/*! 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 28 A total clectric charge of \(3.5... [FREE SOLUTION] | 91Ó°ÊÓ

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A total clectric charge of \(3.50 \mathrm{nC}\) is distributed uniformly over the surface of a metal sphere with a radius of \(24.0 \mathrm{~cm}\). If the potential is zero at a point at infinity, find the value of the potential at the following distances from the center of the sphere: (a) \(48.0 \mathrm{~cm}\) (b) \(24.0 \mathrm{~cm}\) (c) \(12.0 \mathrm{~cm}\)

Short Answer

Expert verified
The potential at 48.0 cm is approximately 64.8 V, at 24.0 cm it is approximately 129.6 V and at 12.0 cm it is the same as at the surface of the sphere, approximately 129.6 V.

Step by step solution

01

Calculate the total charge

Given, Total electric charge, Q = \(3.50 \, nC\) = \(3.50 \times 10^{-9} \, C\). This is already provided in the question so we don't need to do any calculations here.
02

Formulate the potential formula

The potential (V) at a distance (r) from the center due to a sphere with total charge Q is given by \( V = \frac{KQ}{r} \) where K = \( 8.99 × 10^{9} \, Nm^2/C^2 \) is Coulomb's constant.
03

Calculate the potential at 48.0 cm

For a distance of 48.0 cm = \(0.48 \, m\), substitute the charge (Q) and this distance (r) into the formula to find the potential \( V = \frac{8.99 × 10^{9} \times 3.50 × 10^{-9}}{0.48} \)
04

Calculate the potential at 24.0 cm

Now switch to 24.0 cm = \(0.24 \, m\), substitute this distance into the formula and calculate the potential \( V = \frac{8.99 × 10^{9} \times 3.50 × 10^{-9}}{0.24} \)
05

Calculate the potential within the sphere

For distance of 12.0 cm = \(0.12 \, m\) which is inside the sphere, potential should be same as it is on the surface (24 cm point) due to it being a conductor, which implies potential is same at every point. So, no calculation is needed for this step. Potential will be same as it for 24 cm.

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Key Concepts

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

Electric Charge
Electric charge is a fundamental property of matter that exhibits electrostatic attraction or repulsion in the presence of other charges. It comes in discrete quantities and is typically measured in coulombs (C). Charges can be positive or negative, and the interaction between them follows certain physical laws. Understanding the distribution of electric charge is crucial for a variety of applications, and this is particularly pertinent when the charge is distributed over objects like spheres. In our exercise, a total charge of 3.50 nanocoulombs (nC), or \(3.50 \times 10^{-9} C\), is uniformly spread across the surface of a metal sphere, a common scenario in electrostatics.
Coulomb's Law
Coulomb's Law is the cornerstone of electrostatics, describing how the electrostatic force between two point charges is directly proportional to the product of the magnitude of the charges and inversely proportional to the square of the distance between them. Mathematically, the law is stated as \( F = k \frac{|q_1 q_2|}{r^2} \), where \( F \) is the force, \( q_1 \) and \( q_2 \) are the charges, \( r \) is the distance between the charges, and \( k \) is Coulomb's constant (approximately \( 8.99 \times 10^{9} \, \text{Nm}^2/\text{C}^2 \)). While our textbook problem concerns electric potential, not force, Coulomb's Law is the underlying principle that governs the interaction between charges, thus influencing the potential.
Electric Potential Calculation
Electric potential, denoted as \( V \), at a point in space is the amount of electric potential energy that a unit charge placed at that point would have. It's vital for understanding the work done by an electric field in moving a charge from one point to another. For a sphere with a uniformly distributed surface charge, the potential \( V \) at a distance \( r \) from the center can be calculated using the formula \( V = \frac{KQ}{r} \), where \( K \) is Coulomb's constant, \( Q \) is the total charge and \( r \) is the radial distance from the center. This formula simplifies the process of determining the potential at different distances from the sphere, as demonstrated in our textbook exercise for distances of 48 cm, 24 cm, and 12 cm.
Conductor Properties
Conductors are materials that allow the free movement of electric charge, which in metals, is due to the free electrons. One of the key properties of conductors in electrostatic equilibrium is that the electric potential is constant throughout the conductor. This is why, in our problem, the electric potential inside the sphere is the same as on its surface. When a conductor is charged, the excess charge resides entirely on its surface, contributing to the uniform potential within. This concept of equal potential across a conductor's volume is why no calculation was needed when the sphere was probed at a point inside its boundary.

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Most popular questions from this chapter

In a certain region of space, the electric potential is \(V(x, y, z)=A x y-B x^{2}+C y,\) where \(A, B,\) and \(C\) are positive constants. (a) Calculate the \(x-y^{-},\) and \(z\) -components of the electric field. (b) At which points is the electric field equal to zero?

A very long insulating cylindrical shell of radius \(6.00 \mathrm{~cm}\) carries charge of linear density \(8.50 \mu \mathrm{C} / \mathrm{m}\) spread uniformly over its outer surface. What would a voltmeter read if it were connected between (a) the surface of the cylinder and a point \(4.00 \mathrm{~cm}\) above the surface, and (b) the surface and a point \(1.00 \mathrm{~cm}\) from the central axis of the cylinder?

A small particle has charge \(-5.00 \mu\) C and mass \(2.00 \times 10^{-4} \mathrm{~kg} .\) It moves from point \(A,\) where the electric potential is \(V_{A}=+200 \mathrm{~V},\) to point \(B,\) where the electric potential is \(V_{B}=+800 \mathrm{~V}\) The electric force is the only force acting on the particle. The particle has speed \(5.00 \mathrm{~m} / \mathrm{s}\) at point \(A .\) What is its speed at point \(B ?\) Is it moving faster or slower at \(B\) than at \(A ?\) Explain.

A very long insulating cylinder of charge of radius \(2.50 \mathrm{~cm}\) carries a uniform linear density of \(15.0 \mathrm{nC} / \mathrm{m}\). If you put one probe of a voltmeter at the surface, how far from the surface must the other probe be placed so that the voltmeter reads \(175 \mathrm{~V} ?\)

A Geiger counter detects radiation such as alpha particles by using the fact that the radiation ionizes the air along its path. A thin wire lies on the axis of a hollow metal cylinder and is insulated from it (Fig. \(\mathrm{P} 23.62\) ). A large potential difference is established between the wire and the outer cylinder, with the wire at higher potential; this sets up a strong electric field directed radially outward. When ionizing radiation enters the device, it ionizes a few air molecules. The free electrons produced are accelerated by the electric field toward the wire and, on the way there, ionize many more air molecules. Thus a current pulse is produced that can be detected by appropriate electronic circuitry and converted to an audible "click." Suppose the radius of the central wire is \(145 \mu \mathrm{m}\) and the radius of the hollow cylinder is \(1.80 \mathrm{~cm}\). What potential difference between the wire and the cylinder produces an electric field of \(2.00 \times 10^{4} \mathrm{~V} / \mathrm{m}\) at a distance of \(1.20 \mathrm{~cm}\) from the axis of the wire? (The wire and cylinder are both very long in comparison to their radii, so the results of Problem 23.61 apply.)

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