Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 11: Problem 1579

Two charged spheres of radii \(R_{1}\) and \(R_{2}\) having equal surface charge density. The ratio of their potential is ..... (A) \(\left(\mathrm{R}_{2} / \mathrm{R}_{1}\right)\) (B) \(\left(\mathrm{R}_{2} / \mathrm{R}_{1}\right)^{2}\) (C) \(\left(\mathrm{R}_{1} / \mathrm{R}_{2}\right)^{2}\) (D) \(\left(\mathrm{R}_{1} / \mathrm{R}_{2}\right)\)

Short Answer

Expert verified
The ratio of their potential is \(\boxed{\text{(D)}\ \left(\frac{R_{1}}{R_{2}}\right)}\).

Step by step solution

01

Understand surface charge density and electric potential.

Surface charge density is given by the amount of charge per unit area (σ), and electric potential (V) of a charged sphere is the amount of work required to bring a unit positive charge from infinity to the point at a certain distance from the sphere. The relationship between them can be given as \(V = \frac{kQ}{r}\), where k is Coulomb's constant, Q is the total charge on the sphere, and r is the distance from the center of the sphere to the point where electric potential is measured.
02

Write down the given values.

Two charged spheres have equal surface charge densities (σ) and radii R1 and R2, respectively. To calculate the total charge on each sphere, use the formula \(Q = σ × A\), where A is the surface area. Since the spheres have equal surface charge densities, their total charges can be represented as \(Q_{1} = σ × 4πR_{1}^{2}\) and \(Q_{2} = σ × 4πR_{2}^{2}\).
03

Write down the electric potentials for both spheres.

Use the formula \(V = \frac{kQ}{r}\) to write down the electric potentials for both spheres, keeping in mind that the potentials are measured at their surfaces. Therefore, the distances r in the formulas are equal to the radii of the spheres: \(V_{1} = \frac{kQ_{1}}{R_{1}}\) \(V_{2} = \frac{kQ_{2}}{R_{2}}\)
04

Substituting the total charge expressions and finding the ratio of the potentials.

Substitute the total charge expressions from step 2 in the potential formulas: \(V_{1} =\frac{k(σ × 4πR_{1}^{2})}{R_{1}}\) \(V_{2} =\frac{k(σ × 4πR_{2}^{2})}{R_{2}}\) Now, find the ratio of the potentials \(\frac{V_{2}}{V_{1}}\): \(\frac{V_{2}}{V_{1}} = \frac{\frac{k(σ × 4πR_{2}^{2})}{R_{2}}}{\frac{k(σ × 4πR_{1}^{2})}{R_{1}}}\) Simplify the equation and cancel common terms: \(\frac{V_{2}}{V_{1}} = \frac{\frac{(σ × 4πR_{2}^{2})}{R_{2}}}{\frac{(σ × 4πR_{1}^{2})}{R_{1}}} = \frac{R_{1}R_{2}^{2}}{R_{2}R_{1}^{2}}\) \(\frac{V_{2}}{V_{1}} = \frac{R_{1}}{R_{2}}\) The ratio of their potential is: \(\boxed{\text{(D)}\ \left(\frac{R_{1}}{R_{2}}\right)}\)

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

Two uniformaly charged spherical conductors \(\mathrm{A}\) and \(\mathrm{B}\) having radius \(1 \mathrm{~mm}\) and \(2 \mathrm{~mm}\) are separated by a distance of \(5 \mathrm{~cm} .\) If the spheres are connected by a conducting wire then in equilibrium condition, the ratio of the magnitude of the electric fields at the surfaces of spheres \(\mathrm{A}\) and \(\mathrm{B}\) is..... (A) \(4: 1\) (B) \(1: 2\) (C) \(2: 1\) (D) \(1: 4\)

An oil drop of 12 excess electrons is held stationary under a constant electric field of \(2.55 \times 10^{4} \mathrm{Vm}^{-1}\). If the density of the oil is \(1.26 \mathrm{gm} / \mathrm{cm}^{3}\) then the radius of the drop is \(\ldots \ldots \ldots \mathrm{m}\). (A) \(9.75 \times 10^{-7}\) (B) \(9.29 \times 10^{-7}\) (C) \(9.38 \times 10^{-8}\) (D) \(9.34 \times 10^{-8}\)

Three charges \(2 q,-q,-q\) are located at the vertices of an equilateral triangle. At the centre of the triangle. (A) The Field is Zero but Potential is non - zero (B) The Field is non - Zero but Potential is zero (C) Both field and Potential are Zero (D) Both field and Potential are non - Zero

A sphere of radius \(1 \mathrm{~cm}\) has potential of \(8000 \mathrm{v}\), then energy density near its surface will be ...... (A) \(64 \times 10^{5}\left(\mathrm{~J} / \mathrm{m}^{3}\right)\) (B) \(2.83\left(\mathrm{~J} / \mathrm{m}^{3}\right)\) (C) \(8 \times 10^{3}\left(\mathrm{~J} / \mathrm{m}^{3}\right)\) (D) \(32\left(\mathrm{~J} / \mathrm{m}^{3}\right)\)

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) \(\left[(2 \mathrm{Q}) /\left(4 \pi \epsilon_{0} \mathrm{R}\right)\right]\)
See all solutions

Recommended explanations on English Textbooks

Essay Prompts

Read Explanation

Syntax

Read Explanation

Creative Story

Read Explanation

Rhetoric

Read Explanation

Discourse

Read Explanation

Textual Analysis

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