylinder of radius a = 5.3 cm is positioned with its symmetry axis along the z-axis as shown. The cylinder is uniformly charged with a charge density ρ = 45 μC/m3. Concentric with the cylinder is a cylindrical conducting shell of inner radius b = 14.2 cm, and outer radius c = 16.2 cm. The conducting shell has a linear charge density λ = -0.42μC/m.

1)What is Ey(R), the y-component of the electric field at point R, located a distance d = 44 cm from the origin along the y-axis as shown?

2)What is V(P) – V(R), the potential difference between points P and R? Point P is located at (x,y) = (44 cm, 44 cm).



E = Qenc/$\epsilon$
E = [2k($\lambda$(cylinder) + $\lambda$(shell))]/r

$\lambda$c = $\rho$*A
= $\rho$*2$\pi$*0.053*0.044
= 7.0*10^-6 C/m

E = (2*(8.99*10^9)*(7.0*10^6+(-4.2*10^-7)))/0.44m
= 268883 N/C

But that answer is not correct. Please help me! I would appreciate detailed explanation or work, thank you!

 

[diazona]

But that answer is not correct. Please help me! I would appreciate detailed explanation or work, thank you!

Surely you don't mean you want someone here to do the work for you?

Anyway, I see several errors in your calculation:

E = Qenc/$\epsilon$

That formula is not correct, as you could see if you check the units

E = [2k($\lambda$(cylinder) + $\lambda$(shell))]/r

What's k? If you're using $k = \frac{1}{4\pi\epsilon_0}$, then I don't see how you get that formula.

$\lambda$c = $\rho$*A

That line seems right (assuming you meant the c to be a subscript, not a variable on its own)

= $\rho$*2$\pi$*0.053*0.044

but that line is not. Hint: what's the area of a circle? (What does the variable A physically represent?)

 

[ehild]

$\lambda$c = $\rho$*A
= $\rho$*2$\pi$*0.053*0.044
= 7.0*10^-6 C/m

I think A is supposed to be the cross section of the inner cylinder, and ρ its radius. So why ρ*2π? and what is 0.044?

ehild