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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.

Short Answer

Expert verified
(a) \(E_1 = \frac{1}{4\pi\varepsilon_0}\frac{Q_1}{R_1^2}; V_1 = \frac{1}{4\pi\varepsilon_0}\frac{Q_1}{R_1}\). (b) \(Q_1' = \frac{R_1}{R_1 + R_2} Q_1; Q_2' = \frac{R_2}{R_1 + R_2} Q_1\). (c) \(V' = \frac{1}{4\pi\varepsilon_0}\frac{Q_1}{R_1 + R_2}\). (d) \(E_1' = \frac{1}{4\pi\varepsilon_0}\frac{Q_1}{R_1(R_1 + R_2)}; E_2' = \frac{1}{4\pi\varepsilon_0}\frac{Q_1}{R_2(R_1 + R_2)}\).

Step by step solution

01

Electric Field at the Surface of the Sphere

The electric field at the surface of a conducting sphere with charge \(Q_1\) and radius \(R_1\) is given by Coulomb's law: \[E_1 = \frac{1}{4\pi\varepsilon_0}\frac{Q_1}{R_1^2}\] This formula calculates the electric field just outside the sphere, assuming the sphere is conductively isolated.
02

Electric Potential at the Surface of the Sphere

The electric potential \(V\) at the surface of a sphere with radius \(R_1\) is given by: \[V_1 = \frac{1}{4\pi\varepsilon_0}\frac{Q_1}{R_1}\] This results from the integration of the electric field from infinity to the surface of the sphere.
03

Charge Redistribution After Connection

When the two spheres are connected, charge will redistribute until the potentials on both spheres are equal because charge moves to equalize potential. Let the charges be \(Q_1'\) and \(Q_2'\) after equilibrium. The total charge remains conserved: \[Q_1 + 0 = Q_1' + Q_2'\] and the potentials satisfy: \[\frac{1}{4\pi\varepsilon_0}\frac{Q_1'}{R_1} = \frac{1}{4\pi\varepsilon_0}\frac{Q_2'}{R_2}\] Hence: \[\frac{Q_1'}{R_1} = \frac{Q_2'}{R_2}\]
04

Solving for Charges on Each Sphere

Using the conservation of charge, \(Q_1' + Q_2' = Q_1\), and the equality of potentials, we find: \[Q_1' = \frac{R_1}{R_1 + R_2} Q_1\] \[Q_2' = \frac{R_2}{R_1 + R_2} Q_1\] These give the charges on spheres \(1\) and \(2\) respectively, after equilibrium.
05

Electric Potential at Each Sphere's Surface After Equilibrium

Since potentials are equal after equilibrium and given by \[V' = \frac{1}{4\pi\varepsilon_0}\frac{Q_1'}{R_1} = \frac{1}{4\pi\varepsilon_0}\frac{Q_2'}{R_2}\]Plug in \(Q_1' = \frac{R_1}{R_1 + R_2} Q_1\) to find \[V' = \frac{1}{4\pi\varepsilon_0}\frac{Q_1}{R_1 + R_2}\] at both spheres.
06

Electric Field at Each Sphere's Surface After Equilibrium

The electric field for each sphere after charge redistribution can be calculated using the surface charge and size of each sphere. Thus, \[E_1' = \frac{1}{4\pi\varepsilon_0}\frac{Q_1'}{R_1^2} = \frac{1}{4\pi\varepsilon_0}\frac{R_1}{(R_1 + R_2)R_1^2}Q_1\] and \[E_2' = \frac{1}{4\pi\varepsilon_0}\frac{Q_2'}{R_2^2} = \frac{1}{4\pi\varepsilon_0}\frac{R_2}{(R_1 + R_2)R_2^2}Q_1\] giving the electric field for spheres \(1\) and \(2\), respectively.

