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A thin spherical shell with radius \(R_{1}=3.00 \mathrm{cm}\) is concentric with a larger thin spherical shell with radius \(R_{2}=5.00 \mathrm{cm} .\) Both shells are made of insulating material. The smaller shell has charge \(q_{1}=+6.00 \mathrm{nC}\) distributed uniformly over its surface, and the larger shell has charge \(q_{2}=-9.00 \mathrm{nC}\) distributed uniformly over its surface. Take the electric potential to be zero at an infinite distance from both shells. (a) What is the electric potential due to the two shells at the following distance from their common center: (i) \(r=0 ;\) (ii) \(r=4.00 \mathrm{cm} ;\) (iii) \(r=6.00 \mathrm{cm} ?\) (b) What is the magnitude of the potential difference between the surfaces of the two shells? Which shell is at higher potential: the inner shell or the outer shell?

Short Answer

Expert verified
(i) 179.8 V; (ii) 1348.5 V; (iii) -449.5 V. Inner shell is at higher potential by 719.2 V.

Step by step solution

01

Electric Potential at the Center (r = 0 cm)

For a point inside a uniformly charged spherical shell, by Gauss's Law, the electric field is zero. Therefore, the electric potential at the center due to each shell is constant throughout the sphere. Since the potential at infinity is zero, the potential at the center is simply the sum of potentials due to each shell:- Potential due to first shell:\[ V_1 = k \frac{q_1}{R_1} = (8.99 \times 10^9) \frac{6 \times 10^{-9}}{0.03} = 1798 \text{ V}\]- Potential due to second shell:\[ V_2 = k \frac{q_2}{R_2} = (8.99 \times 10^9) \frac{-9 \times 10^{-9}}{0.05} = -1618.2 \text{ V}\]- Total potential at the center:\[ V = V_1 + V_2 = 1798 - 1618.2 = 179.8 \text{ V}\]
02

Electric Potential at r = 4 cm

At point r = 4 cm, we are outside the smaller shell but inside the larger shell. The potential is constant as it depends only on the total charge within radius r.Since only the smaller shell contributes within this region:- The potential contribution from the smaller shell is:\[ V_1 = k \frac{q_1}{r} = (8.99 \times 10^9) \frac{6 \times 10^{-9}}{0.04} = 1348.5 \text{ V}\]- The larger shell contributes no potential because it's external to the point considered. The total potential at r = 4 cm is:\[ V = V_1 = 1348.5 \text{ V}\]
03

Electric Potential at r = 6 cm

For r = 6 cm, which is outside both shells, the potential depends on the net charge of both shells treated as a point charge:- The net charge is:\[ q_{net} = q_1 + q_2 = 6 \times 10^{-9} - 9 \times 10^{-9} = -3 \times 10^{-9} \text{ C} \]- The potential at r = 6 cm is:\[ V = k \frac{q_{net}}{r} = (8.99 \times 10^9) \frac{-3 \times 10^{-9}}{0.06} = -449.5 \text{ V}\]
04

Potential Difference Between Shell Surfaces

To find the potential difference between the surfaces of the two shells, we should calculate the potential at each shell's surface and then find their difference.- Potential at the surface of the inner shell (r = 3 cm):\[ V_{inner} = k \frac{q_1}{3 \times 10^{-2}} + k \frac{q_2}{5 \times 10^{-2}} = 1798 - 1618.2 = 179.8 \text{ V}\]- Potential at the surface of the outer shell (r = 5 cm):\[ V_{outer} = k \left( \frac{6 \times 10^{-9}}{5 \times 10^{-2}} + \frac{-9 \times 10^{-9}}{5 \times 10^{-2}} \right) = -539.4 \text{ V} \]- Potential difference: \[ \Delta V = V_{inner} - V_{outer} = 179.8 - (-539.4) = 719.2 \text{ V} \]The inner shell is at a higher potential than the outer shell by 719.2 V.

