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Chapter 24: Problem 11

A spherical capacitor contains a charge of \(3.30 \mathrm{nC}\) when connected to a potential difference of \(220 \mathrm{~V}\). If its plates are separated by vacuum and the inner radius of the outer shell is \(4.00 \mathrm{~cm}\), calculate: (a) the capacitance; (b) the radius of the inner sphere; (c) the electric field just outside the surface of the inner sphere.

Short Answer

Expert verified
The capacitance of the spherical capacitor is \(15 \mathrm{pF}\). The radius of the inner sphere is \(2.45 \mathrm{cm}\). The electric field just outside the surface of the inner sphere is \(1.74 \times 10^{5} \mathrm{N/C}\).

Step by step solution

01

Calculate the Capacitance

The capacitance \(C\) of a capacitor is given by the formula \(C= Q/V\), where \(Q\) is the charge and \(V\) is the potential difference. Substituting the given values, we get \(C = 3.30 \mathrm{nC}/220 \mathrm{V} =15 \mathrm{pF}\).
02

Calculate the Radius of the Inner Sphere

The formula for the capacitance of a spherical capacitor is \(C = 4蟺蔚鈧�*R鈧丷鈧�/(R鈧� - R鈧�)\), where \(R鈧乗) and \(R鈧俓) are the radii of the inner and outer spheres, and \(蔚鈧�\) is the permittivity of free space. The permittivity of free space \(蔚鈧�\) is approximately \(8.85 \times 10^{-12}\) square meters per newton per square coulomb. Simplifying for \(R鈧乗), we get \(R鈧� = R鈧�*C / (4蟺蔚鈧�*C + R鈧�)\). Substituting the known values, \(R鈧� 鈮� 2.45 \mathrm{cm}\).
03

Calculate the Electric Field

The electric field \(E\) produced by a charge \(Q\) on the surface of a sphere is given by \(E = Q / 4蟺蔚鈧�R鈧伮瞈), where \(R鈧乗) is the radius of the sphere, and \(蔚鈧�\) is the permittivity of free space. Substituting the given and calculated values, we get \(E 鈮� 1.74 \times 10^{5} \mathrm{N/C}\).

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Key Concepts

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

Capacitance
The concept of capacitance \(C\) is crucial when dealing with capacitors, especially spherical ones. Capacitance is defined as the ability of a system to store electric charge. In essence, it is the ratio of the charge \(Q\) on one plate to the potential difference \(V\) across the plates. Mathematically, this relationship is expressed as: \[ C = \frac{Q}{V} \] For a given charge and potential difference, the capacitance tells us how much charge the capacitor can hold for that voltage.
  • **Units:** Capacitance is measured in Farads (F), but many practical capacitors have capacitances much smaller than one Farad, so units like picofarads (pF) or nanofarads (nF) are often used.
  • **Spherical Capacitor:** In a spherical capacitor, the charges reside on the surfaces of two concentric spheres, one inside the other.
  • **In the Example:** By using the formula \(C = \frac{3.30 \mathrm{nC}}{220 \mathrm{V}} = 15 \mathrm{pF}\), we calculated the capacitance of the given spherical capacitor.
Electric Field
The electric field \(E\) is a vector field that represents the force experienced by a unit positive charge at any point in space. For a spherical capacitor, understanding the electric field is key because it describes the interaction between charges in the system.
  • **Mathematical Expression:** The electric field just outside the surface of a charged sphere can be calculated using the formula: \[ E = \frac{Q}{4 \pi \varepsilon_0 R_1^2} \] This equation states that the electric field depends on the charge \(Q\), the permittivity of free space \(\varepsilon_0\), and the distance from the center of the sphere \(R_1\).
  • **Spherical Geometry:** In spherical capacitors, the field outside the inner sphere will diminish with the square of the distance from the sphere, emphasizing how geometry plays a significant role.
  • **In the Example:** By substituting the known values, \(E \approx 1.74 \times 10^{5} \mathrm{N/C}\), we calculated the electric field just outside the surface of the inner sphere, giving insight into the presence of charge around the sphere.
Potential Difference
Potential difference, often referred to as voltage, is fundamentally important in understanding capacitors. It is the work done to move a unit charge between two points in an electric field. In the context of a capacitor, it is the voltage across the capacitor plates.
  • **Units and Expression:** It is measured in volts (V). For a capacitor, the relation between charge \(Q\), capacitance \(C\), and potential difference \(V\) is given by: \[ V = \frac{Q}{C} \]
  • **Role in Capacitance:** Knowing the potential difference helps determine how much charge a capacitor can store at a specific voltage, effectively reflecting the 'pressure' pushing the charges apart.
  • **In the Example:** Given \(Q = 3.30 \, ext{nC}\) and \(V = 220 \, ext{V}\), it helped in calculating the capacitance \(C\) of the spherical capacitor using the relationship \(C = \frac{Q}{V}\).
By understanding these concepts, students can better appreciate the interdependencies between capacitance, electric field, and potential difference in various capacitor configurations.

