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Chapter 24: Problem 45

A metal sphere of radius \(30 \mathrm{~cm}\) has a net charge of \(3.0 \times 10^{-8}\) C. (a) What is the electric field at the sphere's surface? (b) If \(V=0\) at infinity, what is the electric potential at the sphere's surface? (c) At what distance from the sphere's surface has the electric potential decreased by \(500 \mathrm{~V} ?\)

Short Answer

Expert verified
(a) The electric field is approximately 2997 N/C. (b) The electric potential is 899.9 V. (c) The distance is 0.375 m.

Step by step solution

01

Calculate the Electric Field at the Surface

The electric field at the surface of a charged sphere can be found using the formula for the electric field due to a point charge, since a charged sphere behaves as if all its charge is concentrated at its center. The formula is given by:\[E = \frac{k \cdot Q}{r^2}\]where:- \( E \) is the electric field,- \( k \) is the Coulomb's constant \( (8.99 \times 10^9 \mathrm{~N} \cdot \mathrm{m}^2/\mathrm{C}^2) \),- \( Q \) is the charge \( (3.0 \times 10^{-8} \mathrm{~C}) \),- \( r \) is the radius of the sphere \( (0.30 \mathrm{~m}) \).Substituting these values in:\[E = \frac{(8.99 \times 10^9) \cdot (3.0 \times 10^{-8})}{(0.30)^2} \approx 2.997 \times 10^3 \mathrm{~N/C}\]
02

Calculate the Electric Potential at the Surface

The electric potential \( V \) at the surface of the sphere is calculated using the formula:\[V = \frac{k \cdot Q}{r}\]Using the given values:- \( k = 8.99 \times 10^9 \mathrm{~N} \cdot \mathrm{m}^2/\mathrm{C}^2 \),- \( Q = 3.0 \times 10^{-8} \mathrm{~C} \),- \( r = 0.30 \mathrm{~m} \),Substitute these into the formula:\[V = \frac{(8.99 \times 10^9) \cdot (3.0 \times 10^{-8})}{0.30} \approx 899.9 \mathrm{~V}\]
03

Determine the Distance for a Potential Decrease of 500 V

We need to find the distance from the surface where the potential decreases by 500 V, so:\[\Delta V = 899.9 \mathrm{~V} - 500 \mathrm{~V} = 399.9 \mathrm{~V}\]The potential at a distance \( r + d \) from the center is:\[V' = \frac{k \cdot Q}{r+d}\]Set this equal to 399.9 V:\[\frac{(8.99 \times 10^9) \cdot (3.0 \times 10^{-8})}{r+d} = 399.9\]Solve for \( r+d \):\[\begin{align*}399.9 \times (r + d) &= (8.99 \times 10^9) \cdot (3.0 \times 10^{-8}) \r + d &= \frac{(8.99 \times 10^9) \cdot (3.0 \times 10^{-8})}{399.9} \r + d &\approx 0.675 \mathrm{~m}\end{align*}\]The distance from the surface is \( d \):\[d = r + d - r = 0.675 - 0.30 = 0.375 \mathrm{~m}\]

