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Chapter 6: Problem 1

Find the potential and electric field at points on the axis of symmetry of a uniformly charged flat circular disc of charge \(q\) and radius \(a\). What happens to the field if we keep the charge per unit area \(\sigma=q / \pi a^{2}\) fixed, and let \(a \rightarrow \infty\) ?

Short Answer

Expert verified
#Question# Find the potential and electric field at any point on the axis of symmetry of a uniformly charged flat circular disc with charge \(q\) and radius \(a\). Analyze the electric field as the radius \(a\) approaches infinity while keeping the charge per unit area constant. #Answer# The potential at a point located at distance \(z\) from the center of the disc is given by: \(V = \dfrac{\sigma}{2\epsilon_{0}}z \left(-\dfrac{2}{\sqrt{1 + a^{2}/z^{2}}}+\dfrac{2}{\sqrt{1}}\right)\), where \(\sigma = \dfrac{q}{\pi a^{2}}\) is the charge per unit area. The electric field at the same point in the z-direction is given by: \(E_{z} = -\dfrac{\sigma}{2\epsilon_{0}}\left(\dfrac{-2a^{2}}{(a^{2}+z^{2})^{\frac{3}{2}}} + 1 \right)\). As the disc radius approaches infinity while keeping \(\sigma\) constant, the electric field converges to a constant value: \(E_{z} = \dfrac{\sigma}{2\epsilon_{0}}\).

Step by step solution

01

Identify the coordinate system and define variables

Since the charge distribution is symmetrical, we can use cylindrical coordinates. Let's define a point P on the axis of symmetry, located at a distance \(z\) from the center of the disc. We will find the potential and electric field at point P.
02

Determine the contribution of elemental charges

We can consider the disc to be made up of small elemental rings with radius \(r\) and thickness \(dr\). The area of this elemental ring is given by \(dA = 2\pi r dr\). The elemental charge on the ring will be \(dq = \sigma dA = \sigma 2\pi r dr\). Now, we want to find the electric potential and field due to this elemental charge.
03

Calculate the electric potential at point P due to the elemental ring

The distance from the elemental ring to point P is \(\sqrt{r^2+z^2}\). Using Coulomb's law, the electric potential at point P due to the elemental charge \(dq\) is \(dV =\dfrac{1}{4\pi\epsilon_{0}}\dfrac{dq}{\sqrt{r^{2} + z^{2}}}\).
04

Integrate to find the total electric potential at point P

To find the total electric potential at point P due to the entire disc, we need to integrate the electric potential from the center to the edge of the disc: \(V = \int_{0}^{a} dV = \dfrac{\sigma}{2\epsilon_{0}}\int_{0}^{a}\dfrac{r dr}{\sqrt{r^{2} + z^{2}}}\). Using a substitution \(u = \dfrac{r^{2}}{z^{2}}\), we get: \(V = \dfrac{\sigma}{2\epsilon_{0}}z \int_{0}^{a} \dfrac{du}{(u+1)^{\frac{3}{2}}}\), \(V = \dfrac{\sigma}{2\epsilon_{0}}z \left[-\dfrac{2}{\sqrt{u+1}}\right]_{0}^{a}\), \(V = \dfrac{\sigma}{2\epsilon_{0}}z \left(-\dfrac{2}{\sqrt{1 + a^{2}/z^{2}}}+\dfrac{2}{\sqrt{1}}\right)\).
05

Calculate the electric field at point P

By definition, the electric field is the negative gradient of the electric potential. Since we are only interested in the z-component on the axis of symmetry: \(E_{z} = -\dfrac{dV}{dz}\), \(E_{z} = -\dfrac{\sigma}{2\epsilon_{0}}\left(\dfrac{-2a^{2}}{(1+a^{2}/z^{2})^{\frac{3}{2}}z^{3}} + \dfrac{dz}{dz} \right) = -\dfrac{\sigma}{2\epsilon_{0}}\left(\dfrac{-2a^{2}}{(a^{2}+z^{2})^{\frac{3}{2}}} + 1 \right)\).
06

