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Chapter 23: Problem 81

Two metal spheres of different sizes are charged such that the electric potential is the same at the surface of each. Sphere A has a radius three times that of sphere \(B\) . Let \(Q_{A}\) and \(Q_{B}\) be the charges on the two spheres, and let \(E_{A}\) and \(E_{B}\) be the electric-field magnitudes at the surfaces of the two spheres. What are (a) the ratio \(Q_{B} / Q_{A}\) and (b) the ratio \(E_{B} / E_{A} ?\)

Short Answer

Expert verified
(a) \( \frac{Q_B}{Q_A} = \frac{1}{3} \); (b) \( \frac{E_B}{E_A} = 3 \).

Step by step solution

01

Identify Given Information and Formulas

The problem gives us that Sphere A has a radius three times that of Sphere B, i.e., \( R_A = 3R_B \). We know that the electric potential at the surface of a sphere is given by \( V = \frac{kQ}{R} \), where \( k \) is Coulomb's constant, \( Q \) is the charge, and \( R \) is the radius. As both spheres have the same potential, \( V_A = V_B \), we set up the equation: \( \frac{kQ_A}{R_A} = \frac{kQ_B}{R_B} \).
02

Solve for Charge Ratio

Using the equation for electric potential equality, \( \frac{Q_A}{R_A} = \frac{Q_B}{R_B} \), substitute \( R_A = 3R_B \). This gives \( \frac{Q_A}{3R_B} = \frac{Q_B}{R_B} \). Simplify by canceling \( R_B \): \( \frac{Q_A}{3} = Q_B \). Thus, the charge ratio is \( \frac{Q_B}{Q_A} = \frac{1}{3} \).
03

Solve for Electric Field Magnitude Ratio

The electric field at the surface of a sphere is given by \( E = \frac{kQ}{R^2} \). For Sphere A: \( E_A = \frac{kQ_A}{R_A^2} = \frac{kQ_A}{(3R_B)^2} = \frac{kQ_A}{9R_B^2} \), and for Sphere B: \( E_B = \frac{kQ_B}{R_B^2} \). Substitute \( Q_B = \frac{Q_A}{3} \) into \( E_B = \frac{k(Q_A/3)}{R_B^2} = \frac{kQ_A}{3R_B^2} \). Ratio \( \frac{E_B}{E_A} = \frac{\frac{kQ_A}{3R_B^2}}{\frac{kQ_A}{9R_B^2}} = 3 \).
04

Conclusion

The ratio \( \frac{Q_B}{Q_A} \) is \( \frac{1}{3} \) and the ratio \( \frac{E_B}{E_A} \) is \( 3 \).

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Key Concepts

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

Charged Spheres
Imagine two metal spheres sitting side by side, each holding a particular amount of electric charge. These are known as charged spheres. The amount of charge one sphere holds affects various properties such as electric potential and electric field strength. In this scenario, we have Sphere A and Sphere B. Sphere A is larger, with a radius three times that of Sphere B. Despite their different sizes, they have the same electric potential on their surfaces. This is quite intriguing because, usually, larger objects can hold more charge. This exercise explores precisely the relationship between charge, size, and electric potential on these charged spheres. Given these two distinctively sized spheres with identical potential, an essential concept emerges: the charge distribution. To maintain the same electric potential on the surface, the smaller Sphere B has to have proportionally less charge than the larger Sphere A. This forms the basis for calculating the ratio of charges between the two spheres. Ultimately, Sphere B has only one-third the charge of Sphere A, ensuring their surface potentials remain equal.
Electric Field
The electric field represents the force felt by a charge placed within a given space. In simpler terms, it tells us how strong the force from a charged object will be at a certain point. On the surface of our charged spheres, the electric field can be quite different for each, owing to their size variations.For a sphere, the electric field at its surface is calculated as:
  • For Sphere A, the formula is given by: \( E_A = \frac{kQ_A}{R_A^2} \)
  • For Sphere B, it is \( E_B = \frac{kQ_B}{R_B^2} \)
It is noteworthy that the field strength depends on the amount of charge and the square of the radius. Since Sphere A has more charge but also a larger radius, its field would naturally lose strength over a greater surface distance. Conversely, a smaller sphere with less charge will exert a relatively higher electric field at its surface. Ultimately, this calculation shows Sphere B has an electric field three times stronger than that of Sphere A.
Coulomb's Law
Central to understanding these phenomena is Coulomb's Law. This principle governs the force between charged objects, stating that it depends on the product of their charges and inversely on the square of the distance separating them.Coulomb's Law is expressed as:\[ F = k \frac{Q_1 Q_2}{r^2} \]where:
  • \( F \) is the force between the charges
  • \( Q_1 \) and \( Q_2 \) are the magnitudes of the charges
  • \( r \) is the distance between the centers of the two charges
  • \( k \) is Coulomb's constant, typically \( 8.99 \times 10^9 \ Nm^2/C^2 \)
In the case of charged spheres, although both have their charges and belong in the electric field's domain, Coulomb's Law highlights the inner relationship between the charges. It also shows how the electric fields are formed and influenced by the charge amount and the spatial distances defined by the spheres' radii. By understanding Coulomb's Law, we can better calculate the electric forces around and between these charged spheres, illustrating key behaviors of electric potential and electric fields.

