PHYS 212: Electricity and Magnetism

Electric Charge

• Electric charges cause objects to be repelled or attracted to each other.

Electric Charge - Borneo Amber Example

• Borneo amber, mined in Sabah, Malaysia, gains electrons when rubbed with fur, acquiring a net negative charge.
• The fur loses electrons and becomes positively charged.
• This charge separation occurs because materials have different affinities for electrons.
• Initially, both amber and cloth are neutral with equal positive and negative charges.
• Rubbing transfers a tiny fraction of charge, resulting in net charges.
• The absolute value of net positive and negative charges remains equal after separation.

Observations of the Electric Force

• The electric force acts without physical contact.
• The force can be attractive or repulsive:
  • Like charges repel.
  • Opposite charges attract.
  • These are called electrostatic repulsion and attraction.
• Not all objects are affected by this force.
• The force magnitude decreases rapidly with increasing separation distance.

Properties of Electric Charge

• Charge is quantized:
  • The smallest possible charge is e = 1.602 × 10^{-19} C.
  • Macroscopic charged objects have charge q = ne, where n is an integer.
• Charge magnitude is independent of type:
  • Electrons: -1.602 × 10^{-19} C
  • Protons: +1.602 × 10^{-19} C
  • These values are exactly equal.
• Charge is conserved in closed systems:
  • Charge is neither created nor destroyed; it is only transferred.

Source of Charges: The Structure of the Atom

• A simplified hydrogen atom model shows a positively charged nucleus (proton) surrounded by an electron cloud.
• The electron cloud's charge is equal and opposite to the nucleus's charge.
• The electron's location is not definite, represented as a cloud.
• Macroscopic matter contains immense numbers of atoms/molecules and even greater numbers of individual charges.

Conductors and Insulators

• Conductors:
  • Charge flows freely (metals, water, human body).
• Insulators:
  • Almost no charge flows (wood, glass, plastic, rubber).

Conductors and Insulators - Power Adapter Example

• Power adapters use metal wires and connectors for electrical conduction.
• Conducting wires allow electrons to move freely through cables, shielded by rubber and plastic.
• Insulators prevent electric charge from escaping outward.

Induced Polarization in a Conductor

• A positively charged glass rod near a neutral conducting sphere attracts negative charge, leaving the other side positively charged.
• The sphere remains electrically neutral overall but has a charge distribution.
• This charge distribution allows it to exert electric force on nearby charges and be attracted to the glass rod.

Induced Polarization in an Insulator

• Both positively and negatively charged objects attract a neutral object by polarizing its molecules.
• A positive object near a neutral insulator polarizes its molecules, causing a slight electron shift.
• Unlike charges are brought nearer, and like charges moved away.
• The electrostatic force decreases with distance, resulting in a net attraction.
• A negative object produces the opposite polarization but again attracts the neutral object.

Charging by Induction

• Two neutral metal spheres in contact are insulated from the world.
• A positively charged glass rod near the left sphere attracts negative charge, leaving the other sphere positively charged.
• The spheres are separated before removing the charged rod, separating charges.
• The spheres retain net charges after removing the charged rod, without direct contact.

Charging by Induction via Ground Connection

• A positively-charged rod polarizes a neutral metal sphere.
• The sphere is grounded, attracting electrons from Earth.
• The ground connection is disconnected.
• The positively-charged rod is removed, leaving an induced negative charge on the sphere.

Coulomb's Law

• The magnitude of the electrostatic force between two point charges is:
  • Directly proportional to the product of charge magnitudes.
  • Inversely proportional to the square of the distance between them.
• F \propto \frac{q1 q2}{r^2}
• The force acts along the line joining them:
  • Like signs: repulsive force.
  • Different signs: attractive force.