Analyze the electric field as \(a \rightarrow \infty\) while keeping \(\sigma\) constant

As the disc radius approaches infinity, the charged disc becomes an infinite plane with uniform charge density. In this situation, the electric field becomes constant: \(E_{z} = -\dfrac{\sigma}{2\epsilon_{0}}\left(\lim_{a \rightarrow \infty}\dfrac{-2a^{2}}{(a^{2}+z^{2})^{\frac{3}{2}}} + 1 \right)\), \(E_{z} = -\dfrac{\sigma}{2\epsilon_{0}}\left(-2 + 1 \right)\). Thus, the electric field converges to a constant value of \(E_{z} = \dfrac{\sigma}{2\epsilon_{0}}\), which is the result expected for an infinite uniformly charged plane.

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Key Concepts

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

Gauss's Law
Gauss's Law is central to understanding the behavior of electric fields when it comes to symmetrical charge distributions. It relates the flux of the electric field through a closed surface to the charge enclosed by that surface. In mathematical terms, the law is written as:
\[ \Phi_E = \oint_S \mathbf{E} \cdot d\textbf{A} = \frac{q_{\text{enc}}}{\epsilon_0} \]
where \( \Phi_E \) is the electric flux, \( \mathbf{E} \) is the electric field, \( d\textbf{A} \) is a vector representing an infinitesimal area on the closed surface, \( q_{\text{enc}} \) is the enclosed charge, and \( \epsilon_0 \) is the permittivity of free space. For a uniformly charged flat circular disc, Gauss's Law simplifies the process of finding the electric field at points along the axis of symmetry since the symmetry makes the field calculations tractable.
Electric Potential
The electric potential, often denoted by \( V \), is a scalar quantity that represents the potential energy per unit charge at a point in space. In relation to a charged disc, the potential at any point is the work done in bringing a unit positive charge from infinity to that point against the electric field. The electric potential due to a point charge is given by Coulomb's formula:
\[ V = \frac{1}{4\pi\epsilon_{0}}\frac{q}{r} \]
where \( q \) is the charge and \( r \) is the distance from the charge. For a continuous charge distribution, such as a disc, you must sum (integrate) the potential contributions of all infinitesimal elements of charge.
Cylindrical Coordinates
Cylindrical coordinates are a natural choice when dealing with problems involving radial symmetry, such as a charged disc. This coordinate system extends polar coordinates into three dimensions with variables \( (r, \theta, z) \), where \( r \) is the radial distance, \( \theta \) is the azimuthal angle, and \( z \) is the height. In the context of a charged disc, the symmetry enables us to focus on the radial \( (r) \) and height \( (z) \) dimensions while integrating to find the potential and field, thus simplifying the calculations.
Coulomb's Law
Coulomb's Law describes the force between two point charges. The law states that the electric force \( F \) between two charges is directly proportional to the product of the charges \( q_1 \) and \( q_2 \), and inversely proportional to the square of the distance \( r \) between them:
\[ F = k \frac{q_1 q_2}{r^2} \]
where \( k = \frac{1}{4\pi\epsilon_{0}} \) is the Coulomb's constant. For electric fields and potentials, Coulomb's Law helps determine the contribution of each infinitesimal charge element to the total effect.
Charge Density
Charge density is a measure of how much charge is distributed over a given area or volume. For surface charge density \( \sigma \), it is defined as the amount of charge per unit area. In the case of the charged disc, the problem specifies a uniform charge density, allowing us to express an elemental charge \( dq \) on a ring of radius \( r \) and thickness \( dr \) as:
\[ dq = \sigma dA = \sigma 2\pi r dr \]
This expression is crucial in setting up the integral for calculating both the electric field and potential caused by the disc's charge distribution.
Electric Field Calculations
To calculate the electric field produced by a charged disc, we start by considering the contribution of each elemental ring of charge, then integrate across the whole disc. The z-component of the electric field at a point on the axis is found by taking the negative derivative of the potential with respect to \( z \), which is derived using Coulomb's Law for infinitesimal charges:
\[ E_z = -\frac{dV}{dz} \]
For points away from the charge distribution or along its symmetry axis, these calculations often involve integral calculus to sum the contributions from each differential charge element.

