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Chapter 23: Problem 40

(a) How much excess charge must be placed on a copper sphere \(25.0 \mathrm{~cm}\) in diameter so that the potential of its center is \(3.75 \mathrm{kV} ?\) Take the point where \(V=0\) to be infinitely far from the sphere, (b) What is the potential of the sphere's surface?

Short Answer

Expert verified
The excess charge on the copper sphere is approximately \(5.2 \times 10^{-8} \, C\) and the potential of the sphere's surface is \(3750V\).

Step by step solution

01

Calculation of Excess Charge

Electric potential \( V \) due to a point charge is given by \( V = \frac{kQ}{r} \), where \( Q \) is the charge, \( k \) is the electrostatic constant (\( 8.99 \times 10^9 \, N.m^2/C^2 \)), and \( r \) is the distance from the charge (which is the radius of the sphere in this case). Given the radius \( r = 25cm/2 = 12.5cm = 0.125m \), and \( V = 3750V (3.75kV) \), we can rearrange the formula to solve for \( Q \): \( Q = \frac{V \times r}{k} \).
02

Substitute values into the formula

Substitute the known values \( V = 3750V \), \( r = 0.125m \), and \( k = 8.99 \times 10^9 N.m^2/C^2 \) into the rearranged formula: \( Q = \frac{3750 \times 0.125}{8.99 \times 10^9} \).
03

Calculate the Excess Charge

Calculate the value of \( Q \) to get the excess charge on the sphere: \( Q \approx 5.2 \times 10^{-8} \, C \)
04

Surface Potential Calculations

As the electric potential is equal throughout the sphere, the potential at the surface is same as the potential at the center. Therefore, the surface potential \( V_s = 3750V (3.75kV) \).

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Key Concepts

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

Excess Charge Calculation
Understanding how to calculate excess charge on an object is fundamental in electrostatics. Excess charge refers to the additional amount of electric charge placed on an object, altering its electric potential. In the context of the provided exercise, the goal is to determine how much charge must be added to a copper sphere to achieve a specified electric potential at its center.

The calculation requires us to use the electric potential formula for a point charge, which is simplified for a sphere since we can consider the charge as being at the center. We're given the formula \( V = \frac{kQ}{r} \), where \(V\) is the potential due to the charge \(Q\), \(k\) is the Coulomb's constant, and \(r\) is the radius from the charge to the point of calculation鈥攊n this case, the radius of the sphere. By rearranging the formula to solve for \(Q\), we can input the values of \(V\) and \(r\) provided in the problem statement to arrive at the excess charge necessary.

It's important to bear in mind that the result obtained from this calculation is approximate and assumes that the distribution of excess charge is uniform over the surface of the sphere, a common approximation in such problems.
Electrostatics
Electrostatics is the study of electric charges at rest. It inherently involves the analysis of forces, fields, and potentials resulting from static charges. A key concept within electrostatics is the electrostatic force, described by Coulomb's law, which quantifies the force between two point charges.

Within the exercise, electrostatic principles are applied to compute the potential due to a stationary charge distribution on a solid sphere. The idea that electric potential is uniform across a conductor in electrostatic equilibrium, which allows us to state that the potential at the surface of the sphere is the same as that at its center, is a critical concept from electrostatics.

The behavior of electric fields and potentials around conductors is dictated by the materials' ability to allow for free movement of charges until an equilibrium state is achieved. In the case of our copper sphere, free electrons will redistribute in response to the excess charge until the electric potential is equalized across the entire surface.
Electric Potential Formula
The electric potential formula is a mathematical representation of the electric potential energy per unit charge at a given point in space due to an electric field. The potential \(V\) at a distance \(r\) from a point charge \(Q\) is given by the formula \( V = \frac{kQ}{r} \), where \(k\) is the Coulomb's constant (approximately \(8.99 \times 10^9 \, N\cdot m^2/C^2\)).