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Key Concepts

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

Spherical Shells
Spherical shells are three-dimensional objects that feature concentric layers or surfaces, shaped like a sphere, surrounding a common center. In physics, when dealing with charges, these shells are often considered in problems concerning electrostatics and electric fields. A critical factor about spherical shells is that they can be conductive or insulating. In the case of insulating shells, charges are strictly confined to the surface and don't move across it. This exercise demonstrates a system composed of two thin concentric spherical shells, each possessing a distinct uniform charge distribution over their surface.
Key characteristics of these shells include:
  • Each shell has its respective radius (like the smaller shell with radius 3.00 cm and the larger one with radius 5.00 cm).
  • The charge distribution could be positive or negative, as indicated by the charges of the shells.
  • The interactions of these charges and their effects on electric potential are computed using fundamental electrostatic principles.
Gauss's Law
Gauss's Law is a pivotal principle in electromagnetism that defines the relationship between the electric field and the charge distribution within a given volume. The law states that the total electric flux through a closed surface is directly proportional to the enclosed charge. Mathematically, it's expressed as:\[ \Phi = \int \mathbf{E} \cdot d\mathbf{A} = \frac{Q_{enc}}{\varepsilon_0} \]where \(\Phi\) is the electric flux, \(\mathbf{E}\) is the electric field, \(d\mathbf{A}\) is a differential area on the closed surface, \(Q_{enc}\) is the enclosed charge, and \(\varepsilon_0\) is the permittivity of free space.

For spherical shells, Gauss’s Law helps simplify calculations by indicating that the electric field inside a charged spherical shell is zero when observed from a point within the material. This principle is used to determine potentials for different regions within and around the shells. For instance, at a point inside the smaller shell but outside the larger shell, only the smaller shell contributes to the electric potential since its electric field is uniform.
Potential Difference
Potential difference, also known as voltage, is the difference in electric potential between two points in space. It indicates how much work is required to move a charge from one point to another. For spherical shells, potential difference is significant as it determines the relative energy states between the charges on the shells.

In the exercise, we calculated the potential difference between the surfaces of two concentric spherical shells by subtracting the potentials calculated at each shell's surface. The formula for potential due to a charge \(q\) at a distance \(r\) from it is:\[ V = k \frac{q}{r} \]where \(k\) is Coulomb's constant. The potential difference is found as:\[ \Delta V = V_{inner} - V_{outer} \]This allows us to observe that the inner shell exhibited a higher potential than the outer shell, a result of their specific charge values and separation.
Insulating Material
Insulating materials are substances that do not allow the free flow of electric charge. In the context of spherical shells, insulating materials act as barriers that hold charge on their surfaces without allowing it to move through the shell or interact with internal charges.

The significance of using insulating materials in these shells is reducing interaction and charge redistribution within the material, sticking the charges strictly to the surfaces. This makes calculations of electric potential and fields more straightforward as each surface charge distribution maintains its potential defined by its own geometry and the distance to any point of interest. Understanding insulation helps clarify why electric fields and potential can be considered as originating primarily from surface charges, particularly in such electrostatic conditions.

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Most popular questions from this chapter

Four electrons are located at the corners of a square 10.0 \(\mathrm{nm}\) on a side, with an alpha particle at its midpoint. How much work is needed to move the alpha particle to the midpoint of one of the sides of the square?

The Millikan Oil-Drop Experiment. The charge of an electron was first measured by the American physicist Robert Millikan during \(1909-1913 .\) In his experiment, oil is sprayed in very fine drops (around \(10^{-4} \mathrm{mm}\) in diameter) into the space between two parallel horizontal plates separated by a distance \(d . \mathrm{A}\) potential difference \(V_{A B}\) is maintained between the parallel plates, causing a downward electric field between them. Some of the oil drops acquire a negative charge because of frictional effects or because of ionization of the surrounding air by \(x\) rays or radioactivity. The drops are observed through a microscope. (a) Show that an oil drop of radius \(r\) at rest between the plates will remain at rest if the magnitude of its charge is $$q=\frac{4 \pi}{3} \frac{\rho r^{3} g d}{V_{A B}}$$ where \(\rho\) is the density of the oil. (Ignore the buoyant force of the air.) By adjusting \(V_{A B}\) to keep a given drop at rest, the charge on that drop can be determined, provided its radius is known. (b) Millikan's ooil drops were much too small to measure their radii directly. Instead, Millikan determined \(r\) by cutting off the electric field and measuring the terminal speed \(v_{t}\) of the drop as it fell. (We discussed the concept of terminal speed in Section \(5.3 . )\) The viscous force \(F\) on a sphere of radius \(r\) moving with speed \(v\) through a fluid with viscosity \(\eta\) is given by Stokes's law: \(F=6 \pi \eta r v .\) When the drop is falling at \(v_{1}\) , the viscous force just balances the weight \(w=m g\) of the drop. Show that the magnitude of the charge on the drop is $$q=18 \pi \frac{d}{V_{A B}} \sqrt{\frac{\eta^{3} v_{t}^{3}}{2 \rho g}}$$ Within the limits of their experimental error, every one of the thousands of drops that Millikan and his coworkers measured had a charge equal to some small integer multiple of a basic charge \(e\) . That is, they found drops with charges of \(\pm 2 e, \quad \pm 5 e,\) and so on, but none with values such as 0.76\(e\) or 2.49\(e .\) A drop with charge \(-e\) has acquired one extra electron; if its charge is \(-2 e,\) it has acquired two extra electrons, and so on. (c) A charged oil drop in a Millikan oil-drop apparatus is observed to fall 1.00 mm at constant speed in 39.3 s if \(V_{A B}=0 .\) The same drop can be held at rest between two plates separated by 1.00 \(\mathrm{mm}\) if \(V_{A B}=9.16 \mathrm{V} .\) How many excess electrons has the drop acquired, and what is the radius of the drop? The viscosity of air is \(1.81 \times 10^{-5} \mathrm{N} \cdot \mathrm{s} / \mathrm{m}^{2},\) and the density of the oil is 824 \(\mathrm{kg} / \mathrm{m}^{3} .\)

