/*! 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 73 Charge \(Q=+4.00 \mu \mathrm{C}\... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Charge \(Q=+4.00 \mu \mathrm{C}\) is distributed uniformly over the volume of an insulating sphere that has radius \(R=5.00 \mathrm{cm} .\) What is the potential difference between the center of the sphere and the surface of the sphere?

Short Answer

Expert verified
The potential difference is zero.

Step by step solution

01

Identify Key Information

The problem involves a charge, a sphere, and requires the potential difference between two points. Given data: charge \( Q = +4.00 \, \mu C \) and radius \( R = 5.00 \, \text{cm} \).
02

Recognize Gadget Required

We need to find the potential difference \( V \) between the center and the surface of the sphere. This requires us to find the electric potential \( V \) at both points.
03

Calculate Potential at the Center

Inside a uniformly charged sphere at the center, the potential is given by \( V_{center} = \frac{kQ}{R} \) where \( k \) is Coulomb's constant \( 8.99 \times 10^9 \, \text{N} \cdot \text{m}^2/\text{C}^2 \).
04

Calculate Potential at the Surface

The electric potential at the surface of the sphere is the same as it is outside at the distance \( R \) from the center: \( V_{surface} = \frac{kQ}{R} \).
05

Determine Potential Difference

Potential difference \( \Delta V \) between the center and surface is computed as \( V_{center} - V_{surface} = \frac{kQ}{R} - \frac{kQ}{R} = 0 \).
06

Arrive at Conclusion

Since the potential difference is zero, this implies uniform distribution within a sphere does not generate a difference from center to surface in potential.

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.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Uniform Charge Distribution
In the realm of electromagnetism, a uniform charge distribution means that the charge is spread out evenly over a designated volume or area. In our problem, the charge is distributed uniformly over the volume of the sphere. This means that every tiny part of the sphere has the same amount of charge when compared per unit volume.
In practical terms, if you could divide the sphere into tiny equal pieces, each piece will hold an identical amount of charge relative to its size. This attribute simplifies calculations, as the charge distribution does not vary within the sphere, making it easier to apply laws like Gauss's Law for determining fields and potentials inside and outside the sphere. Just remember, uniformity assumes everything is constant and even, from small internal points to larger sections across the sphere.
Insulating Sphere
An insulating sphere is a sphere made of a material that does not allow electric charges to move freely. Unlike conductors, where charges can redistribute themselves quickly, insulators keep their charges fixed in place. This characteristic is crucial in our problem because it means the charge remains where it is placed within or on the sphere.
The insulating nature prevents any migration of charge to the surface or any change in the charge distribution once it is set. That's why when we talk about the sphere, the charge stays uniformly distributed across its entire volume, maintaining the potential calculations simple and preventing any sudden surprises in potential measurements.
Coulomb's Constant
Coulomb's constant, also known as the electrostatic force constant, is a vital part of understanding electrical interactions. It is denoted by the symbol \( k \) and has a value of approximately \( 8.99 \times 10^9 \, \text{N} \cdot \text{m}^2/\text{C}^2 \). This constant appears in Coulomb's law, which describes the force between two point charges.
Within potential calculations, Coulomb's constant helps to determine the electric potential due to the charges involved. By multiplying this constant with a charge and dividing by a distance (such as a radius), we can easily calculate the electric potential at a particular point in space. This simplification allows us to focus solely on the geometry and charge details specific to the problem, while knowing the constant provides a reliable framework for accurate computation of forces and potentials in electric fields.
Electric Potential Calculation
Electric potential calculation involves determining how much work is needed to move a charge within an electric field. In this problem, we seek the potential difference between the center and the surface of a sphere. When working with uniformly charged insulating spheres, the inside calculations become quite convenient.
To find the potential at a specific point like the center or the surface of a sphere, we use the formula: \( V = \frac{kQ}{r} \), where \( k \) is Coulomb's constant, \( Q \) is the total charge, and \( r \) is the radius we're concerned with. Interestingly, for this particular exercise, the potential difference between the center and surface results in zero. Why, you ask? Because inside a uniformly charged sphere, due to the distribution, potentials at the center and surface are equal. No work is required to move the charge, illustrating the elegant symmetry within electrostatics.

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 particle with charge \(+7.60 \mathrm{nC}\) is in a uniform electric field directed to the left. Another force, in addition to the electric force, acts on the particle so that when it is released from rest, it moves to the right. After it has moved \(8.00 \mathrm{cm},\) the additional force has done \(6.50 \times 10^{-5} \mathrm{J}\) of work and the particle has \(4.35 \times 10^{-5} \mathrm{J}\) of kinetic energy. (a) What work was done by the electric force? (b) What is the potential of the starting point with respect to the end point? (c) What is the magnitude of the electric field?

A small metal sphere, carrying a net charge of \(q_{1}=\) \(-2.80 \mu \mathrm{C},\) is held in a stationary position by insulating supports. A second small metal sphere, with a net charge of \(q_{2}=-7.80 \mu \mathrm{C} \quad\) and \(\operatorname{mass}\) \(1.50 \mathrm{g},\) is projected toward \(q_{1}\). When the two spheres are 0.800 \(\mathrm{m}\) apart, \(q_{2}\) is moving toward \(q_{1}\) with speed 22.0 \(\mathrm{m} / \mathrm{s}\) (Fig. E23.5). Assume that the two spheres can be treated as point charges. You can ignore the force of gravity. (a) What is the speed of \(q_{2}\) when the spheres are 0.400 \(\mathrm{m}\) apart? (b) How close does \(q_{2}\) get to \(q_{1}\) ?

A metal sphere with radius \(R_{1}\) has a charge \(Q_{1}\) . Take the electric potential to be zero at an infinite distance from the sphere. (a) What are the electric field and electric potential at the surface of the sphere? This sphere is now connected by a long, thin conduct- ing wire to another sphere of radius \(R_{2}\) that is several meters from the first sphere. Before the connection is made, this second sphere is uncharged. After electrostatic equilibrium has been reached, what are (b) the total charge on each sphere; (c) the electric potential at the surface of each sphere; (d) the electric field at the surface of each sphere? Assume that the amount of charge on the wire is much less than the charge on each sphere.

Three point charges, which initially are infinitely far apart, are placed at the corners of an equilateral triangle with sides \(d .\) Two of the point charges are identical and have charge \(q .\) If zero net work is required to place the three charges at the corners of the triangle, what must the value of the third charge be?

A very large plastic sheet carries a uniform charge density of \(-6.00 \mathrm{nC} / \mathrm{m}^{2}\) on one face. (a) As you move away from the sheet along a line perpendicular to it, does the potential increase or decrease? How do you know, without doing any calculations? Does your answer depend on where you choose the reference point for potential? (b) Find the spacing between equipotential surfaces that differ from each other by 1.00 V. What type of surfaces are these?

See all solutions

Recommended explanations on Physics Textbooks

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.