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Most popular questions from this chapter

Four electrons are located at the corners of a square 10.0 \(\mathrm{nm}\) on a side, with an alpha particle at its midpoint. How much work is needed to move the alpha particle to the midpoint of one of the sides of the square?

Three equal \(1.20-\mu \mathrm{C}\) point charges are placed at the corners of an equilateral triangle whose sides are 0.500 \(\mathrm{m}\) long. What is the potential energy of the system? (Take as zero the potential energy of the three charges when they are infinitely far apart.)

(a) An electron is to be accelerated from 3.00 \(\times 10^{6} \mathrm{m} / \mathrm{s}\) to \(8.00 \times 10^{6} \mathrm{m} / \mathrm{s} .\) Through what potential difference must the electron pass to accomplish this? (b) Through what potential difference must the electron pass if it is to be slowed from \(8.00 \times 10^{6} \mathrm{m} / \mathrm{s}\) to a halt?

In a certain region, a charge distribution exists that is spherically symmetric but nonuniform. That is, the volume charge density \(\rho(r)\) depends on the distance \(r\) from the center of the distribution but not on the spherical polar angles \(\theta\) and \(\phi .\) The electric potential \(V(r)\) due to this charge distribution is $$V(r)=\left\\{\begin{array}{l}{\frac{\rho_{0} a^{2}}{18 \epsilon_{0}}\left[1-3\left(\frac{r}{a}\right)^{2}+2\left(\frac{r}{a}\right)^{3}\right] \text { for } r \leq a} \\ {0} & {\text { for } r \geq a}\end{array}\right.$$ where \(\rho_{0}\) is a constant having units of \(\mathrm{C} / \mathrm{m}^{3}\) and \(a\) is a constant having units of meters. (a) Derive expressions for \(\vec{E}\) for the regions \(r \leq a\) and \(r \geq a .[\)Hint\(:\) Use Eq. \((23.23) .]\) Explain why \(\vec{\boldsymbol{E}}\) has only a radial component. (b) Derive an expression for \(\rho(r)\) in each of the two regions \(r \leq a\) and \(r \geq a .[\)Hint t: Use Gauss's law for two spherical shells, one of radius \(r\) and the other of radius \(r+d r .\) The charge contained in the infinitesimal spherical shell of radius \(d r\) is \(d q=4 \pi r^{2} \rho(r) d r . ](c)\) Show that the net charge contained in the volume of a sphere of radius greater than or equal to \(a\) is zero. [Hint: Integrate the expressions derived in part (b) for \(\rho(r)\) over a spherical volume of radius greater than or equal to \(a.\)) Is this result consistent with the electric field for \(r>a\) that you calculated in part (a)?

At a certain distance from a point charge, the potential and electric-field magnitude due to that charge are 4.98 \(\mathrm{V}\) and \(12.0 \mathrm{V} / \mathrm{m},\) respectively. (Take the potential to be zero at infinity.) (a) What is the distance to the point charge? (b) What is the magnitude of the charge? (c) Is the electric field directed toward or away from the point charge?
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