Electric field of non-conducting cylinder

In summary: True, thanks for spotting. Keyboard always trickier than pen & paper.No need for integrals hereIn summary, the author calculated the electric field within a thin cylinder and found that 106.73 N/C is present.
  • #1
MahalMuh
8
0
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.
 
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  • #2
MahalMuh said:
E = λ / (2πεx^2) = 106.73 N/C.
Are you sure that calculation yields N/C? Check the dimensions.
 
  • #3
haruspex said:
Are you sure that calculation yields N/C? Check the dimensions.

My checking says yes it does but I'm probably wrong.
 
  • #4
MahalMuh said:
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?
 
  • #5
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.
 
  • #6
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.
 
  • #7
MahalMuh said:
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
MahalMuh said:
E = λ / (2πεx^2)
which is wrong, but to get the answer you got you must have used the correct version:
E = λ / (2πεx)
 
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  • #8
haruspex said:
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.
 
  • #9
MahalMuh said:
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.
 

1. What is the electric field of a non-conducting cylinder?

The electric field of a non-conducting cylinder is the force per unit charge that would be experienced by a test charge placed at any point outside the cylinder. It is a vector quantity and is measured in units of Newtons per Coulomb (N/C).

2. How is the electric field calculated for a non-conducting cylinder?

The electric field of a non-conducting cylinder can be calculated using the formula E = kλ/r, where E is the electric field, k is the Coulomb's constant, λ is the linear charge density of the cylinder, and r is the distance from the center of the cylinder to the point where the electric field is being measured.

3. What is the direction of the electric field for a non-conducting cylinder?

The electric field for a non-conducting cylinder is always directed radially outwards from the center of the cylinder. This means that the electric field lines are perpendicular to the surface of the cylinder at all points.

4. How does the electric field of a non-conducting cylinder vary with distance?

The electric field of a non-conducting cylinder decreases with distance from the center of the cylinder. This is because the electric field is inversely proportional to the distance from the source of the electric charge. This means that as the distance increases, the electric field strength decreases.

5. What is the relationship between the electric field and the charge of a non-conducting cylinder?

The electric field of a non-conducting cylinder is directly proportional to the charge of the cylinder. This means that as the charge of the cylinder increases, the electric field strength also increases. Similarly, if the charge of the cylinder decreases, the electric field strength will also decrease.

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