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Chapter 3: Problem 57

A conducting sphere of radius \(R\) has two spherical cavities of radius \(a\) and \(b\). The cavities have charges \(q_{a}\) and \(q_{b}\) respectively at their centres. Find : distance \(r\) (a) The electric field and electric potential at a (i) \(r\) (distance from \(O\), the centre of sphere \(>R\) ) (ii) \(r\) (distance from \(B\), the centre of cavity \(b)

Short Answer

Expert verified
(a) For \( r > R \), electric field is \( E = \frac{k(q_a + q_b)}{r^2} \); potential \( V = \frac{k(q_a + q_b)}{r} \). For \( r < b \), \( E = 0 \). (b) \( \sigma_R = \frac{(q_a + q_b)}{4\pi R^2} \), \( \sigma_a = -\frac{q_a}{4\pi a^2} \), \( \sigma_b = -\frac{q_b}{4\pi b^2} \). (c) Forces on \( q_a \) and \( q_b \) are zero.

Step by step solution

01

Introduction and Assumptions

We have a conducting sphere with radius \( R \), and two cavities containing point charges \( q_a \) and \( q_b \) located at their centers. We need to determine the electric field (E) and potential (V) at various points and calculate surface charge densities and forces on these charges.
02

Analyze the Problem

Since the sphere is a conductor, the electric field inside the conductor itself is zero, except in regions where there are cavities with charges. The charges resided in each cavity center will induce charge distributions on the surfaces.
03

Electric Field and Potential for r > R

For points \( r > R \), we treat the sphere as a point charge with total charge \( q_a + q_b \). Thus, the electric field is given by \( E = \frac{k(q_a + q_b)}{r^2} \) and the electric potential by \( V = \frac{k(q_a + q_b)}{r} \), where \( k = \frac{1}{4\pi\varepsilon_0} \).
04

Electric Field and Potential Inside Cavity b

For \( r < b \), inside cavity \( b \) but outside the charge \( q_b \), the electric field is zero and the potential is constant, influenced only by external charges. Since the cavity and its charge don't affect surrounding distribution, the electric potential \( V \) at this distance still depends on charges at \( q_a \) and \( q_b \).
05

Surface Charge Density at Radius R

The surface charge density \( \sigma_R \) on the conducting sphere of radius \( R \) can be found using the formula \( \sigma_R = \frac{(q_a + q_b)}{4\pi R^2} \). This is because the charges in cavities induce equal but opposite charges on the inside surfaces, resulting in total charge \( q_a + q_b \) on the sphere's outer surface.
06

Surface Charge Densities on Cavities

For the cavity walls of radius \( a \) and \( b \), the charges on these surfaces exactly counterbalance the charges at their centers, so \( \sigma_a = -\frac{q_a}{4\pi a^2} \) and \( \sigma_b = -\frac{q_b}{4\pi b^2} \).
07

Forces on Charges qa and qb

The net force on \( q_a \) and \( q_b \) is zero as they are surrounded by conductor material. Charges in cavities do not experience force due to their own field or the induced charges.

