Log out Go to App
Go to App

Learning Materials

Features

Discover

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: \(\vec{E} = \vec{E}_{x} + \vec{E}_{y} = -\dfrac{2kQ}{R}\hat{j}\)

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

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.

A conducting solid sphere of radius \(20.0 \mathrm{~cm}\) is located with its center at the origin of a three-dimensional coordinate system. A charge of \(0.271 \mathrm{nC}\) is placed on the sphere. a) What is the magnitude of the electric field at point \((x, y, z)=\) \((23.1 \mathrm{~cm}, 1.1 \mathrm{~cm}, 0 \mathrm{~cm}) ?\) b) What is the angle of this electric field with the \(x\) -axis at this point? c) What is the magnitude of the electric field at point \((x, y, z)=\) \((4.1 \mathrm{~cm}, 1.1 \mathrm{~cm}, 0 \mathrm{~cm}) ?\)

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).

A solid nonconducting sphere of radius \(a\) has a total charge \(+Q\) uniformly distributed throughout its volume. The surface of the sphere is coated with a very thin (negligible thickness) conducting layer of gold. A total charge of \(-2 Q\) is placed on this conducting layer. Use Gauss's Law to do the following. a) Find the electric field \(E(r)\) for \(ra\) (outside the coated sphere, beyond the sphere and the gold layer). c) Sketch the graph of \(E(r)\) versus \(r\). Comment on the continuity or discontinuity of the electric field, and relate this to the surface charge distribution on the gold layer.

A uniform sphere has a radius \(R\) and a total charge \(+Q,\) uniformly distributed throughout its volume. It is surrounded by a thick spherical shell carrying a total charge \(-Q,\) also uniformly distributed, and having an outer radius of \(2 R\). What is the electric field as a function of \(R ?\)
See all solutions

Recommended explanations on Physics Textbooks

Rotational Dynamics

Read Explanation

Kinematics Physics

Read Explanation

Magnetism

Read Explanation

Solid State Physics

Read Explanation

Radiation

Read Explanation

Fields in Physics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free