Coulomb's Law - Formula

• The magnitude of the electric force F{12} exerted on charge q1 due to q2 at a distance r{12} is:
  • |F{12}| = k \frac{|q1 q2|}{r{12}^2}
  • Where k = \frac{1}{4 \pi \epsilon0} ≈ 8.99 × 10^9 Nm^2/C^2 and \epsilon0 = 8.85 × 10^{-12} C^2/Nm^2
• The direction is along the axis connecting the two charges.
  • \hat{r} is a unit vector along the axis.

Coulomb's Law with Multiple Source Charges

• The net electric force on a test charge Q due to N fixed source charges q_i is found by superposition:
  • FQ = F{Q1} + F{Q2} + … + F{QN} = \sum{i=1}^{N} F{Qi}
  • FQ = kQ \sum{i=1}^{N} \frac{qi}{ri^2} \hat{r_i}
  • Where \vec{ri} = ri \hat{r_i} are the displacements.
• Q is referred to as a test charge, used to "test" the net force.

Problem Solving with Coulomb's Law and Multiple Source Charges

• Solving for the resultant force on a test charge due to multiple source charges:
  • Read the prompt carefully.
  • Draw a diagram and label all of the charges and distances.
  • Draw a Free Body Diagram (FBD) for the test charge.
  • Use Coulomb's Law to compute the magnitudes and directions of each force vector on the test charge.
  • Determine the net horizontal and vertical force acting on the test charge by decomposing each vector into its x and y components, and adding them accordingly:
    • F{Qx} = \sum{i=1}^{N} F{Qix} = F{Q1x} + … + F_{QNX}
    • F{Qy} = \sum{i=1}^{N} F{Qiy} = F{Q1y} + … + F_{QNY}
  • Determine the resultant force vector magnitude and direction:
    • |FQ| = \sqrt{(F{Qx})^2 + (F_{Qy})^2}
    • \theta = tan^{-1} \frac{F{Qy}}{F{Qx}}

Electric Field with Multiple Source Charges

• We have N fixed source charges that apply N electrostatic forces on a test charge Q. The net force on Q is:
  • FQ = F{Q1} + F{Q2} + … + F{QN} = kQ \sum{i=1}^{N} \frac{qi}{ri^2} \hat{ri}
• The expression in parentheses is independent of test charge Q, so we can define it as the electric field E:
  • F_Q = QE
• The electric field E(r) at any position r, due to N fixed source charges, is given by:
  • E(r) = k \sum{i=1}^{N} \frac{qi}{ri^2} \hat{ri}
  • Where the vectors \vec{ri} = ri \hat{r_i} are the displacements from the position of the ith source charge to the position r = (x, y, z). that the field is being evaluated at.

Electric Field with Multiple Source Charges - Superposition

• Electric field due to N fixed source charges also follows the principle of superposition:
  • E(r) = E1(r) + E2(r) + … + EN(r) = \sum{i=1}^{N} Ei(r) = k \sum{i=1}^{N} \frac{qi}{ri^2} \hat{r_i}
• The electric field E is a vector field; i.e., a vector quantity that is a function of position, relative to the source of the field.
• By convention, all electric fields E point away from positive source charges and point toward negative source charges.

Electric Fields of Continuous Charge Distributions

• The charge differential elements are in the configuration of:
  • line charge
  • sheet of charge
  • volume of charge
• Some components of the total electric field cancel out, with the remainder resulting in a net electric field.

Electric Fields of Continuous Charge Distributions (Densities)

• Linear Charge Density: \lambda = \frac{Q}{L} = \frac{dq}{dl} ⇒ dq = \lambda dl
• Surface Charge Density: \sigma = \frac{Q}{A} = \frac{dq}{dA} ⇒ dq = \sigma dA
• Volume Charge Density: \rho = \frac{Q}{V} = \frac{dq}{dv} ⇒ dq = \rho dv

Electric Fields of Continuous Charge Distributions - How to Solve

• Problem-Solving Process:
  • Sketch the applicable diagram
  • Select a representative small section of the continuous charge distribution and label the section as dq.
  • Express the differential charge dq in terms of the appropriate charge density; e.g., dq = \lambda dx for a line of charge, or dq = \lambda R d\theta for a ring of charge.
  • Draw a line from the dq to the field evaluation point (mark the point P).
  • At point P, draw a differential electric field vector caused by the small element dq: dE = k \frac{dq}{r^2}
  • Utilize the right triangle that includes the charge element, the point P and the perpendicular distance away from the object, label one of the acute angles as theta.
  • Replace r and any \cos \theta or \sin \theta with their equivalents in terms of x or y.
  • Set up the integral and solve.