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Key Concepts

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

Electric Potential
Electric potential is a measure of the potential energy that a unit positive charge would have at a specific point in space due to the presence of electric fields. It is usually denoted by the symbol \( V \), and it is measured in volts (V).
  • Electric potential energy per unit charge is what defines this potential, and it reflects how much work would be required to move a charge to that position from a point with zero potential, such as infinity.
  • The formula for electric potential, especially for a spherical charge distribution, is \( V = \frac{k \cdot Q}{r} \), where \( k \) is Coulomb's constant, \( Q \) is the charge, and \( r \) is the distance from the center of the charge.
For a charged sphere, the electric potential can be seen as the work needed to bring a unit positive charge from infinity to the surface of the sphere. Therefore, at the surface of the sphere, the potential is calculated based on the total charge enclosed within that sphere鈥檚 radius.
Coulomb's Law
Coulomb's Law is fundamental in understanding electric interactions. It describes the force between two charged objects. The law states that the electric force \( F \) between two charges is directly proportional to the product of the magnitude of the two charges and inversely proportional to the square of the distance between their centers. The law is expressed by the equation:
\[ F = k \frac{|q_1 \cdot q_2|}{r^2}\]
  • Here, \( F \) is the electrostatic force, \( q_1 \) and \( q_2 \) are the charges, \( r \) is the distance separating the charges, and \( k \) is the Coulomb's constant \( (8.99 \times 10^9 \mathrm{~N} \cdot \mathrm{m}^2/\mathrm{C}^2) \).
  • This principle underlies the relationship between charges and the resulting electric fields.
By applying Coulomb's Law to a charged sphere, we can deduce the behavior of electric fields around the sphere. The law allows us to predict how charges will interact with one another based on their separation.
Charged Sphere
A charged sphere, in physics, is often treated as if all its charge is concentrated at its center. This simplifies calculations involving electric fields and potentials. For a uniformly charged sphere, the electric field outside the sphere behaves as if the charge were concentrated at a single point at its center.
  • The sphere can be thought of as a point charge when calculating the electric field outside it using the formula \( E = \frac{k \cdot Q}{r^2} \).
  • Inside the sphere, however, different rules apply due to uniform charge distribution.
This approach allows us to use simpler equations normally applicable to point charges, facilitating calculations in electrostatics. Understanding how a charged sphere behaves helps in the practical application of physical laws like Coulomb's Law to determine effects outside the sphere.
Electric Field Equation
The electric field equation is crucial in determining the influence of charges in space. The electric field \( E \) generated by a point charge is calculated using the formula
\[E = \frac{k \cdot Q}{r^2}\]
This formula shows:
  • \( E \) is the electric field directly related to the charge \( Q \) and inversely related to the square of the distance \( r \) from the charge.
  • \( k \) is the proportionality constant, Coulomb's constant \( (8.99 \times 10^9 \mathrm{~N} \cdot \mathrm{m}^2/\mathrm{C}^2) \).
For a charged sphere, even though the charge is distributed over its surface, we can effectively use this formula to find the electric field at any point outside the sphere. Applying this concept helps to clearly display the intensity and direction of the force a charge would experience within the electric field.

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Most popular questions from this chapter

The ammonia molecule \(\mathrm{NH}_{3}\) has a permanent electric dipole moment equal to \(1.47 \mathrm{D}\), where \(1 \mathrm{D}=1\) debye unit \(=\) \(3.34 \times 10^{-30} \mathrm{C} \cdot \mathrm{m}\). Calculate the electric potential due to an ammonia molecule at a point \(103 \mathrm{~nm}\) away along the axis of the dipole. (Set \(V=0\) at infinity.)

A nonuniform linear charge distribution given by \(\lambda=b x\), where \(b\) is a constant, is located along an \(x\) axis from \(x=0\) to \(x=0.20 \mathrm{~m}\). If \(b=15 \mathrm{nC} / \mathrm{m}^{2}\) and \(V=0\) at infinity, what is the electric potential at (a) the origin and (b) the point \(y=0.15 \mathrm{~m}\) on the \(y\) axis?

The electric potential \(V\) in the space between two flat parallel plates 1 and 2 is given (in volts) by \(V=1500 x^{2}\), where \(x\) (in meters) is the perpendicular distance from plate 1 . At \(x=1.8 \mathrm{~cm}\), (a) what is the magnitude of the electric field and (b) is the field directed toward or away from plate \(1 ?\)

An infinite nonconducting sheet with a uniform surface charge density sets up parallel equipotential surfaces. Any pair of surfaces differing by \(25.0 \mathrm{~V}\) are separated by \(8.80 \mathrm{~mm}\). (a) What is the magnitude of the surface charge density? (b) If an electron is released near the sheet, does it tend to move from higher to lower potential or vice versa?

Identical \(50 \mu \mathrm{C}\) charges are fixed on an \(x\) axis at \(x=\pm 2.0 \mathrm{~m} .\) 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?
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