Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 11: Problem 1602

A thin spherical conducting shell of radius \(\mathrm{R}\) has a charge q. Another charge \(Q\) is placed at the centre of the shell. The electrostatic potential at a point p a distance \((\mathrm{R} / 2)\) from the centre of the shell is ..... (A) \(\left[(q+Q) /\left(4 \pi \epsilon_{0}\right)\right](2 / R)\) (B) $\left[\left\\{(2 Q) /\left(4 \pi \epsilon_{0} R\right)\right\\}-\left\\{(2 Q) /\left(4 \pi \epsilon_{0} R\right)\right]\right.$ (C) $\left[\left\\{(2 Q) /\left(4 \pi \in_{0} R\right)\right\\}+\left\\{q /\left(4 \pi \epsilon_{0} R\right)\right]\right.$ (D) \(\left[(2 \mathrm{Q}) /\left(4 \pi \epsilon_{0} \mathrm{R}\right)\right]\)

Short Answer

Expert verified
The short answer to the problem is: \(V = \frac{2(q+Q)}{4\pi\epsilon_{0}R}\)

Step by step solution

01

Finding potential due to charge q

: To determine the electrostatic potential due to charge q at point p, we need to find the distance between point p and the surface of the shell. Since point p is at a distance R/2 from the center of the shell, and the shell has a radius of R, the distance between point p and the surface of the shell (r) is R/2. Calculate the potential as follows: \(V_{q} = \frac{q}{4\pi\epsilon_{0}(R/2)} = \frac{2q}{4\pi\epsilon_{0}R}\)
02

Finding potential due to charge Q

: Next, find the electrostatic potential due to charge Q at point p. Since point p is at a distance R/2 from the center of the shell, the distance between charge Q and point p is also R/2. Calculate the potential as follows: \(V_{Q} = \frac{Q}{4\pi\epsilon_{0}(R/2)} = \frac{2Q}{4\pi\epsilon_{0}R}\)
03

Finding total potential at point p

: Finally, add the potentials due to charges q and Q to find the total potential at point p: \(V = V_{q}+V_{Q} = \frac{2q}{4\pi\epsilon_{0}R} + \frac{2Q}{4\pi\epsilon_{0}R}\) Combine the terms: \(V = \frac{2(q+Q)}{4\pi\epsilon_{0}R}\) Comparing this expression to the given options, we can see that it matches with option (A): \(\left[(q+Q) /\left(4 \pi \epsilon_{0}\right)\right](2 / R)\)

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

An inclined plane making an angle of \(30^{\circ}\) with the horizontal is placed in an uniform electric field \(E=100 \mathrm{Vm}^{-1}\). A particle of mass \(1 \mathrm{~kg}\) and charge \(0.01 \mathrm{c}\) is allowed to slide down from rest from a height of \(1 \mathrm{~m} .\) If the coefficient of friction is \(0.2\) the time taken by the particle to reach the bottom is $\ldots \ldots . .$ sec (A) \(2.337\) (B) \(4.337\) (C) 5 (D) \(1.337\)

At what angle \(\theta\) a point \(P\) must be located from dipole axis so that the electric field intensity at the point is perpendicular to the dipole axis? (A) \(\tan ^{-1}(1 / \sqrt{2})\) (B) \(\tan ^{-1}(1 / 2)\) (C) \(\tan ^{-1}(2)\) (C) \(\tan ^{-1}(\sqrt{2})\)

A ball of mass \(1 \mathrm{gm}\) and charge \(10^{-8} \mathrm{c}\) moves from a point \(\mathrm{A}\), where the potential is 600 volt to the point \(B\) where the potential is zero. Velocity of the ball of the point \(\mathrm{B}\) is $20 \mathrm{~cm} / \mathrm{s}\(. The velocity of the ball at the point \)\mathrm{A}$ will be \(\ldots \ldots\) (A) \(16.8(\mathrm{~m} / \mathrm{s})\) (B) \(22.8(\mathrm{~cm} / \mathrm{s})\) (C) \(228(\mathrm{~cm} / \mathrm{s})\) (D) \(168(\mathrm{~m} / \mathrm{s})\)

A small conducting sphere of radius \(r\) is lying concentrically inside a bigger hollow conducting sphere of radius \(R\). The bigger and smaller sphere are charged with \(\mathrm{Q}\) and \(\mathrm{q}(\mathrm{Q}>\mathrm{q})\) and are insulated from each other. The potential difference between the spheres will be \(\ldots \ldots\) (A) \(\left[1 /\left(4 \pi \epsilon_{0}\right)\right][(q / r)-(Q / R)]\) (B) \(\left[1 /\left(4 \pi \epsilon_{0}\right)\right][(q / r)-(q / R)]\) (C) \(\left[1 /\left(4 \pi \epsilon_{0}\right)\right][(Q / R)+(q / r)]\) (D) $\left[1 /\left(4 \pi \epsilon_{0}\right)\right][(\mathrm{q} / \mathrm{R})-(\mathrm{Q} / \mathrm{r})]$

Two positive point charges of \(12 \mu \mathrm{c}\) and \(8 \mu \mathrm{c}\) are placed \(10 \mathrm{~cm}\) apart in air. The work done to bring them $4 \mathrm{~cm}$ closer is (A) Zero (B) \(4.8 \mathrm{~J}\) (C) \(3.5 \mathrm{~J}\) (D) \(-5.8 \mathrm{~J}\)
See all solutions

Recommended explanations on English Textbooks

Listening and Speaking

Read Explanation

5 Paragraph Essay

Read Explanation

Key Concepts in Language and Linguistics

Read Explanation

Text Comparison

Read Explanation

Language Acquisition

Read Explanation

Essay Writing Skills

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