Electric Field Vectors

• The electric field of a positive point charge. A large number of field vectors are shown.
• Like all vector arrows, the length of each vector is proportional to the magnitude of the field at each point.

Electric Field Lines - Properties

• Electric field lines convey essential information about the field:
  • The direction of the field at every point is simply the direction of the field vector at that same point. In other words, at any point in space, the field vector at each point is tangent to the field line at that same point. The arrowhead placed on a field line indicates its direction.
  • The magnitude of the field is indicated by the field line density - the number of field lines per unit area passing through a small cross-sectional area perpendicular to the electric field. This field line density is drawn to be proportional to the magnitude of the field at that cross-section.

Electric Field Lines - Examples

• Two typical electric field diagrams:
  • a positive point charge
  • a dipole.
• In both diagrams, the magnitude of the field is indicated by the field line density.
• Field vectors (not shown) are everywhere tangent to the field lines.

Electric Field Lines - Guideline

• Drawing Electric Field Lines
  • Electric field lines either originate on positive charges or come in from infinity, and either terminate on negative charges or extend out to infinity.
  • The number of field lines originating or terminating at a charge is proportional to the magnitude of that charge. A charge of 2q will have twice as many lines as a charge of q.
  • At every point in space, the field vector at that point is tangent to the field line at that same point.
  • The field line density at any point in space is proportional to (and therefore is representative of) the magnitude of the field at that point in space.
  • Field lines can never cross. Since a field line represents the direction of the field at a given point, if two field lines crossed at some point, that would imply that the electric field was pointing in two different directions at a single point. This in turn would suggest that the (net) force on a test charge placed at that point would point in two different directions. Since this is obviously impossible, it follows that field lines must never cross.

Electric Field Lines - Typical Diagrams

• Three typical electric field diagrams:
  • A dipole.
  • Two identical charges.
  • Two charges with opposite signs and different magnitudes (+3q, -1q).

Electric Dipoles

• For an electric dipole ±q in an external electric field E, although the net force on the dipole is zero, the net torque \tau is not. If d is the distance (vector) between the charges (from negative to positive), the dipole rotates to become aligned with the external field with torque:
  • \tau = d × F = d × qE = qd × E = p × E
    • Where the vector p = qd is the dipole moment of the dipole.

Electric Dipoles - Induced Dipole

• Induced dipole: When a neutral atom is placed in an external electric field, the external field causes oppositely directed forces on the positive nucleus of the atom versus the negative electrons that surround the nucleus, resulting in a new charge distribution of the atom, and therefore, an induced dipole moment that is aligned with the external field.

Electric Dipoles - Net Electric Field

• The net electric field is the vector sum of the field of the dipole plus the external field: once the alignment of the dipole with the field is complete, the net effect is a decrease in the total field in the regions between the dipole charges.

Electric Flux - Introduction

• The concept of flux describes how much of something goes through a given area. More formally, it is the dot product of a vector field with an area.
• The numerical value of the electric flux thus depends on the magnitudes of the electric field and the area, as well as the relative orientation of the area with respect to the direction of the electric field.
• Electric flux can thus be conceptualized as a measure of the number of electric field lines passing through an area: orientation matters
  • The larger the area, the more field lines go through it and, hence, the greater the flux; similarly, the stronger the electric field is (represented by a greater density of lines), the greater the flux.
  • On the other hand, if the area rotated so that the plane is aligned with the field lines, none will pass through and there will be no flux.

