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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) Electric field: \( E = \frac{kQ_1}{R_1^2} \), potential: \( V = \frac{kQ_1}{R_1} \). (b) Charges: \( Q_1 = \frac{Q R_2}{R_1+R_2} \), \( Q_2 = \frac{Q R_1}{R_1+R_2} \). (c) Same potential: \( V = \frac{kQ}{R_1+R_2} \). (d) Fields: \( E_1 = \frac{kQ_1}{R_1^2} \), \( E_2 = \frac{kQ_2}{R_2^2} \).

Step by step solution

01

Determine the Electric Field at the Surface

For a charged conducting sphere, the electric field on the surface is given by the formula:\[ E = \frac{kQ_1}{R_1^2} \]where \(k\) is Coulomb's constant \((8.99 \times 10^9 \, \text{N m}^2/\text{C}^2)\), \(Q_1\) is the charge on the sphere, and \(R_1\) is the radius of the sphere.
02

Determine the Electric Potential at the Surface

The electric potential \(V\) at the surface of a charged sphere is given by:\[ V = \frac{kQ_1}{R_1} \]This is because the potential is constant over the surface of a conductor. The reference potential is zero at an infinite distance.
03

Establish Charge Redistribution in Equilibrium

When a charged sphere is connected to an uncharged sphere by a conducting wire, charge will flow until the potential at the surface of both spheres becomes equal.If \(Q_2\) is the charge on the second sphere after redistribution, then:\[ V_1 = V_2 \rightarrow \frac{kQ_1}{R_1} = \frac{kQ_2}{R_2} \]
04

Solve for Charges on Each Sphere at Equilibrium

Because charge is conserved, the total charge \(Q = Q_1 + Q_2\) remains constant. Using \(Q = Q_1 + Q_2\) and Step 3, solve for \(Q_1\) and \(Q_2\):1. Express \(Q_2\) in terms of \(Q_1\) using \(\frac{Q_1}{R_1} = \frac{Q_2}{R_2}\).2. Combine with \(Q = Q_1 + Q_2\) to find:\[ Q_1 = \frac{QR_2}{R_1+R_2}, \quad Q_2 = \frac{QR_1}{R_1+R_2} \]
05

Calculate the Electric Potential for Each Sphere

At equilibrium, both spheres have the same potential:\[ V = \frac{kQ_1}{R_1} = \frac{kQ_2}{R_2} \]Each calculation should yield \( V = \frac{kQ}{R_1+R_2} \).
06

Determine the Electric Field for Each Sphere

Use the formula for electric field at the surface for each sphere:\[ E_1 = \frac{kQ_1}{R_1^2}, \quad E_2 = \frac{kQ_2}{R_2^2} \]Plug the charges found in Step 4 into these equations to get specific values.

