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

Gauss's Law - Cylindrical Symmetry

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 $2\pi R^2+2\pi RL$ .
In cylindrical Gauss's Law problems, the endcaps often have zero electric flux through them, so the surface area of interest is typically $2\pi RL$ 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):

\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}

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 $\lambda$ . The charge enclosed is found by multiplying the length by the linear charge density:

Q_{enc}=\lambda L

Now, we write down Gauss's law:

\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 $\vec E$ and $d \vec A$ is 90 degrees, so the dot product is always zero).

As for the curved part of the cylinder, $\vec E$ and $d \vec A$ 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:

\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:

\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 $\hat r$ is used to show that the direction of the electric field vector is radially symmetric:

\boxed{\vec E=\frac{\lambda }{2\pi r \epsilon_o}\hat r}

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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 $+\lambda$ on the inner cylinder and a linear charge density of $-\lambda$ on the outer shell. The radius of the inner cylinder is $a$ , and the inner and outer radii of the cylindrical shell are $b$ and $c$ , respectively.

Illustration for practice question: Concentric Cylindrical Shells

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Practice: Electric Field of an Insulating Cylindrical Shell

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

Illustration for practice question: Electric Field of an Insulating Cylindrical Shell

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Gauss's Law - Cylindrical Symmetry

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Gauss's Law with Cylindrical Symmetries

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Example: Electric Field from an Infinite Line of Charge

Practice: Concentric Cylindrical Shells

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Practice: Electric Field of an Insulating Cylindrical Shell

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