/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 33 A thin glass rod is bent into a ... [FREE SOLUTION] | 91影视

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Chapter 22: Problem 33

A thin glass rod is bent into a semicircle of radius \(R\). A charge \(+Q\) is uniformly distributed along the upper half, and a charge \(-Q\) is uniformly distributed along the lower half as shown in the figure. Find the magnitude and direction of the electric field \(\vec{E}\) (in component form) at point \(P\), the center of the semicircle.

Short Answer

Expert verified
Answer: The electric field at the center of the semicircle is \(\vec{E} = -\dfrac{2kQ}{R}\hat{j}\), where \(k\) is the Coulomb constant, \(Q\) is the total charge on each half of the semicircle, and \(R\) is the radius of the semicircle.

Step by step solution

01

Define a small charge element on the semicircle

Let's define a small charge element \(dq\) at an angle \(\theta\) on the semicircle. For the positive charges on the upper half, the angle \(\theta\) goes from \(0\) to \(\pi\) radians. For the negative charges on the lower half, the angle \(\theta\) goes from \(\pi\) to \(2\pi\) radians.
02

Determine the electric field contribution due to a small charge element

The electric field contribution from a small charge element \(dq\) is given by \(d\vec{E} = k \dfrac{dq}{r^2}\hat{r}\), where \(k\) is the Coulomb constant, \(r = R\) (distance from charge element \(dq\) to point \(P\)), and \(\hat{r}\) is the unit vector pointing from \(dq\) to \(P\).
03

Determine the components of the electric field due to the small charge element

To determine the \(X\) and \(Y\) components of the electric field, let's express \(d\vec{E}\) in component form: \(d\vec{E}_x = d\vec{E}\cos\theta = k \dfrac{dq}{R^2}\cos\theta\hat{i}\) \(d\vec{E}_y = d\vec{E}\sin\theta = k \dfrac{dq}{R^2}\sin\theta\hat{j}\)
04

Integrate the electric field components over the entire semicircle

Now we need to integrate these components over the entire semicircle. We begin with the positive charges on the upper half \((0 \leq \theta \leq \pi)\): \(\vec{E}_{+x} = \int_{0}^{\pi} k \dfrac{dq}{R^2}\cos\theta\hat{i}\) \(\vec{E}_{+y} = \int_{0}^{\pi} k \dfrac{dq}{R^2}\sin\theta\hat{j}\) Then, we perform the same integration for the negative charges on the lower half \((\pi \leq \theta \leq 2\pi)\): \(\vec{E}_{-x} = \int_{\pi}^{2\pi} k \dfrac{dq}{R^2}\cos\theta\hat{i}\) \(\vec{E}_{-y} = \int_{\pi}^{2\pi} k \dfrac{dq}{R^2}\sin\theta\hat{j}\)
05

Add the contributions from the positive and negative charges

To find the total electric field at point \(P\), we add the contributions from the positive and negative charges: \(\vec{E}_{x} = \vec{E}_{+x} + \vec{E}_{-x}\) \(\vec{E}_{y} = \vec{E}_{+y} + \vec{E}_{-y}\)
06

Find the final expression for the electric field

After completing the integration, we find that the electric field components are: \(\vec{E}_{x} = 0\hat{i}\) \(\vec{E}_{y} = -\dfrac{2kQ}{R}\hat{j}\) Thus, the total electric field at point \(P\) is: \(\vec{E} = \vec{E}_{x} + \vec{E}_{y} = -\dfrac{2kQ}{R}\hat{j}\)

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Most popular questions from this chapter

A solid nonconducting sphere has a volume charge distribution given by \(\rho(r)=(\beta / r) \sin (\pi r / 2 R) .\) Find the total charge contained in the spherical volume and the electric field in the regions \(rR\). Show that the two expressions for the electric field equal each other at \(r=R\).

A thin, flat washer is a disk with an outer diameter of \(10.0 \mathrm{~cm}\) and a hole in the center with a diameter of \(4.00 \mathrm{~cm} .\) The washer has a uniform charge distribution and a total charge of \(7.00 \mathrm{nC}\). What is the electric field on the axis of the washer at a distance of \(30.0 \mathrm{~cm}\) from the center of the washer?

A hollow conducting spherical shell has an inner radius of \(8.00 \mathrm{~cm}\) and an outer radius of \(10.0 \mathrm{~cm} .\) The electric field at the inner surface of the shell, \(E_{\mathrm{i}}\), has a magnitude of \(80.0 \mathrm{~N} / \mathrm{C}\) and points toward the center of the sphere, and the electric field at the outer surface, \(E_{\infty}\) has a magnitude of \(80.0 \mathrm{~N} / \mathrm{C}\) and points away from the center of the sphere (see the figure). Determine the magnitude of the charge on the inner surface and the outer surface of the spherical shell.

Two infinite, uniformly charged, flat nonconducting surfaces are mutually perpendicular. One of the surfaces has a charge distribution of \(+30.0 \mathrm{pC} / \mathrm{m}^{2}\), and the other has a charge distribution of \(-40.0 \mathrm{pC} / \mathrm{m}^{2}\). What is the magnitude of the electric field at any point not on either surface?

A long conducting wire with charge distribution \(\lambda\) and radius \(r\) produces an electric field of \(2.73 \mathrm{~N} / \mathrm{C}\) just outside the surface of the wire. What is the magnitude of the electric field just outside the surface of another wire with charge distribution \(0.81 \lambda\) and radius \(6.5 r ?\)
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