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Two metal spheres, each of radius \(3.0 \mathrm{~cm}\), have a center-to-center separation of \(2.0 \mathrm{~m}\). Sphere 1 has charge \(+1.0 \times 10^{-8} \mathrm{C} ;\) sphere 2 has charge \(-3.0 \times 10^{-8} \mathrm{C}\). Assume that the separation is large enough for us to say that the charge on each sphere is uniformly distributed (the spheres do not affect each other). With \(V=0\) at infinity, calculate (a) the potential at the point halfway between the centers and the potential on the surface of (b) sphere 1 and (c) sphere 2 .

Short Answer

Expert verified
(a) -179.8 V, (b) 299.7 V, (c) -899.1 V.

Step by step solution

01

Understand the Problem

We have two charged spheres with given charges and separation distances. We need to calculate the electric potential at specific points using the given information. The electric potential due to a point charge is given by the formula \( V = \frac{kQ}{r} \), where \( k \) is Coulomb's constant \( 8.99 \times 10^9 \, \text{Nm}^2/ ext{C}^2 \), \( Q \) is the charge, and \( r \) is the distance from the charge.
02

Calculate Potential at Midpoint Between Spheres

First, we find the potential at a point halfway between the two spheres, which is \( 1.0 \, \text{m} \) from both centers. The potential at this point, due to both spheres, is the sum of the potentials from each:\[ V = V_1 + V_2 = \frac{kQ_1}{r_1} + \frac{kQ_2}{r_2} \]Substituting the known values:\[ V = \frac{8.99 \times 10^9 \times 1.0 \times 10^{-8}}{1.0} + \frac{8.99 \times 10^9 \times (-3.0 \times 10^{-8})}{1.0} \]Calculating each term gives:\[ V = 89.9 - 269.7 = -179.8 \, \text{V} \]
03

Potential on the Surface of Sphere 1

Next, calculate the potential on the surface of sphere 1 due to its own charge.\[ V = \frac{kQ_1}{3.0 \times 10^{-2}} \]Substituting the values:\[ V = \frac{8.99 \times 10^9 \times 1.0 \times 10^{-8}}{3.0 \times 10^{-2}} \]\[ V = 299.7 \, \text{V} \]
04

Potential on the Surface of Sphere 2

Finally, calculate the potential on the surface of sphere 2 due to its own charge.\[ V = \frac{kQ_2}{3.0 \times 10^{-2}} \]Substituting the values:\[ V = \frac{8.99 \times 10^9 \times (-3.0 \times 10^{-8})}{3.0 \times 10^{-2}} \]\[ V = -899.1 \, \text{V} \]

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Key Concepts

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

Charge Distribution
In electrostatics, charge distribution refers to how electric charge is arranged in space on or within an object. When we talk about uniformly distributed charge, it means the charge is spread out evenly over the entire surface of the object, like a metal sphere in this case.

When two metal spheres are described as having uniformly distributed charges, it means that whatever measurement of charge they carry is spread evenly around their surfaces. This simplification allows us to treat each sphere as if all its charge were concentrated at its center, which is vital for calculating the electric potential from a distance. This assumption is particularly handy with spherical objects because of their symmetry.

This uniform charge distribution is essential when applying Coulomb's law to find the potential at various points in the vicinity of the spheres. By considering the charge to be concentrated at the center of each sphere, we simplify the calculations significantly without losing accuracy for points outside the sphere.
Coulomb's Law
Coulomb's law is a fundamental principle used to describe the interaction between two charged objects. It states that the electric force between two charges is directly proportional to the product of the magnitudes of charges and inversely proportional to the square of the distance between them.

Mathematically, this is expressed as:\[F = k \frac{|Q_1 Q_2|}{r^2}\]where:
  • \( F \) is the magnitude of the force between the charges.
  • \( k \) is Coulomb's constant, approximately \( 8.99 \times 10^9 \ \text{Nm}^2/\text{C}^2 \).
  • \( Q_1 \) and \( Q_2 \) are the quantities of the charges.
  • \( r \) is the distance between the centers of the two charges.
While Coulomb's law is primarily focused on calculating forces, it is also inherently connected to finding potentials. By integrating this law, we arrive at the expression for electric potential due to a point charge, which is what we use in our step-by-step analysis to find the potential at different points: \( V = \frac{kQ}{r} \). This relationship gives insight into how energy is spread in the field around a charge.
Electric Field
An electric field is a region around a charged object where other charges experience a force. It is a vector field and is defined as the force per unit positive charge. The direction of the electric field is the direction of the force it would exert on a positive test charge.

Electric fields are vital in understanding how charges influence each other from a distance and are closely linked to Coulomb's law. While Coulomb’s law provides the force between two specific charges, the electric field allows us to study the effect of charge distribution over space, even when no other specific charges are involved.

For a point charge, the electric field \( E \) can be calculated from the equation:\[E = \frac{kQ}{r^2}\]When dealing with multiple charges, like in the case of two spheres, the net electric field at a point is a vector sum of the fields due to each charge, considering both their magnitudes and directions. Understanding electric fields helps answer questions about the influence of the charged spheres at specific points, like at the midpoint between them, by considering how the fields overlap and superimpose on one another.

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Most popular questions from this chapter

Consider a particle with charge \(q=1.50 \times 10^{-8} \mathrm{C},\) and take \(V=0\) at infinity. (a) What are the shape and dimensions of an equipotential surface having a potential of \(30.0 \mathrm{~V}\) due to \(q\) alone? (b) Are surfaces whose potentials differ by a constant amount \((1.0 \mathrm{~V},\) say \()\) evenly spaced?

Identical \(50 \mu \mathrm{C}\) charges are fixed on an \(x\) axis at \(x=\pm 3.0 \mathrm{~m} .\) A particle of charge \(q=-15 \mu \mathrm{C}\) is then released from rest at a point on the positive part of the \(y\) axis. Due to the symmetry of the situation, the particle moves along the \(y\) axis and has kinetic energy \(1.2 \mathrm{~J}\) as it passes through the point \(x=0, y=4.0 \mathrm{~m}\) (a) What is the kinetic energy of the particle as it passes through the origin? (b) At what negative value of \(y\) will the particle momentarily stop?

Two charged, parallel, flat conducting surfaces are spaced \(d=1.00 \mathrm{~cm}\) apart and produce a potential difference \(\Delta V=625 \mathrm{~V}\) between them. An electron is projected from one surface directly toward the second. What is the initial speed of the electron if it stops just at the second surface?

What is the magnitude of the electric field at the point \((3.00 \hat{\mathrm{i}}-2.00 \mathrm{j}+4.00 \mathrm{k}) \mathrm{m}\) if the electric potential in the region is given by \(V=2.00 x y z^{2},\) where \(V\) is in volts and coordinates \(x, y,\) and \(z\) are in meters?

A charge of \(1.50 \times 10^{-8} \mathrm{C}\) lies on an isolated metal sphere of radius \(16.0 \mathrm{~cm} .\) With \(V=0\) at infinity, what is the electric potential at points on the sphere's surface?

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