Electric field of non-conducting cylinder

[MahalMuh]

Homework Statement

We have a very long and non-conducting cylinder with radius r=0.021 m and even charge distribution with volume density ρ=0.18 μC/m3.

a) What is the electric field outside the cylinder with x=0.042 m from the axis of cylinder?
b) What is the electric field inside the cylinder with y=0.0087 m from the axis of cylinder?

Relevant Equations

Q=λl = ρAl = lρπr^2 (charge caused) (1)
E = λ / (2πεx^2) (electric field of cylinder) (2)
ε = 8.8542*10^-12 C^2/(Nm^2)

a) I have calculated (1) λ = ρA = ρπr^2 = 2.49 * 10^-10 C/m and placed it into (2) yielding E = λ / (2πεx^2) = 106.73 N/C.

This doesn't seem to be correct by the feedback, however.

b) Here just to consider the proportion of the cylinder mass constrained by y.

 

[haruspex]

E = λ / (2πεx^2) = 106.73 N/C.

Are you sure that calculation yields N/C? Check the dimensions.

 

[MahalMuh]

Are you sure that calculation yields N/C? Check the dimensions.

My checking says yes it does but I'm probably wrong.

 

[haruspex]

My checking says yes it does but I'm probably wrong.

$[\rho]=QL^{-3}$
$[r^2]=L^2$
$[x^{-2}]=L^{-2}$
$[\frac1{\epsilon_0}]=ML^3T^{-2}Q^{-2}$
From which I get $MT^{-2}Q^{-1}$, not $MLT^{-2}Q^{-1}$.

Where do you think you might have lost a dimension L?

 

[Delta2]

Your mistake is in No.2 of the equations in the list of relevant equations. Check again your theory and how we do apply Gauss's law in integral form to determine the electric field in this system.

 

[MahalMuh]

Actually I had it right, just had it rounded wrong. No need for integrals here. In b) the proportion was opposite / complement of what my intuition said.

 

[haruspex]

Actually I had it right, just had it rounded wrong. No need for integrals here. In b) the proportion was opposite / complement of what my intuition said.

You fooled us both by misquoting an equation. You wrote

E = λ / (2πεx^2)

which is wrong, but to get the answer you got you must have used the correct version:
E = λ / (2πεx)

 

[MahalMuh]

You fooled us both by misquoting an equation. You wrote

which is wrong, but to get the answer you got you must have used the correct version:
E = λ / (2πεx)

True, thanks for spotting. Keyboard always trickier than pen & paper.

 

[Delta2]

No need for integrals here

No actually we don't do an integral because the E-field has azimuthal and z symmetry (cylindrical symmetry in one word), however we apply the integral form of Gauss's law to correctly calculate the E-field, inside and outside of the thin cylinder.