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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 describes the force per unit charge experienced by a small test charge placed in the vicinity of another charge. In the case of a conducting sphere with charge, this field can be specifically calculated using Coulomb's law. Just outside the surface of a sphere of radius \(R_1\) with charge \(Q_1\), the electric field \(E_1\) is represented by:
  • \(E_1 = \frac{1}{4\pi\varepsilon_0}\frac{Q_1}{R_1^2}\)
This equation springs from the observation that the electric field created by a uniformly charged sphere behaves as if the entire charge is concentrated at its center. The field strength decreases with the square of the distance from the center, hence the \(R_1^2\) in the denominator. After two spheres are connected, charges adjust till the potential is the same on both, affecting how fields are calculated around each sphere. Understanding the electric field is key to describing how charges influence one another.
Electric Potential
Electric potential is the work done to bring a unit positive charge from infinity to a point in space in an electric field, without acceleration. For a conducting sphere, this depends only on the charge and the radius. The potential \(V_1\) at the surface of a sphere of radius \(R_1\) and charge \(Q_1\) is:
  • \(V_1 = \frac{1}{4\pi\varepsilon_0}\frac{Q_1}{R_1}\)
This formula results from the integration of the electric field from infinity to the point of interest. When two spheres at different initial potentials are connected, the charge rearranges so that the potential is uniform across both surfaces. This balance and uniformity throughout connected conducting equipment simplify the analysis of complex charge systems.
Charge Distribution
Charge distribution refers to how electric charge is spread over a structure. In electrostatics, charges will move around until they reach equilibrium, particularly in conductors. Upon connecting the charged sphere and an initially uncharged one with a conducting wire, charge redistributes between them:
  • Total charge stays constant: \(Q_1 + 0 = Q_1' + Q_2'\)
  • Equalize potentials: \(\frac{Q_1'}{R_1} = \frac{Q_2'}{R_2}\)
Using these relations, the final charges \(Q_1'\) and \(Q_2'\) are found:
  • \(Q_1' = \frac{R_1}{R_1 + R_2} Q_1\)
  • \(Q_2' = \frac{R_2}{R_1 + R_2} Q_1\)
This shows how charge equalizes across two connected spheres based on their sizes, ensuring both surfaces have identical electric potentials after the charge transfer.
Conservation of Charge
The principle of conservation of charge is a fundamental concept in physics, stating that the total charge in an isolated system remains constant. This is crucial in electrostatic interactions like those involving two spheres connected by a wire. When these spheres are connected, charge moves to equalize potential across the spheres, but the total amount of charge does not change:
  • Start with \(Q_1\) on the first sphere and 0 on the second.
  • After equilibrium, \(Q_1' + Q_2' = Q_1\).
Even as charges shuffle between them, the sum remains conserved. Understanding conservation of charge helps track how electricity behaves and redistributes in connected systems, foundational to circuit design and electrostatic applications.

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Most popular questions from this chapter

Coaxial Cylinders. A long metal cylinder with radius a is supported on an insulating stand on the axis of a long, hollow, metal tube with radius \(b\). The positive charge per unit length on the inner cylinder is \(\lambda\), and there is an equal negative charge per unit length on the outer cylinder. (a) Calculate the potential \(V(r)\) for (i) \(r < a\); (ii) \(a < r < b\); (iii) \(r > b\). (\(Hint:\) The net potential is the sum of the potentials due to the individual conductors.) Take \(V = 0\) at \(r = b\). (b) Show that the potential of the inner cylinder with respect to the outer is $$V^{ab} = \frac{\lambda} {2\pi\epsilon_0} ln \frac{b} {a}$$ (c) Use Eq. (23.23) and the result from part (a) to show that the electric field at any point between the cylinders has magnitude $$E(r) = \frac{V_{ab}} {ln(b/a)} \frac{1} {r}$$ (d) What is the potential difference between the two cylinders if the outer cylinder has no net charge?

Two large, parallel, metal plates carry opposite charges of equal magnitude. They are separated by 45.0 mm, and the potential difference between them is 360 V. (a) What is the magnitude of the electric field (assumed to be uniform) in the region between the plates? (b) What is the magnitude of the force this field exerts on a particle with charge \(+\)2.40 nC? (c) Use the results of part (b) to compute the work done by the field on the particle as it moves from the higher-potential plate to the lower. (d) Compare the result of part (c) to the change of potential energy of the same charge, computed from the electric potential.

The electric field at the surface of a charged, solid, copper sphere with radius 0.200 m is 3800 N\(/\)C, directed toward the center of the sphere. What is the potential at the center of the sphere, if we take the potential to be zero infinitely far from the sphere?

A helium ion (He\(^{++}\)) that comes within about 10 fm of the center of the nucleus of an atom in the sample may induce a nuclear reaction instead of simply scattering. Imagine a helium ion with a kinetic energy of 3.0 MeV heading straight toward an atom at rest in the sample. Assume that the atom stays fixed. What minimum charge can the nucleus of the atom have such that the helium ion gets no closer than 10 fm from the center of the atomic nucleus? (1 fm = 1 \(\times\) 10\(^{-15}\) m, and \(e\) is the magnitude of the charge of an electron or a proton.) (a) 2\(e\); (b) 11\(e\); (c) 20\(e\); (d) 22\(e\).

A very large plastic sheet carries a uniform charge density of \(-\)6.00 nC\(/\)m\(^2\) on one face. (a) As you move away from the sheet along a line perpendicular to it, does the potential increase or decrease? How do you know, without doing any calculations? Does your answer depend on where you choose the reference point for potential? (b) Find the spacing between equipotential surfaces that differ from each other by 1.00 V. What type of surfaces are these?

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