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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?

Short Answer

Expert verified
The surface charge density (\( \sigma \)) of the sphere is approximately \( 5.55 \times 10^{-6} \ C/m^2 \) and the capacitance (C) of the sphere is approximately \( 1.06 \times 10^{-10} \ F \).

Step by step solution

01

Calculate the charge on the sphere

Using Coulomb's law for a sphere, the electric field (E) at a distance (r) from its center is given by the formula: \( E = \frac{kQ}{r^2} \). Solve for \( Q \), the charge on the sphere, by rearranging the formula: \( Q = \frac{E \cdot r^2}{k} \), where \( k = 8.99 \times 10^9 \ N \cdot m^2 / C^2 \) is Coulomb's constant, \( E = 4.90 \times 10^4 \ N/C \), and \( r = 0.21 \ m \) is the distance from the center of the sphere where the electric field is measured.
02

Calculate surface charge density

The surface charge density (\( \sigma \)) is defined as the charge per unit area on the sphere. Use the formula: \( \sigma = \frac{Q}{4 \pi r_{s}^2} \), where \( Q \) is the charge found in Step 1, and \( r_{s} = 0.12 \ m \) is the radius of the sphere. By substituting the charge \( Q \) from Step 1 into this formula, you can find the surface charge density \( \sigma \).
03

Calculate capacitance of the sphere

The capacitance (C) of an isolated sphere is given by the formula: \( C = \frac{Q}{V} \), where \( Q \) is the charge on the sphere and \( V \) is the potential. An important property to remember is that for a conducting sphere, the potential (V) is the same across the surface and is given by \( V = \frac{kQ}{r_{s}} \). Rearrange the capacitance formula to find the capacitance in terms of known quantities: \( C = \frac{4 \pi \epsilon_0 r_{s}}{1} \), where \( \epsilon_0 = 8.85 \times 10^{-12} \ C^2 / (N \cdot m^2) \) is the vacuum permittivity and \( r_{s} \) is the radius of the conducting sphere.

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Key Concepts

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

Coulomb's Law
Coulomb's law is the cornerstone of electrostatics. It describes the force between two point charges as being directly proportional to the product of the charges and inversely proportional to the square of the distance between them. Mathematically, it's expressed as
\[ F = k \frac{|q_1 q_2|}{r^2} \]
where \( F \) is the electrostatic force, \( q_1 \) and \( q_2 \) are the point charges, \( r \) is the distance between charges, and \( k \) is Coulomb's constant (\( 8.99 \times 10^9 \ N \cdot m^2 / C^2 \) in SI units).

This fundamental law not only applies to static electric fields but also plays a crucial role in computing the electric field around charged objects. In the case of the charged sphere from our exercise, we use a form of Coulomb's law adapted for spherical symmetry to calculate the charge on the sphere using the electric field and distance from the sphere's surface.
Surface Charge Density
Surface charge density is a measure of how much electric charge resides on a unit area of a surface. For a spherical object, the formula to determine surface charge density, \( \sigma \), is given by
\[ \sigma = \frac{Q}{4 \pi r_{s}^2} \]
where \( Q \) stands for the total charge distributed uniformly across the sphere's surface, and \( r_{s} \) is the radius of the sphere. The concept of surface charge density is particularly useful for understanding how charges distribute themselves on conductors.
Capacitance of a Sphere
The capacitance of a sphere can be thought of as its ability to hold charge per unit of electric potential. It is a scalar quantity that gives us an idea about how much charge can be stored on the sphere's surface. For an isolated spherical conductor, capacitance (\( C \)) is directly proportional to the radius of the sphere (\( r_{s} \)) and is expressed as
\[ C = 4 \pi \epsilon_0 r_{s} \]
where \( \epsilon_0 \) is the vacuum permittivity. It's important to understand that the capacitance only depends on the physical dimensions of the sphere and the properties of the surrounding medium (vacuum in this case), not on the charge or potential itself.
Vacuum Permittivity
Vacuum permittivity (\( \epsilon_0 \)) is a physical constant that describes how an electric field influences and is influenced by a vacuum. Also known as the electric constant, its value is approximated as \( 8.85 \times 10^{-12} \ C^2 / (N \cdot m^2) \) in the International System of Units (SI). This constant appears in many key equations in electrodynamics, including the one used to derive the capacitance of a sphere. It provides information on the amount of resistance that space offers against the formation of an electric field, and hence it is crucial for understanding the electrical properties of materials and the behavior of charged particles in a vacuum.

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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.

A parallel-plate capacitor has a charge \(Q\) and plates of area \(A .\) What force acts on one plate to attract it toward the other plate? Because the electric field between the plates is \(E=Q / A \epsilon_{0},\) you might think that the force is \(F=Q E=Q^{2} / A \epsilon_{0} .\) This is wrong, because the field \(E\) includes contributions from both plates, and the field created by the positive plate cannot exert any force on the positive plate. Show that the force exerted on each plate is actually \(F=Q^{2} / 2 \epsilon_{0} A\) . (Suggestion: Let \(C=\epsilon_{0} A / x\) for an arbitrary plate separation \(x ;\) then require that the work done in separating the two charged plates be \(W=\int F d x\) . The force exerted by one charged plate on another is sometimes used in a machine shop to hold a workpiece stationary.

Two capacitors, \(C_{1}=25.0 \mu \mathrm{F}\) and \(C_{2}=5.00 \mu \mathrm{F}\) , are connected in parallel and charged with a \(100-\mathrm{V}\) power supply. (a) Draw a circuit diagram and calculate the total energy stored in the two capacitors. (b) What If? What potential difference would be required across the same two capacitors connected in series in order that the combination stores the same amount of energy as in (a)? Draw a circuit diagram of this circuit.

A wafer of titanium dioxide \((\kappa=173)\) of area 1.00 \(\mathrm{cm}^{2}\) has a thickness of 0.100 \(\mathrm{mm}\) . Aluminum is evaporated on the parallel faces to form a parallel-plate capacitor. (a) Calculate the capacitance. (b) When the capacitor is charged with a \(12.0-\mathrm{V}\) battery, what is the magnitude of charge delivered to each plate? (c) For the situation in part \((\mathrm{b}),\) what are the free and induced surface charge densities? (d) What is the magnitude of the electric field?

An air-filled capacitor consists of two parallel plates, each with an area of \(7.60 \mathrm{cm}^{2},\) separated by a distance of \(1.80 \mathrm{mm} .\) A \(20.0-\mathrm{V}\) potential difference is applied to these plates. Calculate (a) the electric field between the plates, (b) the surface charge density, (c) the capacitance, and (d) the charge on each plate.

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