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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

A metal sphere with radius \(r_a\) is supported on an insulating stand at the center of a hollow, metal, spherical shell with radius \(r_b\) . There is charge \(+\)q on the inner sphere and charge \(-\)q on the outer spherical shell. (a) Calculate the potential \(V(r)\) for (i) \(r < r_a ;\) (ii) \(r_a < r < r_b ;\) (iii) \(r > r_b .\) (\(Hint\): The net potential is the sum of the potentials due to the individual spheres.) Take \(V\) to be zero when \(r\) is infinite. (b) Show that the potential of the inner sphere with respect to the outer is $$V_{ab} = \frac{q} {4\pi\epsilon_0} ( \frac{1} {r_a} - \frac{1} {r_b} )$$ (c) Use Eq. (23.23) and the result from part (a) to show that the electric field at any point between the spheres has magnitude $$E(r) = \frac {V_{ab}} {(1/r_a - 1/r_b)}\frac {1} {r_2}$$ (d) Use Eq. (23.23) and the result from part (a) to find the electric field at a point outside the larger sphere at a distance \(r\) from the center, where \(r > r_b\) . (e) Suppose the charge on the outer sphere is not \(-q\) but a negative charge of different magnitude, say \(-Q\). Show that the answers for parts (b) and (c) are the same as before but the answer for part (d) is different.

A point charge \(q_1 =\) 4.00 nC is placed at the origin, and a second point charge \(q_2 = -\)3.00 nC is placed on the \(x\)-axis at \(x = +\)20.0 cm. A third point charge \(q_3 =\) 2.00 nC is to be placed on the \(x\)-axis between \(q_1\) and \(q_2\) . (Take as zero the potential energy of the three charges when they are infinitely far apart.) (a) What is the potential energy of the system of the three charges if \(q_3\) is placed at \(x = +\)10.0 cm? (b) Where should \(q_3\) be placed to make the potential energy of the system equal to zero?

A positive charge \(q\) is fixed at the point \(x = 0, y = 0\), and a negative charge \(-2_q\) is fixed at the point \(x = a, y = 0\). (a) Show the positions of the charges in a diagram. (b) Derive an expression for the potential \(V\) at points on the \(x\)-axis as a function of the coordinate \(x\). Take \(V\) to be zero at an infinite distance from the charges. (c) At which positions on the \(x\)-axis is \(V = 0\)? (d) Graph \(V\) at points on the \(x\)-axis as a function of \(x\) in the range from \(x = -2a\) to \(x = +2a\). (e) What does the answer to part (b) become when \(x \gg a\)? Explain why this result is obtained.

Two plastic spheres, each carrying charge uniformly distributed throughout its interior, are initially placed in contact and then released. One sphere is 60.0 cm in diameter, has mass 50.0 g, and contains \(-\)10.0 \(\mu\)C of charge. The other sphere is 40.0 cm in diameter, has mass 150.0 g, and contains \(-\)30.0 \(\mu\)C of charge. Find the maximum acceleration and the maximum speed achieved by each sphere (relative to the fixed point of their initial location in space), assuming that no other forces are acting on them. (\(Hint:\) The uniformly distributed charges behave as though they were concentrated at the centers of the two spheres.)

Charge \(Q =\) 5.00 mC is distributed uniformly over the volume of an insulating sphere that has radius \(R =\) 12.0 cm. A small sphere with charge \(q = +\)3.00 \(\mu\)C and mass 6.00 \(\times 10^{-5}\) kg is projected toward the center of the large sphere from an initial large distance. The large sphere is held at a fixed position and the small sphere can be treated as a point charge. What minimum speed must the small sphere have in order to come within 8.00 cm of the surface of the large sphere?

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