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Key Concepts

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

Electric Field
Understanding the concept of the electric field around a conducting sphere with cavities is crucial. When we discuss electric fields, we're referring to the region around a charged object where other charges can feel a force. In this problem, we have a conducting sphere with two spherical cavities containing charges.
The electric field inside a perfect conductor is zero. However, when dealing with cavities in a conducting sphere:
  • For any point outside the sphere (\(r > R\)), we treat the whole setup like a single point charge with a total charge of\(q_a + q_b\). The electric field at any point outside is calculated as:\(E = \frac{k(q_a + q_b)}{r^2}\), where\(k = \frac{1}{4\pi\varepsilon_0}\).
  • Inside each cavity, for example inside cavity\(b\) (\(r < b\)), the electric field is zero. This is because the only charge in the cavity is\(q_b\) and it does not contribute to the field outside itself.Instead, the charge induces a charge distribution on the surface of the cavity that cancels out any internal electric field.
Understanding these points helps in visualizing how charged conductors behave and how electric fields develop around them.
Electric Potential
Electric potential is another key concept when analyzing problems with conductors. It represents the potential energy per unit charge at a certain point in space.
When considering electric potential around our conducting sphere with cavities:
  • At points outside the sphere (\(r > R\)), the potential is influenced by the total charge\(q_a + q_b\). The formula is:\(V = \frac{k(q_a + q_b)}{r}\). This reflects the potential created by a point charge distribution at a given distance.
  • Inside cavity\(b\) (\(r < b\)), the potential is constant despite the charge\(q_b\) being at its center. This is because the inside region doesn't contribute to potential changes, but it represents the influence of outer charges. Recall, the electric field here is zero, so the potential remains the same throughout the cavity.
Comprehending electric potential aids in determining the work needed to move charges within these regions.
Surface Charge Density
Surface charge density gives insight into how charges distribute over surfaces in electrostatics. It's defined as the amount of charge per unit area.
For our problem with the conducting sphere, three key surfaces to consider are the outer surface and the two cavity surfaces.
  • On the outer surface of the sphere (\(r = R\)), the surface charge density\(\sigma_R\) is linked to the total charge on the sphere. It's calculated as:\(\sigma_R = \frac{q_a + q_b}{4\pi R^2}\). This represents the induced charges on this surface balancing out the cavity charges.
  • On the surface of cavity\(a\), the charge density\(\sigma_a\) works to neutralize the charge inside. Therefore, we have:\(\sigma_a = -\frac{q_a}{4\pi a^2}\).
  • Similarly, for cavity\(b\), the surface charge density is\(\sigma_b = -\frac{q_b}{4\pi b^2}\).
These densities are essential for understanding how different parts of the conductor respond to internal charges.
Cavity Charges
Charges placed in cavities within a conducting material have unique properties. Here's how they affect the sphere.
Consider point charges\(q_a\) and\(q_b\) in their respective cavities.
  • These charges create their own electric fields within the cavities, but because the conductor shells around these cavities are perfect, they block any effect from the internal charges reaching outside spaces beyond the conductor.
  • Moreover, these charges induce equal and opposite charges on the surface of the cavities, leading to zero net electric field within the conductor material itself.
  • Due to shielding by the conductor, external electric fields or potential differences do not affect these charges. Consequently, the force on any charge within the cavity is zero since it only acts upon the charge itself, not the induced charges.
Understanding these unique properties of cavity charges helps clarify the protective nature of conductors and field interactions inside cavities.

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Most popular questions from this chapter

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?

* Three equal \(1.20-\mu \mathrm{C}\) point charges are placed at the corners of an equilateral triangle whose sides are \(0.500 \mathrm{~m}\) long. What is the potential energy of the system? (Take as zero the potential energy of the three charges when they are infinitely far apart.)

The electric potential \(V\) in a region of space is given by $$ V(x, y, z)=A\left(x^{2}-3 y^{2}+z^{2}\right) $$ where \(A\) is a constant. (a) Derive an expression for the electric field \(\overrightarrow{\boldsymbol{E}}\) at any point in this region. (b) The work done by the field when a \(1.50-\mu \mathrm{C}\) test charge moves from the point \((x, y, z)=\) \((0,0,0.250 \mathrm{~m})\) to the origin is measured to be \(6.00 \times 10^{-5} \mathrm{~J}\). Determine \(A\). (c) Determine the electric field at the point \((0,0\), \(0.250 \mathrm{~m}\) ). (d) Show that in every plane parallel to the \(x z\)-plane the equipotential contours are circles. (e) What is the radius of the equipotential contour corresponding to \(V=1280 \mathrm{~V}\) and \(\mathrm{y}=2.00 \mathrm{~m}\) ?

A vacuum tube diode consists of concentric cylindrical electrodes, the negative cathode and the positive anode. Because of the accumulation of charge near the cathode, the electric potential between the electrodes is not a linear function of the position, even with planar geometry, but is given by $$ V(x)=C x^{4 / 3} $$ where \(x\) is the distance from the cathode and \(C\) is a constant, characteristic of a particular diode and operating conditions. Obtain a formula for the electric field between the electrodes as a function of \(x\)

A point charge \(q_{1}\) is held stationary at the origin. A second charge \(q_{2}\) is placed at point \(a\), and the electric potential energy of the pair of charges is \(+5.4 \times 10^{-8} \mathrm{~J}\). When the second charge is moved to point \(b\), the electric force on the charge does \(-1.9 \times 10^{-8} \mathrm{~J}\) of work. What is the electric potential energy of the pair of charges when the second charge is at point \(b\) ?
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