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Chapter 23: Problem 42

Derive an expression for electric potential along the axis (the \(x\) -axis) of a disk with a hole in the center, as shown in the figure, where \(R_{1}\) and \(R_{2}\) are the inner and outer radii of the disk. What would the potential be if \(R_{1}=0 ?\)

Short Answer

Expert verified
Question: Determine the electric potential along the x-axis of a disk with a hole in the center with inner and outer radii R1 and R2 respectively. Answer: The electric potential along the x-axis of a disk with a hole in the center is given by: \(V(x)= \frac{\sigma}{4\epsilon_0}\left[R_2\sqrt{R_2^2+x^2}-R_1\sqrt{R_1^2+x^2}\right]\) When R1 = 0, the electric potential is: \(V(x) = \frac{\sigma R_2}{4\epsilon_0} \sqrt{R_2^2+x^2}\)

Step by step solution

01

Identify the given information and visualize the problem

We have a disk with a hole, where R1 and R2 are the inner and outer radii. The x-axis is running along the axis of the disk. We want to find the electric potential along the x-axis.
02

Define the variables

Let \(V(x)\) be the electric potential along the x-axis, \(\sigma\) be the surface charge density, \(\epsilon_0\) be the vacuum permittivity, and let \(dx\) be a distance element along the x-axis. The disk can be broken down into infinitesimally small rings. Let \(r\) be the distance from the center of the ring to the x-axis and \(dr\) be the thickness of the ring.
03

Define the differential ring

Consider a differential ring at a distance r from the center and with thickness dr. The area of the differential ring, dA, can be described as: \(dA = 2 \pi r dr\)
04

Calculate the electric potential contribution from each differential ring

The electric potential contributed by the differential ring, dV, at a point on the x-axis, can be calculated using the formula for electric potential due to a surface charge: \(dV = \frac{1}{4 \pi \epsilon_0} \frac{\sigma dA}{\sqrt{x^2 + r^2}}\) Substituting the expression for dA, we have: \(dV = \frac{1}{4 \pi \epsilon_0} \frac{2 \pi \sigma r dr}{\sqrt{x^2 + r^2}}\) Simplifying, \(dV = \frac{\sigma r dr}{2 \epsilon_0 \sqrt{x^2 + r^2}}\)
05

Integrate over the entire disk to get the electric potential

To find the total electric potential along the x-axis, integrate the expression for dV from R1 to R2: \(V(x) = \int_{R_1}^{R_2} \frac{\sigma r dr}{2 \epsilon_0 \sqrt{x^2 + r^2}}\) This integral can be solved via a substitution method or by carefully reviewing a table of integrals. The result yields: \(V(x)= \frac{\sigma}{4\epsilon_0}\left[R_2\sqrt{R_2^2+x^2}-R_1\sqrt{R_1^2+x^2}\right]\)
06

Find the potential when R1 = 0

Now, we need to determine the electric potential when R1 = 0: \(V(x) = \frac{\sigma}{4\epsilon_0}\left[R_2\sqrt{R_2^2+x^2}-0\right]\) Simplifying, we have: \(V(x) = \frac{\sigma R_2}{4\epsilon_0} \sqrt{R_2^2+x^2}\) Therefore, the potential along the x-axis of a disk with a hole in the center is given by: \(V(x)= \frac{\sigma}{4\epsilon_0}\left[R_2\sqrt{R_2^2+x^2}-R_1\sqrt{R_1^2+x^2}\right]\) And when R1 = 0, the electric potential is: \(V(x) = \frac{\sigma R_2}{4\epsilon_0} \sqrt{R_2^2+x^2}\)

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