/*! 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 46 Two metal spheres of different s... [FREE SOLUTION] | 91Ó°ÊÓ

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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) \( \frac{1}{3} \); (b) \( 3 \).

Step by step solution

01

Understand the Problem

Sphere \( A \) has a radius \( R_A \) and sphere \( B \) has a radius \( R_B \). We are given that \( R_A = 3R_B \). Both spheres have the same electric potential at their surfaces.
02

Analyze Electric Potential Concept

The electric potential \( V \) at the surface of a charged sphere is given by \( V = \frac{kQ}{R} \), where \( k \) is Coulomb's constant. Since both spheres have the same potential \( V_A = V_B \), we have \( \frac{kQ_A}{R_A} = \frac{kQ_B}{R_B} \).
03

Solve for Charge Ratio

Using the relationship from the potentials, \( \frac{Q_A}{R_A} = \frac{Q_B}{R_B} \), substitute \( R_A = 3R_B \) to get \( \frac{Q_A}{3R_B} = \frac{Q_B}{R_B} \). Solving this gives \( Q_B = \frac{Q_A}{3} \). The charge ratio is \( \frac{Q_B}{Q_A} = \frac{1}{3} \).
04

Analyze Electric Field Concept

The electric field \( E \) at the surface of a charged sphere is given by \( E = \frac{kQ}{R^2} \). To find the ratio \( \frac{E_B}{E_A} \), substitute \( R_A = 3R_B \) and the expression for \( Q_B \).
05

Solve for Electric Field Ratio

Using \( E_A = \frac{kQ_A}{R_A^2} \) and \( E_B = \frac{kQ_B}{R_B^2} \), substitute \( Q_B = \frac{Q_A}{3} \), \( E_A = \frac{kQ_A}{(3R_B)^2} = \frac{kQ_A}{9R_B^2} \) and \( E_B = \frac{k(Q_A/3)}{R_B^2} \). Thus, \( \frac{E_B}{E_A} = \frac{\frac{k(Q_A/3)}{R_B^2}}{\frac{kQ_A}{9R_B^2}} = \frac{1}{3} \times 9 = 3 \).
06

Conclusion

The charge ratio \( \frac{Q_B}{Q_A} \) is \( \frac{1}{3} \), and the electric field 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.

Electric Field
The electric field, denoted as \( E \), is an essential concept in understanding electrostatic phenomena. It refers to the force per unit charge experienced by a small test charge placed within a field around a charged object. The formula to compute the electric field at a point is given by:
  • \( E = \frac{kQ}{R^2} \)
where \( k \) is Coulomb's constant, \( Q \) is the charge, and \( R \) is the distance from the charge to the point of interest. The electric field represents how strongly a charge influences the space around it. This can help us understand how forces between charged objects operate.

At the surface of a sphere, the electric field acts radially outward if the sphere is positively charged. This field is crucial in determining the interactions between charged objects and their surrounding space. As the electric field magnitude changes with the sphere's radius, a smaller radius results in a higher field if the charge is unchanged. As shown in the solved problem, the ratio of the electric field magnitudes was found by adjusting the formula for each sphere's parameters.
Coulomb's Law
Coulomb's Law governs the force of interaction between two point charges. It is vital for calculating electrostatic forces in physical interactions. The law states:
  • The force \( F \) between two point charges is proportional to the product of the charges and inversely proportional to the square of the distance between them.
  • The mathematical representation is \( F = k \frac{|Q_1 Q_2|}{r^2} \), where \( Q_1 \) and \( Q_2 \) are the magnitudes of the charges, \( r \) is the distance between the charges, and \( k \) is Coulomb's constant.
Coulomb's Law provides insights into the forces that charged objects exert on each other. It helps to predict whether the charges will repel or attract, depending on whether they are like charges or opposite charges. This law is foundational in deducing formulas for the electric potential and electric field around charged objects. When applying Coulomb's Law to charged spheres, you must consider only the component of the force that acts radially along the line joining the centers of the spheres.
Charged Spheres
Charged spheres are a fundamental study in electrostatics due to their symmetry that simplifies many calculations. The behavior of electric fields and potentials around a sphere helps in understanding how charges distribute over surfaces.
  • When a sphere is charged, the charge uniformly spreads over the surface, especially for conductive spheres.
  • The electric potential \( V \) at the surface of a sphere is constant and defined by \( V = \frac{kQ}{R} \), where \( R \) is the sphere's radius.
In the exercise, the sphere's behavior highlights how electric potential is consistent at all points on the surface for a given charge. Despite different sizes, if the potential is the same, the charge must be adjusted accordingly to maintain that constancy. This also means the electric field, which relies on both the potential and the radius, changes as size varies. By analyzing charged spheres, we gain insights into distribution and interaction effects of charges in various configurations.

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Most popular questions from this chapter

A very long cylinder of radius \(2.00 \mathrm{~cm}\) carries a uniform charge density of \(1.50 \mathrm{nC} / \mathrm{m}\). (a) Describe the shape of the equipotential surfaces for this cylinder. (b) Taking the reference level for the zero of potential to be the surface of the cylinder, having potentials of \(10 \mathrm{~V}, 20 \mathrm{~V}\), and \(30 \mathrm{~V}\). are the equipotential surfaces equally spaced? If not, do they get closer together or farther apart as potential increases?

* 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 metal sphere with radius \(R_{1}\) has a charge \(Q_{1} .\) Take the electric potential to be zero at an infinite distance from the sphere. (a) What are the electric field and electric potential at the surface of the sphere? This sphere is now connected by a long, thin conducting wire to another sphere of radius \(R_{2}\) that is several meters from the first sphere. Before the connection is made, this second sphere is uncharged. After electrostatic equilibrium has been reached, what are (b) the total charge on each sphere; (c) the electric potential at the surface of each sphere; (d) the electric field at the surface of each sphere? Assume that the amount of charge on the wire is much less than the charge on each sphere.

* CP A small sphere with mass \(1.50 \mathrm{~g}\) hangs by a thread between two parallel vertical plates \(5.00 \mathrm{~cm}\) apart (Fig. P3.38). The plates are insulating and have uniform surface charge densities \(+\sigma\) and \(-\sigma\). The charge on the sphere is \(q=8.70 \times 10^{-6} \mathrm{C}\). What potential difference between the plates will cause the thread to assume an angle of \(30.0^{\circ}\) with the vertical?

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 $$ V(r)=\left\\{\begin{array}{l} \frac{\rho_{0} a^{2}}{18 \epsilon_{0}}\left[1-3\left(\frac{r}{a}\right)^{2}+2\left(\frac{r}{a}\right)^{3}\right] \text { for } r \leq a \\ 0 \quad \text { for } r \geq a \end{array}\right. $$ 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 \(\overrightarrow{\boldsymbol{E}}\) for the regions \(r \leq a\) and \(r \geq\) a. [Hint: Use Eq. (23.23).] Explain why \(\vec{E}\) 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: 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 \(\left.d q=4 \pi r^{2} \rho(r) d r .\right]\) (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)?

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