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Most popular questions from this chapter

BIO Potential in Human Cells. Some cell walls in the human body have a layer of negative charge on the inside surface and a layer of positive charge of equal magnitude on the outside surface. Suppose that the charge density on either surface is \(\pm 0.50 \times 10^{-3} \mathrm{C} / \mathrm{m}^{2},\) the cell wall is \(5.0 \mathrm{nm}\) thick, and the cell-wall material is air. (a) Find the magnitude of \(\vec{E}\) in the wall between the two layers of charge. (b) Find the potential difference between the inside and the outside of the cell. Which is at the higher potential? (c) A typical cell in the human body has a volume of \(10^{-16} \mathrm{~m}^{3}\). Estimate the total electric- field energy stored in the wall of a cell of this size. (Hint: Assume that the cell is spherical, and calculate the volume of the cell wall.) (d) In reality, the cell wall is made up, not of air, but of tissue with a dielectric constant of 5.4 . Repeat parts (a) and (b) in this case.

CP Dielectric elastomers are used to create voltage-dependent capacitors. These elastomers contract when subject to the stresses induced by Coulomb forces between capacitor plates. Consider a pair of parallel plates with area \(A\) separated by distance \(d_{0}\) when no voltage is applied. A material with dielectric constant \(K\) and Young's modulus \(Y\) fills the space between the plates. (a) What compressive force is applied to the dielectric when the plates have opposite charges \(Q\) and \(-Q ?\) (b) The distance between the plates is \(d=d_{0}-s Q^{2},\) where \(s\) is a squeezing coefficient. Determine \(s\) in terms of the parameters specified. (c) A capacitor of this sort has plate area \(1.00 \mathrm{~cm}^{2}\) and noncharged separation distance \(d_{0}=0.400 \mathrm{~mm}\). The plates are separated by a silicone dielectric elastomer with dielectric constant \(K=3.00\) and Young's modulus \(0.0100 \mathrm{GPa}\). What is the squeezing coefficient in this case? (d) If each plate has charge of magnitude \(0.700 \mu \mathrm{C}\), what is the applied voltage? (e) What new voltage would double the amount of charge on the plates?

A parallel-plate capacitor is made from two plates \(12.0 \mathrm{~cm}\) on each side and \(4.50 \mathrm{~mm}\) apart. Half of the space between these plates contains only air, but the other half is filled with Plexiglas of dielectric constant 3.40 (Fig. \(\mathbf{P} 24.64)\). An \(18.0 \mathrm{~V}\) battery is connected across the plates. (a) What is the capacitance of this combination? (Hint: Can you think of this capacitor as equivalent to two capacitors in parallel?) (b) How much energy is stored in the capacitor? (c) If we remove the Plexiglas but change nothing else, how much energy will be stored in the capacitor?

Capacitance of a Thundercloud. The charge center of a thundercloud, drifting \(3.0 \mathrm{~km}\) above the earth's surface, contains \(20 \mathrm{C}\) of negative charge. Assuming the charge center has a radius of \(1.0 \mathrm{~km}\), and modeling the charge center and the earth's surface as parallel plates, calculate: (a) the capacitance of the system; (b) the potential difference between charge center and ground; (c) the average strength of the electric field between cloud and ground; (d) the electrical energy stored in the system.

After combing your hair on a dry day, some of your hair stands up, forced away from your head by electrostatic repulsion. (a) Estimate the length \(L\) of your hair. (b) Using the average linear mass density of hair, which is \(65 \mu \mathrm{g} / \mathrm{cm},\) estimate the mass \(m\) of one of your hairs. (c) Estimate the number \(N\) of hairs that stand after combing. (d) Assume that the comb has taken away a charge \(-2 Q,\) and that your hair has therefore gained an amount of charge \(2 Q .\) Assume further that half of this charge resides next to your head and the other half is distributed at the ends of the \(N\) strands that stand up. Assume the electrostatic force that lifted a hair was twice its weight. Show that this leads to \(2 m g=\frac{1}{4 \pi \epsilon_{0}} \frac{Q^{2} / N}{L^{2}}\) Use this equation to estimate the charge \(Q\) that resides on your head. (e) If your head were a sphere with radius \(R\), it would have a capacitance of \(4 \pi \epsilon_{0} R .\) Estimate the radius of your head; then use your estimate to determine your head's capacitance. (f) Use your result to estimate the potential attained due to combing. (The surprising result illustrates an interesting feature of static electricity.)
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