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Two conducting spheres with diameters of \(0.400 \mathrm{m}\) and \(1.00 \mathrm{m}\) are separated by a distance that is large compared with the diameters. The spheres are connected by a thin wire and are charged to \(7.00 \mu \mathrm{C}\). (a) How is this total charge shared between the spheres? (Ignore any charge on the wire.) (b) What is the potential of the system of spheres when the reference potential is taken to be \(V=0\) at \(r=\infty ?\)

Short Answer

Expert verified
The charges on the two spheres are \(2.80 \mu C\) and \(4.20 \mu C\), and the potential of the system is \(1.26 x 10^7 V\).

Step by step solution

01

Calculate the radius of each sphere

The diameters of the two spheres are given as \(0.400 \mathrm{m}\) and \(1.00 \mathrm{m}\). Therefore, the radii of the spheres are half of these values: \(0.200 \mathrm{m}\) and \(0.500 \mathrm{m}\) respectively.
02

Determine the relation between charges

The spheres are in equilibrium, which means the potentials of each sphere will be the same. The potential \(V\) of a sphere with charge \(Q\) and radius \(r\) is given by the formula \(V = kQ/r\), where \(k = 9.00 x 10^9 Nm^2/C^2\) is Coulomb's constant. Given two spheres 1 and 2 with charges \(Q_1\) and \(Q_2\), and radii \(r_1\) and \(r_2\), we can set up the equation \(kQ_1/r_1 = kQ_2/r_2\). Because \(k\) is a constant, it can be dropped from the equation, leading to \(Q_1/r_1 = Q_2/r_2\). This equation will be used in the next step.
03

Calculate the charges on each sphere

We know the total charge \(Q_t = Q_1 + Q_2 = 7.00 \mu C\) and the relation \(Q_1/r_1 = Q_2/r_2\). Solve these two equations simultaneously to find the values of \(Q_1\) and \(Q_2\). You will find that \(Q_1 = 2.80 \mu C\) and \(Q_2 = 4.20 \mu C\).
04

Calculate the potential of the system

The potential \(V\) of each sphere is \(V = kQ/r\), where \(Q\) and \(r\) are the charge and radius of the sphere. Because the potential at infinity is taken to be zero, the potential of the system is just the sum of the potentials of the two spheres: \(V_t = V_1 + V_2 = kQ_1/r_1 + kQ_2/r_2\). Substituting the charges and radii obtained in step 3, you will find that \(V_t = 1.26 x 10^7 V\).

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Key Concepts

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

Coulomb's Law
Understanding Coulomb's Law is essential for analyzing charge interactions. It is a quantitative description of the force between two point charges. The law states that the electric force (\f\(F_e\f\)) between two charges (\f\(q_1\f\) and \f\(q_2\f\)) is directly proportional to the product of the two charges and inversely proportional to the square of the distance (\f\(r\f\)) between them. Mathematically, it's represented as \f\(F_e = k \frac{q_1q_2}{r^2}\f\), where \f\(k\f\) is Coulomb's constant (\f\(8.987 \times 10^9 \text{Nm}^2/\text{C}^2\f\)).

For example, when two conducting spheres are charged, the forces they exert on one another can be understood through this law. In the case of the textbook exercise, understanding how the spheres share the total charge is rooted in knowing how electric forces and potentials interact between the spheres in accordance with Coulomb’s Law.
Electric Potential
Electric potential is a concept crucial for comprehending the distribution of charge in a system. It represents the potential energy per unit charge. If an external agent moves a charge \f\(q\f\) within an electric field, the work done on the charge changes its electric potential. The potential (\f\(V\f\)) at a point in space is defined by \f\(V = k \frac{Q}{r}\f\), where \f\(Q\f\) is the charge creating the electric field, \f\(r\f\) is the distance from the charge to the point in question, and \f\(k\f\) again represents Coulomb’s constant.

In our exercise, the spheres reach a common potential when connected by a wire; this is why solving for their charges requires equating their potentials. As the spheres are conductors, the charge distributes itself until the potential over both spheres is uniform. The system's total potential, considered at a point infinitely far away, is calculated through the addition of the potentials due to individual spheres, as established in the step-by-step solution.
Electrostatic Equilibrium
Electrostatic equilibrium occurs when charges on a conductor distribute themselves in such a way that there is no net motion of charge within the conductor. This state is characterized by an electric field that is zero everywhere inside the conductor and a surface charge distribution that creates an electric field just outside the conductor's surface, perpendicular to the surface.

In the context of the problem, when the two conducting spheres are connected by a wire and then charged, they must reach electrostatic equilibrium. This will lead to a situation where their electric potentials are equalized because the charge can flow freely through the wire until the potential difference is eliminated. At electrostatic equilibrium, the charge distribution on the surface of the spheres depends on their radii, demonstrating an inverse relationship between surface charge density and radius.

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Most popular questions from this chapter

A detector of radiation called a Geiger tube consists of a closed, hollow, conducting cylinder with a fine wire along its axis. Suppose that the internal diameter of the cylinder is \(2.50 \mathrm{cm}\) and that the wire along the axis has a diameter of \(0.200 \mathrm{mm} .\) The dielectric strength of the gas between the central wire and the cylinder is \(1.20 \times 10^{6} \mathrm{V} / \mathrm{m} .\) Calculate the maximum potential difference that can be applied between the wire and the cylinder before breakdown occurs in the gas.

An isolated capacitor of unknown capacitance has been charged to a potential difference of \(100 \mathrm{V}\). When the charged capacitor is then connected in parallel to an uncharged \(10.0-\mu \mathrm{F}\) capacitor, the potential difference across the combination is \(30.0 \mathrm{V}\). Calculate the unknown capacitance.

An isolated charged conducting sphere of radius \(12.0 \mathrm{cm}\) creates an electric field of \(4.90 \times 10^{4} \mathrm{N} / \mathrm{C}\) at a distance \(21.0 \mathrm{cm}\) from its center. (a) What is its surface charge density? (b) What is its capacitance?

The general form of Gauss's law describes how a charge creates an electric field in a material, as well as in vacuum. It is $$\oint \mathbf{E} \cdot d \mathbf{A}=\frac{q}{\epsilon}$$ where \(\epsilon=\kappa \epsilon_{0}\) is the permittivity of the material. (a) \(\mathrm{A}\) sheet with charge \(Q\) uniformly distributed over its area \(A\) is surrounded by a dielectric. Show that the sheet creates a uniform clectric ficld at nearby points, with magnitude \(E=Q / 2 A \epsilon\). (b) Two large sheets of area \(A\), carrying opposite charges of equal magnitude \(Q,\) are a small distance \(d\) apart. Show that they create uniform electric field in the space between them, with magnitude \(E=Q / A \epsilon\) (c) Assume that the negative plate is at zero potential. Show that the positive plate is at potential Qd/Ae. (d) Show that the capacitance of the pair of plates is \(A \epsilon / d=\kappa A \epsilon_{0} / d\).

(a) How much charge is on each plate of a \(4.00-\mu \mathrm{F}\) capacitor when it is connected to a \(12.0-\mathrm{V}\) battery? (b) If this same capacitor is connected to a \(1.50-\mathrm{V}\) battery, what charge is stored?

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