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Chapter 22: Problem 35

A uniformly charged rod of length \(L\) with total charge \(Q\) lies along the \(y\) -axis, from \(y=0\) to \(y=L\). Find an expression for the electric field at the point \((d, 0)\) (that is, the point at \(x=d\) on the \(x\) -axis).

Short Answer

Expert verified
The electric field at point (d, 0) due to the uniformly charged rod is given by: $$\vec{E} = \frac{\lambda d}{4\pi\epsilon_0} \bigg(\frac{1}{d} - \frac{1}{\sqrt{d^2 + L^2}}\bigg) \hat{x}$$ where $\lambda$ is the charge density, $L$ is the length of the rod, and $\epsilon_0$ is the vacuum permittivity.

Step by step solution

01

Begin by calculating the charge density

Since the rod is uniformly charged, we can find the charge density using the following formula: $$\lambda = \frac{Q}{L}$$ where \(\lambda\) is the charge density, \(Q\) is the total charge, and \(L\) is the length of the rod.
02

Find the electric field due to \(\delta q\)

Now we'll consider a small element \(\delta y\) of the rod at distance \(y\) from the reference point and having a charge \(\delta q = \lambda \delta y\). The electric field due to this small element \(\delta q\) at point \((d, 0)\) can be calculated using Coulomb's law: $$\delta \vec{E} = \frac{1}{4\pi\epsilon_0} \frac{\delta q}{r^2} \hat{r}$$ where \(\epsilon_0\) is the vacuum permittivity, \(r\) is the distance between point \((d, 0)\) and the charge element \(\delta q\), and \(\hat{r}\) is the unit vector from the charge element to the point.
03

Calculate the distance and the unit vector

To calculate \(r\) and \(\hat{r}\), we first need to find the coordinates of the charge element: \((0, y)\). The distance between the point \((0, y)\) and point \((d, 0)\) is: $$r = \sqrt{d^2 + y^2}$$ And the unit vector \(\hat{r}\) can be found as: $$\hat{r} = \frac{\vec{r}}{r} = \frac{d\hat{x} + y\hat{y}}{\sqrt{d^2 + y^2}}$$
04

Plug the values into the electric field expression

Substituting the values of \(r\) and \(\hat{r}\) in the expression for the electric field due to the charge element \(\delta q\), we get: $$\delta \vec{E} = \frac{1}{4\pi\epsilon_0} \frac{\lambda \delta y}{d^2 + y^2}(\frac{d\hat{x} + y\hat{y}}{\sqrt{d^2 + y^2}})$$
05

Integrate the electric field expression along the rod

Now we'll integrate this electric field expression along the length of the rod to find the total electric field at point \((d, 0)\). We need to integrate individually for the components along the \(\hat{x}\) and \(\hat{y}\) directions: $$\vec{E} = \int_0^L \delta \vec{E} = \frac{\lambda}{4\pi\epsilon_0} \int_0^L \frac{d\hat{x} + y\hat{y}}{(d^2 + y^2)^{3/2}} \delta y$$ Separate the components and integrate: $$E_x = \frac{\lambda}{4\pi\epsilon_0} \int_0^L \frac{d}{(d^2 + y^2)^{3/2}} \delta y$$ $$E_y = \frac{\lambda}{4\pi\epsilon_0} \int_0^L \frac{y}{(d^2 + y^2)^{3/2}} \delta y$$
06

Solve the integrals

Solve the integration for both components: $$E_x = \frac{\lambda d}{4\pi\epsilon_0} \bigg[-\frac{1}{\sqrt{d^2 + y^2}}\bigg]_0^L = \frac{\lambda d}{4\pi\epsilon_0} \bigg(\frac{1}{d} - \frac{1}{\sqrt{d^2 + L^2}}\bigg)$$ $$E_y = 0$$ (Symmetry along the \(y\)-axis)
07

Write the final expression for the electric field

The electric field at point \((d, 0)\) can now be written as: $$\vec{E} = E_x\hat{x} + E_y\hat{y} = \frac{\lambda d}{4\pi\epsilon_0} \bigg(\frac{1}{d} - \frac{1}{\sqrt{d^2 + L^2}}\bigg) \hat{x}$$ This is the electric field due to the uniformly charged rod at point \((d, 0)\).

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