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Wize University Physics Textbook (Master) > Electrostatic Forces and Electric Fields

Gauss's Law - Cylindrical Symmetry

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Gauss's Law with Cylindrical Symmetries


Cylindrical symmetry allows us to use Gauss's Law, although we cannot forget that cylinders have three faces!
  • Gauss's Law problems have cylindrical symmetry if the charge distribution depends only on the distance from the axis.
  • Examples: line of charge, cylinder, cylindrical cavities
  • The Gaussian surface to use is a cylinder. The total surface area of a cylinder of radius R and length L is 2πR2+2πRL2\pi R^2+2\pi RL.
  • In cylindrical Gauss's Law problems, the endcaps often have zero electric flux through them, so the surface area of interest is typically 2πRL2\pi RL representing only the curved face.
  • In the Gauss's Law integral, the electric field and area vectors are always parallel for the curved face, and are perpendicular for the flat face (giving zero flux contribution):
∮E⃗⋅dA⃗=2∫end capE⃗⋅dA⃗+∫curvedE⃗⋅dA⃗=2(0)+∣E⃗∣∫curveddA=∣E⃗∣2πRL\begin{aligned} &\oint \vec E \cdot d\vec A \\ &=2\int_{end~cap} \vec E \cdot d\vec A + \int_{curved} \vec E \cdot d\vec A \\ &=2(0) + |\vec E| \int_{curved} dA \\ &= |\vec E|2\pi RL \end{aligned}


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Example: Electric Field from an Infinite Line of Charge


Find the electric field at distance r away from an infinite wire with uniform linear charge density λ.


We know that Gauss's law is only useful for planar, spherical, or cylindrical symmetries. For the case of a line of charge, we can draw a Gaussian cylinder that is concentric with the line of charge.

We will draw the cylinder with a radius of r, assuming that we are trying to find the electric field at a distance r from the line of charge. It will have a length of L.

The electric field points radially outward from the line of charge. (This is drawn below as if it was a positive line of charge, but the same concept applies if the charge is negative.)

Finding the enclosed charge: our cylinder covers length L, and the linear charge density of the distribution is λ\lambda. The charge enclosed is found by multiplying the length by the linear charge density:
Qenc=λLQ_{enc}=\lambda L
Now, we write down Gauss's law:
∮E⃗⋅dA⃗=Qencϵo∮E⃗⋅dA⃗=λLϵo\begin{aligned} \oint \vec{E}\cdot d\vec A&=\frac{Q_{enc}}{\epsilon_o}\\ \oint \vec{E}\cdot d\vec A&=\frac{\lambda L}{\epsilon_o}\\ \end{aligned}
Now we solve the integral. Remember that Gauss's law involves closed surfaces, so the Gaussian surface also includes the two flat sides of the cylinder. However, the electric flux through these surfaces is zero (the angle between E⃗\vec E and dA⃗d \vec A is 90 degrees, so the dot product is always zero).

As for the curved part of the cylinder, E⃗\vec E and dA⃗d \vec A are parallel, so the dot product simplifies to a simple multiplication. Since the electric field is constant at all parts of the Gaussian surface, we can pull it out of the integral:
∮E⃗⋅dA⃗=λLϵo2∫S, flatE⃗⋅dA⃗+∫S, curvedE⃗⋅dA⃗=λLϵo2∫flat(0)⋅dA⃗+∫S∣E⃗∣dAcos⁡(0o)=λLϵo2⋅(0)+∣E⃗∣∫SdA=λLϵo\begin{aligned} \oint \vec{E}\cdot d\vec A&=\frac{\lambda L}{\epsilon_o}\\ 2\int_{S,~flat} \vec{E}\cdot d\vec A+\int_{S,~curved}\vec{E}\cdot d\vec A&=\frac{\lambda L}{\epsilon_o}\\ 2\int_{flat} (0)\cdot d\vec A+\int_{S}|\vec{E}|dA\cos(0^o)&=\frac{\lambda L}{\epsilon_o}\\ 2\cdot (0)+|\vec E|\int_{S}dA&=\frac{\lambda L}{\epsilon_o}\\ \end{aligned}
Now, all we need to know to solve this integral is the surface area of a cylinder. The round part of a cylinder has surface area of circumference times length:
∣E⃗∣∫SdA=λLϵo∣E⃗∣(2πrL)=λLϵo∣E⃗∣=λL2πrLϵo∣E⃗∣=λ2πrϵo\begin{aligned} |\vec E|\int_{S}dA&=\frac{\lambda L}{\epsilon_o}\\ |\vec E|(2\pi r L)&=\frac{\lambda L}{\epsilon_o}\\ |\vec E|&=\frac{\lambda L}{2\pi r L\epsilon_o}\\ |\vec E|&=\frac{\lambda }{2\pi r \epsilon_o}\\ \end{aligned}
There we have it - the magnitude of the electric field at a distance of r from a line of charge. The direction points radially inward or outward, depending on if the line of charge is positive or negative.

Sometimes the unit vector r^\hat ris used to show that the direction of the electric field vector is radially symmetric:
E⃗=λ2πrϵor^\boxed{\vec E=\frac{\lambda }{2\pi r \epsilon_o}\hat r}


checklist
Mark Yourself Question
  1. Grab a piece of paper and try this problem yourself.
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Practice: Concentric Cylindrical Shells


Find the electric field for each region for this infinitely long metallic cylinder with a linear charge density of +λ+\lambda on the inner cylinder and a linear charge density of −λ-\lambda on the outer shell. The radius of the inner cylinder is aa, and the inner and outer radii of the cylindrical shell are bb and cc, respectively.



checklist
Mark Yourself Question
  1. Grab a piece of paper and try this problem yourself.
  2. When you're done, check the "I have answered this question" box below.
  3. View the solution and report whether you got it right or wrong.

Practice: Electric Field of an Insulating Cylindrical Shell


A long, hollow plastic cylinder has an inner radius of aa, an outer radius of bb, and a uniform charge density of ρ\rho. Find the electric field at all points in the charge distribution.