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Chapter 2: Problem 30

Curl of a gradient The electric field equals the negative gradient of the potential, that is, \(\mathbf{E}=-\nabla \phi\). Show that this implies that the curl of \(\mathbf{E}\), which we can write as \(\nabla \times \mathbf{E}\), is identically zero. Do this by: (a) calculating \(\nabla \times \nabla \phi\) in Cartesian coordinates; (b) making judicious use of Stokes' theorem.

Short Answer

Expert verified
The curl of the electric field \(\mathbf{E}\), which equals the negative gradient of the potential, is identically zero. This was proven by direct calculation of the curl of the gradient in Cartesian coordinates and by applying Stokes' Theorem.

Step by step solution

01

Direct Calculation of \(\nabla \times \nabla \phi\) in Cartesian coordinates

Begin by expressing the gradient \(\nabla \phi\) and curl operator \(\nabla \times\) in Cartesian coordinates: \(\nabla \phi = (\frac{\partial \phi}{\partial x}, \frac{\partial \phi}{\partial y}, \frac{\partial \phi}{\partial z})\) and \(\nabla \times = (\frac{\partial}{\partial x},\frac{\partial}{\partial y},\frac{\partial}{\partial z}) \times\). Now calculate the curl \(\nabla \times\nabla \phi\) using the determinant of the resulting matrix. As all the second order mixed derivatives are equal (by Clairaut's theorem), the result is zero.
02

Prove Zero Curl Using Stokes' theorem

To use Stokes' theorem, initially define a vector field \(\vec{F}\) as \(\vec{F}=\nabla \phi\). By definition, Stokes' theorem relates the surface integral of the curl to the line integral around its boundary. That is, \(\int (\nabla \times \vec{F}) \cdot d\vec{S} = \oint \vec{F} \cdot d\vec{l}\). Here, \(d\vec{S}\) is the infinitesimally small surface vector and \(d\vec{l}\) is the infinitesimally small line vector. Now calculate the line integral of the gradient, which can be seen as total change in \(\phi\) in a closed loop. According to the fundamental theorem for line integrals, the total change is zero, implying that the curl must also be zero.
03

Zero Curl of the Electric Field

In the final step, transfer the argument made regarding \(\nabla \phi\) to \(\mathbf{E}\), since \(\mathbf{E}= -\nabla \phi\). By the linearity of the curl operator, you can move the negative sign in front: \(\nabla \times \mathbf{E} = \nabla \times (-\nabla \phi) = - \nabla \times \nabla \phi = 0\). Thus, it has been shown that the curl of the electric field \(\mathbf{E}\) is identically zero.

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Key Concepts

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

Electric Field
An electric field is a region around a charged particle where a force would be experienced by other charges. It is often denoted by the vector \(\mathbf{E}\). Electric fields influence how charged particles interact with each other, essentially dictating the direction in which they move.
  • An electric field points away from positive charges and toward negative charges.
  • The strength of the electric field diminishes with distance.
  • Its potential \(\phi\) relates directly to the field, with \(\mathbf{E} = -abla \phi\).
When considering the electric field as the negative gradient of a potential, the minus sign indicates the field direction is from higher to lower potential. This concept is crucial in understanding electric potential energy and the movement of electrons in fields.
Gradient
The gradient is a vector operator that signifies the rate and direction of change in a scalar field. In this exercise, we are dealing with the potential field \(\phi\), and the gradient tells us how this potential changes in space.
  • The gradient of \(\phi\) is calculated as \((\frac{\partial \phi}{\partial x}, \frac{\partial \phi}{\partial y}, \frac{\partial \phi}{\partial z})\).
  • It points in the direction of the steepest ascent of the scalar field.
  • The magnitude of the gradient gives the rate of change.
In terms of electric fields, the gradient shows the direction and rate of increase of electric potential. As shown in vector calculus, the curl of a gradient is always zero, which stems from the fact that mixed partials are equal, making \(abla \times abla \phi = 0\).
Curl
The curl is a vector operator that describes the infinitesimal rotation of a vector field. It helps us understand how a field can circulate around points in space. In this context, we are examining the curl of the electric field \(\mathbf{E}\).
  • The curl is calculated as \(abla \times \mathbf{E}\) using determinant methods.
  • A zero curl indicates the field has vanishing circulation.
  • A non-zero curl suggests the presence of rotational motion around a point.
As a property of gradient fields, the curl being \(0\) is significant in showing that electric fields derived from potentials are conservative; that is, their line integrals depend only on the endpoints and not on the path between them. This reinforces that a change in potential within a conservative field relies only on start and end points, very much like gravitational fields.
Stokes' Theorem
Stokes' theorem provides a powerful relationship between the surface integral of a curl and a line integral over its boundary. It connects the macroscopic circulation around a surface to the microscopic circulation inside a surface. This principle was crucial to prove the curl of a gradient is zero in vector calculus.
  • The theorem states: \(\int (abla \times \vec{F}) \cdot d\vec{S} = \oint \vec{F} \cdot d\vec{l}\).
  • The left side represents the net curl over a surface.
  • The right side relates to the net circulation around a closed loop.
By applying Stokes' theorem to the vector field \(\vec{F} = abla \phi\), we showed that the circulation around any closed path is zero, leading to the deduction that \(abla \times \mathbf{E} = 0\). This validation is pivotal in understanding how theoretical constructions logically align with physical principles, particularly within conservative fields.

