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Chapter 1: Problem 11

Use Gauss's theorem to prove that at the surface of a curved charged conductor, the normal derivative of the electric field is given by $$ \frac{1}{E} \frac{\partial E}{\partial n}=-\left(\frac{1}{R_{1}}+\frac{1}{R_{2}}\right) $$ where \(R_{1}\) and \(R_{2}\) are the principal radii of curvature of the surface.

Short Answer

Expert verified
Using Gauss's theorem, the curvature correlates to the normal derivative of \( E \) at the surface.

Step by step solution

01

Recall Gauss's law

Gauss's law states that the electric flux \( \Phi \) through a closed surface is equal to the integral of the charge enclosed \( Q_{enc} \) divided by the permittivity of free space \( \varepsilon_0 \). Mathematically, it is given by: \[ \oint_S \mathbf{E} \cdot d\mathbf{A} = \frac{Q_{enc}}{\varepsilon_0} \] where \( \mathbf{E} \) is the electric field and \( d\mathbf{A} \) is the differential area vector.
02

Apply Gauss's Law to the Surface of a Conductor

Consider a small Gaussian pillbox at the surface of a conductor. The pillbox is partially inside the conductor and partially outside, with its flat surfaces parallel to the conductor's surface. As the conductor is charged, the electric field just outside is perpendicular to the surface.
03

Evaluate Electric Field at the Surface

On the surface of the conductor, the electric field is \( E \) and points normal to the surface. Inside the conductor, the electric field is zero, because conductors in electrostatic equilibrium do not have any electric field inside. Thus, the electric field contribution is only from the surface facing outward.
04

Express the Electric Field Due to Charge Density

Consider the surface charge density \( \sigma \) at the conductor's surface. The electric field just outside the surface can be related to the surface charge density using \( E = \frac{\sigma}{\varepsilon_0} \), given that the electric field inside is zero.
05

Relate Curvature with Electric Field Derivative

The curvature of the surface affects the change of the electric field. The total curvature at a point is given by \( \frac{1}{R_1} + \frac{1}{R_2} \), where \( R_1 \) and \( R_2 \) are principal radii. The change in the electric field normal to the surface, or \( \frac{\partial E}{\partial n} \), relates to the curvature: \[ \frac{1}{E} \frac{\partial E}{\partial n} = -\left(\frac{1}{R_1} + \frac{1}{R_2} \right) \]. This is derived from the geometric relation between curvature and electric field change on a curved surface.

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Key Concepts

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

Electric Field
The electric field is a fundamental concept in electromagnetism, representing the force per unit charge exerted on a positive test charge in the vicinity of other charges. It is a vector field, meaning it has both magnitude and direction. An electric field exists around charged objects, and its strength is described by the equation\[E = \frac{F}{q}\]where \( E \) is the electric field, \( F \) is the force experienced by a test charge, and \( q \) is the magnitude of the test charge. Fields point away from positive charges and toward negative charges.
  • Electric fields are represented by lines that start from positive charges and end at negative charges.
  • The denser the line distribution, the stronger the electric field.
In the context of curved conductors, the electric field outside the conductor is perpendicular to its surface. This is because charges redistribute themselves to maintain electrostatic equilibrium, meaning no net electric field exists inside a conductor under static conditions.
Curved Conductor
A curved conductor is a conductor whose surface is not flat. The unique geometry affects how electric charges distribute themselves on its surface. When a conductor surface is curved, the electric field just outside the surface is always normal, or perpendicular, to the surface.
  • Curvature causes variations in electric field strength due to changes in surface geometry.
  • Areas with more curvature will often have higher surface charge densities.
This property is critical when applying Gauss's theorem to determine electric fields around charged conductors. The surface curvature will impact how charges behave, notably affecting how much charge accumulates at each point on the surface, following the principle of electrostatic equilibrium.
Charge Density
Charge density \( \sigma \) defines how charge is distributed on a conductor's surface. It's a measure of the quantity of charge per unit area at any point on the surface. In mathematical terms, it is expressed as:\[\sigma = \frac{Q}{A}\]where \( \sigma \) is the surface charge density, \( Q \) is the total charge, and \( A \) is the area. The nature of curved conductors leads to non-uniform charge density, since curvature affects how charges spread.
  • Higher curvature areas typically have higher charge densities.
  • Charge density directly influences the strength of the electric field just outside a conductor.
This relationship is critical in calculating electric fields using Gauss's law, especially when a conductor is not flat.
Understanding charge density helps us predict electric field intensity and potential effects at any given surface point.
Surface Charge
Surface charge refers to electrical charges situated on the surface of a conductor. When a conductor is charged, the charges swiftly move to the surface due to repulsion, redistributing until they reach equilibrium. The surface charge density \( \sigma \) tells us how much charge is on each unit area of the surface. For conductors at electrostatic equilibrium:
  • The inside has no net charge, with all excess charge residing on the surface.
  • The electric field right outside the conductor can be given by the equation \( E = \frac{\sigma}{\varepsilon_0} \).
Surface charge is essential not only for electric field calculations but also in practical applications such as capacitors. In these, surface charge across different plates allows for the storage of electrical energy. Understanding surface charge is crucial for precise predictions regarding how the conductor will interact with other charges or fields it encounters.
Curvature
Curvature refers to how much a geometric surface (such as a conductor) deviates from being flat. In the context of electricity and magnetism, curvature significantly affects how charges distribute and how the electric field behaves near a surface. For a curved conductor with principal radii \(R_1\) and \(R_2\), the total curvature at a point is given by:\[\frac{1}{R_1} + \frac{1}{R_2}\]This influences not only charge distribution but also electric field intensity.
  • Areas with high curvature tend to accumulate more charge.
  • The electric field's derivative (how it changes) is directly related to surface curvature as shown by the equation involving these curvatures.
In electrostatics, understanding curvature assists in predicting how fields change across different parts of a charged body, shedding light on complex interactions in various applications like shielding and capacitance.