This fundamental formula is integral to the solution of the exercise. It encapsulates the inverse relationship between the electric potential and the distance from the charge, as well as the direct relationship with the amount of charge. By rearranging and substituting the desired potential and the radius of the sphere into the formula, students can find the required excess charge.

For a solid conducting sphere, the formula simplifies because the entire charge is assumed to be at the center for purposes of calculation, making it easier to compute the potential at any point outside (or on the surface of) the sphere. In practice, manipulating this fundamental formula is essential for problem-solving in electrostatics.

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Most popular questions from this chapter

A hollow, thin-walled insulating cylinder of radius \(R\) and length \(L\). (like the cardboard tube in a roll of toilet paper) has charge \(Q\) uniformly distributed over its surface. (a) Calculate the electric potential at all points along the axis of the tube. Take the origin to be at the center of the tube, and take the potential to be zero at infinity. (b) Show that if \(L

An alpha particle with kinetic energy \(9.50 \mathrm{MeV}\) (when far away) collides head-on with a lead nucleus at rest. What is the distance of closest approach of the two particles? (Assume that the lead nucleus remains stationary and may be treated as a point charge. The atomic number of lead is \(82 .\) The alpha particle is a helium nucleus, with atomic number \(2 .\) )

A helium nucleus, also known as an \(\alpha\) (alpha) particle, consists of two protons and two neutrons and has a diameter of \(10^{-15} \mathrm{~m}\) \(=1 \mathrm{fm} .\) The protons, with a charge of te, are subject to a repulsive Coulomb force. since the neutrons have zero charge, there must be an attractive force that counteracts the electric repulsion and keeps the protons from flying apart. This so-called strong force plays a central role in particle physics. (a) As a crude model, assume that an \(\alpha\) particle consists of two pointlike protons attracted by a Hooke's-law spring with spring constant \(k,\) and ignore the neutrons. Assume further that in the absence of other forces, the spring has an equilibrium separation of zero. Write an expression for the potential energy when the protons are separated by distance \(d\). (b) Minimize this potential to find the equilibrium separation \(d_{0}\) in terms of \(e\) and \(k .\) (c) If \(d_{0}=1.00 \mathrm{fm}\), what is the value of \(k ?\) (d) How much energy is stored in this system, in terms of electron volts? (e) A proton has a mass of \(1.67 \times 10^{-27} \mathrm{~kg}\). If the spring were to break, the \(\alpha\) particle would disintegrate and the protons would fly off in opposite directions. What would be their ultimate speed?

A thin insulating rod is bent into a semicircular arc of radius \(a,\) and a total electric charge \(Q\) is distributed uniformly along the rod. Calculate the potential at the center of curvature of the are if the potential is assumed to be zero at infinity.

A heart cell can be modeled as a cylindrical shell that is \(100 \mu \mathrm{m}\) long, with an outer diameter of \(20.0 \mu \mathrm{m}\) and a cell wall thickness of \(1.00 \mu \mathrm{m}\), as shown in Fig. \(\mathrm{P} 23.81 .\) Potassium ions move across the cell wall, depositing positive charge on the outer surface and leaving a net negative charge on the inner surface. During the so-called resting phase, the inside of the cell has a potential that is \(90.0 \mathrm{mV}\) lower than the potential on its outer surface. (a) If the net charge of the cell is zero, what is the magnitude of the total charge on either cell wall membrane? Ignore edge effects and treat the cell as a very long cylinder. (b) What is the magnitude of the electric field just inside the cell wall? (c) In a subsequent depolarization event, sodium ions move through channels in the cell wall, so that the inner membrane becomes positively charged. At the cnd of this event, the inside of the cell has a potential that is \(20.0 \mathrm{mV}\) higher than the potential outside the cell. If we model this event by charge moving from the outer membrane to the inner membrane, what magnitude of charge moves across the cell wall during this event? (d) If this were done entirely by the motion of sodium ions, \(\mathrm{Na}^{+}\), how many ions have moved?
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