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Most popular questions from this chapter

\(E\) and \(\phi\) for a slab A rectangular slab with uniform volume charge density \(\rho\) has thickness \(2 \ell\) in the \(x\) direction and infinite extent in the \(y\) and \(z\) directions. Let the \(x\) coordinate be measured relative to the center plane of the slab. For values of \(x\) both inside and outside the slab: (a) find the electric field \(E(x)\) (you can do this by considering the amount of charge on either side of \(x\), or by using Gauss's law); (b) find the potential \(\phi(x)\), with \(\phi\) taken to be zero at \(x=0\); (c) verify that \(\rho(x)=\epsilon_{0} \nabla \cdot \mathbf{E}(x)\) and \(\rho(x)=-\epsilon_{0} \nabla^{2} \phi(x)\).

Dividing the charge We have two metal spheres, of radii \(R_{1}\) and \(R_{2}\), quite far apart from one another compared with these radii. Given a total amount of charge \(Q\) which we have to divide between the spheres, how should it be divided so as to make the potential energy of the resulting charge distribution as small as possible? To answer this, first calculate the potential energy of the system for an arbitrary division of the charge, \(q\) on one sphere and \(Q-q\) on the other. Then minimize the energy as a function of \(q\). You may assume that any charge put on one of these spheres distributes itself uniformly over the surface of the sphere, the other sphere being far enough away so that its influence can be neglected. When you have found the optimum division of the charge, show that with that division the potential difference between the two spheres is zero. (Hence they could be connected by a wire, and there would still be no redistribution. This is a special example of a very general principle we shall meet in Chapter 3: on a conductor, charge distributes itself so as to minimize the total potential energy of the system.)

(a) A ring with radius \(R\) has charge \(Q\) uniformly distributed on it. It lies in the \(x y\) plane, with its center at the origin. Find the electric field at all points on the \(z\) axis. For what value of \(z\) is the field maximum? (b) Make a rough sketch of the equipotential curves everywhere in space (or rather, everywhere in a plane containing the \(z\) axis; you can represent the ring by two dots where it intersects the plane). Be sure to indicate what the curves look like very close to and very far from the ring, and how the transition from close to far occurs. (c) There is a particular \(z\) value (along with its negative) at which the equipotentials make the transition from concave up to concave down. Explain why this \(z\) value equals the \(z\) value you found in part (a). Hint: The divergence of \(\mathbf{E}\) is zero.

E for a line, from a cutoff potential Consider the electric field \(E\) due to an infinite straight wire with uniform linear charge density \(\lambda\). In Section \(1.12\) we found \(E\) by direct integration of Coulomb's law, and again by using Gauss's law. Find \(E\) here by calculating the potential and then taking the derivative. You will find that the potential (relative to infinity) due to an infinite wire diverges. But you can get around this difficulty by instead finding the potential due to a very long but finite wire of length \(2 L\), at a point lying on its perpendicular bisecting plane. Use a Taylor series to simplify your result, and then take the derivative to find \(E\). Explain why this procedure is valid, even though it cuts off an infinite amount from the potential.

Suppose eight protons are permanently fixed at the corners of a cube. A ninth proton floats freely near the center of the cube. There are no other charges around, and no gravity. Is the ninth proton trapped? Can it find an escape route that is all downhill in potential energy? Feel free to analyze this numerically/graphically.
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