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Most popular questions from this chapter

The potential \(\phi(\boldsymbol{r})=\left(q / 4 \pi \epsilon_{0} r\right) \mathrm{e}^{-\mu r}\) may be regarded as representing the effect of screening of a charge \(q\) at the origin by mobile charges in a plasma. Calculate the charge density \(\rho\) (at points where \(r \neq 0\) ) and find the total charge throughout space, excluding the origin.

Show that the work done in bringing two charges \(q_{1}\) and \(q_{2}\), initially far apart, to a separation \(r_{12}\) is \(q_{1} q_{2} / 4 \pi \epsilon_{0} r_{12}\). Write down the corresponding expression for a system of many charges. Show that the energy stored in the charge distribution is $$ V=\frac{1}{2} \sum_{j} q_{j} \phi_{j}\left(\boldsymbol{r}_{j}\right) $$ where \(\phi_{j}\left(\boldsymbol{r}_{j}\right)\) is the potential at \(\boldsymbol{r}_{j}\) due to all the other charges. Why does a factor of \(\frac{1}{2}\) appear here, but not in the corresponding expression for the potential energy in an external potential \(\phi(\boldsymbol{r}) ?\)

*Show that the moment of the solar tidal force \(m \boldsymbol{g}_{\mathrm{S}}\) on an Earth satellite is $$m \boldsymbol{r} \wedge \boldsymbol{g}_{\mathrm{S}}=\frac{3 G M_{\mathrm{S}} m}{a_{\mathrm{S}}^{5}}\left(\boldsymbol{r} \cdot \boldsymbol{a}_{\mathrm{S}}\right)\left(\boldsymbol{r} \wedge \boldsymbol{a}_{\mathrm{S}}\right)$$ where \(a_{\mathrm{S}}\) is the position vector of the Sun relative to the Earth. Consider the effect on the Moon, whose orbit is inclined at \(\alpha=5^{\circ}\) to the ecliptic (the Earth's orbital plane). Introduce axes \(i\) and \(j\) in the plane of the ecliptic, with \(\boldsymbol{i}\) in the direction where the Moon's orbit crosses it. Write down the positions of the Sun and Moon, relative to Earth, as functions of time. By averaging both over a month and over a year (i.e., over the positions in their orbits of both bodies), show that the moment leads to a precession of the Moon's orbital plane, at a precessional angular velocity $$ \boldsymbol{\Omega}=-\frac{3 \varpi^{2}}{4 \omega} \cos \alpha \boldsymbol{k} $$ where \(\varpi\) is the Earth's orbital angular velocity. Compute the precessional period. (Note that this calculation is only approximately correct.)

Find the quadrupole moment of a distribution of charge on the surface of a sphere of radius \(a\) with surface charge density \(\sigma=\sigma_{0}\left(\frac{3}{2} \cos ^{2} \theta-\frac{1}{2}\right)\). Find the total energy stored in this distribution.

Write down the potential energy of a pair of charges, \(q\) at \(\boldsymbol{a}\) and \(-q\) at the origin, in a field with potential \(\phi(\boldsymbol{r})\). By considering the limit \(a \rightarrow 0\), show that the potential energy of a dipole of moment \(d\) is \(V=-\boldsymbol{d} \cdot \boldsymbol{E} .\) If the electric field is uniform, when is this potential energy a minimum? Show that the dipole experiences a net moment, or couple, \(\boldsymbol{G}=\boldsymbol{d} \wedge \boldsymbol{E}\), and that in a non-uniform field there is also a net force, \(\boldsymbol{F}=(\boldsymbol{d} \cdot \boldsymbol{\nabla}) \boldsymbol{E} .\) (Take \(\boldsymbol{d}\) in the \(z\)-direction, and show that \(\boldsymbol{F}=d \partial \boldsymbol{E} / \partial z .)\)
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