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Key Concepts

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

Electrostatic Equilibrium
When two conducting spheres are connected with a long, thin wire, electrostatic equilibrium occurs. This means that charge flows between the spheres until the potential on both surfaces is equal. This is crucial because it ensures that there is no net movement of charge on the wire, which is important for maintaining equilibrium.
At equilibrium, both spheres will have the same electric potential. The equation for this can be written as:
  • \( V_1 = V_2 \) where \( V_1 \) is the potential of the first sphere and \( V_2 \) is the potential of the second sphere.
Using the formula for electric potential of a sphere \( V = \frac{kQ}{R} \), we equate the potentials: \( \frac{kQ_1}{R_1} = \frac{kQ_2}{R_2} \). This equation allows us to determine the distribution of charge across the two spheres after equilibrium is reached.The conservation of charge is another vital principle here. The total charge before and after connecting the spheres remains constant. Thus, if \( Q_1 \) is the initial charge and \( Q \) is the combined charge, we have:
  • \( Q = Q_1 + Q_2 \)
Using this relationship, we can solve for the new charges \( Q_1 \) and \( Q_2 \) once electrostatic equilibrium is established.
Electric Field
The electric field at the surface of a charged sphere is determined by how the charge is distributed over the sphere. The formula for the electric field at the surface is given by:
  • \( E = \frac{kQ}{R^2} \)
In this equation, \( E \) is the electric field, \( k \) is Coulomb's constant, \( Q \) is the charge, and \( R \) is the radius of the sphere.
When the spheres reach electrostatic equilibrium, the charge distribution changes, affecting the electric fields at the surfaces. For connected spheres with charges \( Q_1 \) and \( Q_2 \), the electric fields become:
  • \( E_1 = \frac{kQ_1}{R_1^2} \) for the first sphere
  • \( E_2 = \frac{kQ_2}{R_2^2} \) for the second sphere
The change in charge distribution results in variations in the electric field experienced at each sphere's surface. Accurately calculating these fields is vital for understanding the behavior of charged conductors in electric potential problems.
Charge Redistribution
Charge redistribution is an essential concept when dealing with connected conductors. Initially, one sphere is charged, and the other is uncharged. Once they are connected, charge will move from the higher potential sphere to the uncharged one until both have the same potential.
The equations of equilibrium, \( \frac{Q_1}{R_1} = \frac{Q_2}{R_2} \) and \( Q = Q_1 + Q_2 \), allow us to solve for the final distribution of charge. This can be summarized as:
  • The charge on the first sphere after redistribution, \( Q_1 = \frac{QR_2}{R_1 + R_2} \)
  • The charge on the second sphere after redistribution, \( Q_2 = \frac{QR_1}{R_1 + R_2} \)
With these equations, students can determine how charge distributes between the spheres, depending on their respective radii.
Understanding charge redistribution is crucial because it affects both the electric field and electric potential at the surface of each sphere. This knowledge can be applied to various contexts where conductors are used, such as in networking multiple capacitors in circuitry.

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Most popular questions from this chapter

In the Rohr model of the hydrogen atom, a single electron revolves around a single proton in a circle of radius \(r .\) Assume that the proton remains at rest. (a) By equating the electric force to the electron mass times its acceleration, derive an expression for the electron's speed. (b) Obtain an expression for the electron's kinetic energy, and show that its magnitude is just half that of the electric potential energy. (c) Obtain an expression for the total energy, and evaluate it using \(r=5.29 \times 10^{-11} \mathrm{m} .\) Give your numerical result in joules and in electron volts.

A point charge \(Q=+4.60 \mu \mathrm{C}\) is held fixed at the origin. A second point charge \(q=+1.20 \mu \mathrm{C}\) with mass of \(2.80 \times 10^{-4} \mathrm{kg}\) is placed on the \(x\) -axis, 0.250 \(\mathrm{m}\) from the origin. (a) What is the electric potential energy \(U\) of the pair of charges? (Take \(U\) to be zero when the charges have infinite scparation.) (b) The scond point charge is released from rest. What is its speed when its distance from the origin is (i) \(0.500 \mathrm{m} ;\) (ii) \(5.00 \mathrm{m} ;\) (iii) 50.0 \(\mathrm{m} ?\)

In a certain region of space, the electric potential is \(V(x, y, z)=A x y-B x^{2}+C y,\) where \(A, B,\) and \(C\) are positive constants. (a) Calculate the \(x-y\) - and \(z\) -components of the electric field. (b) At which points is the electric field equal to zero?

For each of the following arrangements of two point charges, find all the points along the line passing through both charges for which the electric potential \(V\) is zero (take \(V=0\) infinitely far from the charges) and for which the electric field \(E\) is zero: (a) charges \(+Q\) and \(+2 Q\) separated by a distance \(d\) , and (b) charges \(-Q\) and \(+2 Q\) separated by a distance \(d\) . (c) Are both \(V\) and \(E\) zero at the same places? Explain.

Two stationary point charges \(+3.00 \mathrm{nC}\) and \(+2.00 \mathrm{nC}\) are separated by a distance of 50.0 \(\mathrm{cm}\) . An electron is released from rest at a point midway between the two charges and moves along the line connecting the two charges. What is the speed of the electron when it is 10.0 \(\mathrm{cm}\) from the \(+3.00-\mathrm{charge}\) ?

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