Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 11: Problem 1636

A thin spherical shell of radius \(R\) has charge \(Q\) spread uniformly over its surface. Which of the following graphs, figure most closely represents the electric field \(\mathrm{E}\) (r) produced by the shell in the range $0 \leq \mathrm{r}<\infty\(, where \)\mathrm{r}$ is the distance from the centre of the shel1.

Short Answer

Expert verified
The electric field E(r) for a thin spherical shell of radius R with charge Q uniformly spread over its surface is given by: \[E(r) = \begin{cases} 0 & \text{for } r < R \\ \frac{Q}{4 \pi \epsilon_0 r^2} & \text{for } r \geq R \end{cases}\] The graph representing this function should show E(r) as zero for \(r<R\) and follow an inverse square relationship (\(\frac{1}{r^2}\)) as \(r\geq R\).

Step by step solution

01

Gauss's Law states that the electric flux through a closed surface is equal to the enclosed charge divided by the permittivity constant, which can be mathematically written as: \[\oint \vec{E} \cdot d\vec{A} = \frac{Q_{enclosed}}{\epsilon_0}\] To apply Gauss's Law, we need to choose an appropriate Gaussian surface. Since we are dealing with a sphere, the Gaussian surface will be a sphere concentric with the given thin shell and with radius r. #Step 2: Determine the enclosed charge inside the Gaussian surface#

We can determine the enclosed charge based on the distance r from the center of the shell: - If r<R: The Gaussian surface lies completely inside the shell, and therefore, there is no charge enclosed. \(Q_{enclosed}=0\) - If r≥R: The Gaussian surface entirely encloses the shell, so the enclosed charge is equal to the total charge on the shell, Q. #Step 3: Calculate the electric field based on the enclosed charge#
02

Now, we can use Gauss's Law to calculate the electric field: - If r

Based on our calculations in Steps 3 and 4, we can write the final electric field function as: \[E(r) = \begin{cases} 0 & \text{for } r < R \\ \frac{Q}{4 \pi \epsilon_0 r^2} & \text{for } r \geq R \end{cases}\] Now we can compare this function to the given graphs to choose the most representative one. The graph should show E(r) as zero while r<R and follow an inverse square relationship (1/r^2) as r≥R.

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 Semicircular rod is charged uniformly with a total charge \(\mathrm{Q}\) coulomb. The electric field intensity at the centre of curvature is $\ldots \ldots \ldots$ (A) \(\left[(2 \mathrm{KQ}) /\left(\pi \mathrm{R}^{2}\right)\right]\) (B) \(\left[(3 \mathrm{KQ}) /\left(\pi \mathrm{R}^{2}\right)\right]\) (C) \(\left[(\mathrm{KQ}) /\left(\pi \mathrm{R}^{2}\right)\right]\) (D) \(\left[(4 \mathrm{KQ}) /\left(\pi \mathrm{R}^{2}\right)\right]\)

A charged particle of mass \(\mathrm{m}\) and charge \(q\) is released from rest in a uniform electric field \(E\). Neglecting the effect of gravity, the kinetic energy of the charged particle after 't' second is ...... (A) $\left[\left(\mathrm{Eq}^{2} \mathrm{~m}\right) /\left(2 \mathrm{t}^{2}\right)\right]$ (B) $\left[\left(\mathrm{E}^{2} \mathrm{q}^{2} \mathrm{t}^{2}\right) /(2 \mathrm{~m})\right]$ (C) \(\left[\left(2 \mathrm{E}^{2} \mathrm{t}^{2}\right) /(\mathrm{qm})\right]\) (D) \([(\mathrm{Eqm}) / \mathrm{t}]\)

Two equal charges \(q\) are placed at a distance of \(2 \mathrm{a}\) and a third charge \(-2 q\) is placed at the midpoint. The potential energy of the system is ....... (A) $\left[\left(9 \mathrm{q}^{2}\right) /\left(8 \pi \epsilon_{0} \mathrm{a}\right)\right]$ (B) \(\left[q^{2} /\left(8 \pi \epsilon_{0} a\right)\right]\) (C) \(-\left[\left(7 q^{2}\right) /\left(8 \pi \epsilon_{0} a\right)\right]\) (D) \(\left[\left(6 q^{2}\right) /\left(8 \pi \in_{0} a\right)\right]\)

Equal charges \(q\) are placed at the vertices \(A\) and \(B\) of an equilateral triangle \(\mathrm{ABC}\) of side \(\mathrm{a}\). The magnitude of electric field at the point \(c\) is \(\ldots \ldots \ldots\) (A) \(\left(\mathrm{Kq} / \mathrm{a}^{2}\right)\) (B) \(\left.(\sqrt{3} \mathrm{Kq}) / \mathrm{a}^{2}\right)\) (C) \(\left.(\sqrt{2} \mathrm{Kq}) / \mathrm{a}^{2}\right)\) (D) $\left[\mathrm{q} /\left(2 \pi \mathrm{t} \varepsilon_{0} \mathrm{a}^{2}\right)\right]$

Two metal plate form a parallel plate capacitor. The distance between the plates is \(\mathrm{d}\). A metal sheet of thickness \(\mathrm{d} / 2\) and of the same area is introduced between the plates. What is the ratio of the capacitance in the two cases? (A) \(4: 1\) (B) \(3: 1\) (C) \(2: 1\) (D) \(5: 1\)
See all solutions

Recommended explanations on English Textbooks

Language Acquisition

Read Explanation

Summary Text

Read Explanation

Listening and Speaking

Read Explanation

Essay Writing Skills

Read Explanation

5 Paragraph Essay

Read Explanation

Essay Prompts

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