Two point charges are moving to the right along the \(x\) -axis. Point charge 1 has charge \(q_{1}=2.00 \mu \mathrm{C},\) mass \(m_{1}=\) \(6.00 \times 10^{-5} \mathrm{kg},\) and speed \(v_{1} .\) Point charge 2 is to the right of \(q_{1}\) and has charge \(q_{2}=-5.00 \mu \mathrm{C},\) mass \(m_{2}=3.00 \times 10^{-5} \mathrm{kg},\) and speed \(v_{2} .\) At a particular instant, the charges are separated by a distance of 9.00 \(\mathrm{mm}\) and have speeds \(v_{1}=400 \mathrm{m} / \mathrm{s}\) and \(v_{2}=1300 \mathrm{m} / \mathrm{s} .\) The only forces on the particles are the forces they exert on each other. (a) Determine the speed \(v_{\mathrm{cm}}\) of the center of mass of the system. (b) The relative energy \(E_{\text { rel }}\) of the system is defined as the total energy minus the kinetic energy contributed by the motion of the center of mass: $$E_{\mathrm{rel}}=E-\frac{1}{2}\left(m_{1}+m_{2}\right) v_{\mathrm{cm}}^{2}$$ where \(E=\frac{1}{2} m_{1} v_{1}^{2}+\frac{1}{2} m_{2} v_{2}^{2}+q_{1} q_{2} / 4 \pi \epsilon_{0} r\) is the total energy of the system and \(r\) is the distance between the charges. Show that \(E_{\mathrm{rel}}=\frac{1}{2} \mu v^{2}+q_{1} q_{2} / 4 \pi \epsilon_{0} r, \quad\) where \(\mu=m_{1} m_{2} /\left(m_{1}+m_{2}\right)\) is called the reduced mass of the system and \(v=v_{2}-v_{1}\) is the relative speed of the moving particles. (c) For the numerical values given above, calculate the numerical value of \(E_{\text { rel. }}\) (d) Based on the result of part (c), for the conditions given above, will the particles escape from one another? Explain. (e) If the particles do escape, what will be their final relative speed when \(r \rightarrow \infty ?\) If the particles do not escape, what will be their distance of maximum separation? That is, what will be the value of \(r\) when \(v=0 ?\) (f) Repeat parts \((c)-(e)\) for \(v_{1}=400 \mathrm{m} / \mathrm{s}\) and \(v_{2}=1800 \mathrm{m} / \mathrm{s}\) when the separation is 9.00 \(\mathrm{mm} .\)

A small metal sphere, carrying a net charge of \(q_{1}=\) \(-2.80 \mu \mathrm{C},\) is held in a stationary position by insulating supports. A second small metal sphere, with a net charge of \(q_{2}=-7.80 \mu \mathrm{C} \quad\) and \(\operatorname{mass}\) \(1.50 \mathrm{g},\) is projected toward \(q_{1}\). When the two spheres are 0.800 \(\mathrm{m}\) apart, \(q_{2}\) is moving toward \(q_{1}\) with speed 22.0 \(\mathrm{m} / \mathrm{s}\) (Fig. E23.5). Assume that the two spheres can be treated as point charges. You can ignore the force of gravity. (a) What is the speed of \(q_{2}\) when the spheres are 0.400 \(\mathrm{m}\) apart? (b) How close does \(q_{2}\) get to \(q_{1}\) ?

The electric field at the surface of a charged, solid, copper sphere with radius 0.200 \(\mathrm{m}\) is 3800 \(\mathrm{N} / \mathrm{C}\) , directed toward the center of the sphere. What is the potential at the center of the sphere, if we take the potential to be zero infinitely far from the sphere?

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