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Most popular questions from this chapter

Two infinite grounded parallel conducting planes are separated by a distance \(d\). \(\mathrm{A}\) point charge \(q\) is placed between the planes. Use the reciprocation theorem of Green to prove that the total induced charge on one of the planes is equal to \((-q)\) times the fractional perpendicular distance of the point charge from the other plane. (Hint: As your comparison electrostatic problem with the same surfaces choose one whose charge densities and potential are known and simple.)

Two-dimensional relaxation calculations commonly use sites on a square lattice with spacing \(\Delta x=\Delta y=h\), and label the sites by \((i, j)\), where \(i, j\) are integers and \(x_{i}=\) \(i h+x_{0}, y_{j}=j h+y_{0} .\) The value of the potential at \((i, j)\) can be approximated by the average of the values at neighboring sites. [Recall the relevant theorem about harmonic functions.] But what average? (a) If \(F(x, y)\) is a well-behaved function in the neighborhood of the origin, but not necessarily harmonic, by explicit Taylor series expansions, show that the "cross" sum $$ S_{e}=F(h, 0)+F(0, h)+F(-h, 0)+F(0,-h) $$ can be expressed as $$ S_{c}=4 F(0,0)+h^{2} \nabla^{2} F+\frac{h^{4}}{12}\left(F_{x \ldots x}+F_{y m y}\right)+O\left(h^{6}\right) $$ (b) Similarly, show that the "square" sum, $$ S_{\mathrm{S}}=F(h, h)+F(-h, h)+F(-h,-h)+F(h,-h) $$ can be expressed as $$ S_{\mathrm{S}}=4 F(0,0)+2 h^{2} \nabla^{2} F-\frac{h^{4}}{3}\left(F_{\mathrm{xxx}}+F_{y y y}\right)+\frac{h^{4}}{2} \nabla^{2}\left(\nabla^{2} F\right)+O\left(h^{6}\right) $$ Here \(F_{\text {mer is the fourth partial derivative of }} F\) with respect to \(x\), evaluated at \(x=0, y=0\), etc. If \(\nabla^{2} F=0\), the averages \(S_{d} / 4\) and \(S_{\sqrt{ }} / 4\) each give the value of \(F(0,0)\), correct to order \(h^{3}\) inclusive. Note that an improvement can be obtained by forming the "improved" average, $$ \langle\langle F(0,0)\rangle\rangle=\frac{1}{5}\left[S_{e}+\frac{1}{4} S_{5}\right] $$ where $$ \langle(F(0,0))\rangle=F(0,0)+\frac{3}{10} h^{2} \nabla^{2} F+\frac{h^{4}}{40} \nabla^{2}\left(\nabla^{2} F\right)+O\left(h^{6}\right) $$ If \(\nabla^{2} F=0\), then \(S\) gives \(F(0,0)\), correct to order \(h^{3}\) inclusive, For Poisson's equation, the charge density and its lowest order Laplacian can be inserted for the same accuracy.

Prove Thomson's theorem: If a number of surfaces are fixed in position and a given total charge is placed on each surface, then the electrostatic energy in the region bounded by the surfaces is an absolute minimum when the charges are placed so that every surface is an equipotential, as happens when they are conductors.

Prove Green's reciprocation theorem: If \(\Phi\) is the potential due to a volume-charge density \(\rho\) within a volume \(V\) and a surface-charge density \(\sigma\) on the conducting surface \(\$$ bounding the volume \)V\(, while \)\Phi^{\prime}\( is the potential due to another charge distribution \)\rho^{\prime}\( and \)\sigma^{\prime}$, then $$ \int_{V} \rho \Phi^{\prime} d^{3} x+\int_{S} \sigma \Phi^{\prime} d a=\int_{V} \rho^{\prime} \Phi d^{3} x+\int_{s} \sigma^{\prime} \Phi d a $$

Prove the mean value theorem: For charge-free space the value of the electrostatic potential at any point is equal to the average of the potential over the surface of any sphere centered on that point.
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