Electric Flux - Illustration

• (a) A planar surface S1 of area A1 is perpendicular to the electric field E \hat{j}. N field lines cross surface S1, and the electric flux \Phi = EA1.
• (b) A surface S2 of area A2 whose projection onto the xz-plane is S1. The same number of field lines cross each surface, and the electric flux is the same for both S1 and S2: \Phi = EA2 \cos \theta = EA1. Defining \hat{n2} as the unit vector normal to S2, the flux can be written as the following dot product: \Phi = E \cdot A2 = EA_2 \cos \theta

Electric Flux - Area Vector

• The area vector A of a surface of area A has a magnitude of A and a direction that is along the normal (perpendicular) to the surface:
  • A = A \hat{n}
    • where \hat{n} is the unit normal vector to the surface.
• An open surface always has two possible directions for the area vector.
• For a closed surface, which encloses a volume, the area vector direction is always chosen to be the outward normal at every point to be consistent with the sign convention for electric charge.

Electric Flux - Uniform Field

• Using the area vector, the electric flux of a uniform field \vec{E} through a flat surface of area \vec{A} becomes the (scalar) dot product: \Phi = E \cdot A = |E||A| \cos \theta = EA \cos \theta
• Electric flux through a cube, placed between two charged plates:
  • \Phi_{top} = E \cdot A = EA (1) > 0
  • \Phi_{bottom} = E \cdot A = EA (-1) < 0
  • \Phi_{sides} = E \cdot A = EA (0) = 0
• The net electric flux through the cube is the sum of fluxes through the six faces, which, in this case, is equal to zero.

Electric Flux - Smooth Non-Flat Surface

• Any smooth, non-flat surface S can be replaced by a collection of tiny, approximately flat surfaces (patches) Si, each with area vector \Delta Ai. If the patch size is made sufficiently small, we may approximate the electric field over any given patch as uniform. If Ei is the average electric field over the ith patch, the electric flux through this patch \Phii = Ei \cdot \Delta Ai, and the net flux through the entire surface is approximately equal to the sum over all N patches:
  • \PhiE ≈ \sum{i=1}^{N} \Phii = \sum{i=1}^{N} Ei \cdot \Delta Ai
• Letting N → ∞ (and \Delta A_i → 0), the sum becomes the following surface integral over S:
  • \Phi = \intS E \cdot \hat{n} dA = \intS E dA
• Additionally, if S is a closed surface, the surface integral is written as follows:
  • \Phi = \oint E dA

Explaining Gauss's Law

• To investigate what happens to the electric flux if there are some charges inside an enclosed volume, consider a familiar example: a closed spherical surface surrounding a point charge q. We know that the electric field at a distance R from the charge is given by:
  • E = \frac{1}{4 \pi \epsilon_0} \frac{q}{R^2} \hat{r} = K \frac{q}{R^2} \hat{r}
  • \Phi = \oint E \cdot dA = \oint E \hat{n} dA = \frac{1}{4 \pi \epsilon_0} \frac{q}{R^2} (\hat{r} \cdot \hat{n}) dA
  • \Phi = \frac{1}{4 \pi \epsilon_0} \frac{q}{R^2} \oint dA
  • \Phi = \frac{1}{4 \pi \epsilon0} \frac{q}{R^2} (4 \pi R^2) = \frac{q}{\epsilon0}
• Therefore, the electric flux is independent of the size of the spherical surface.

Gauss's Law - Electric Flux

• The electric flux through a closed spherical surface is independent of the radius of the surface.
• the electric flux through spherical surfaces of radii R1 and R2 enclosing a charge q are equal, independent of the size of the surface, since all electric field lines that traverse one surface from the inside to outside direction also traverse the other surface in the same direction.

Gauss's Law - Field Lines

• Understanding electric flux in terms of field lines:
  • (a) The electric flux through a closed surface due to a charge outside that surface is zero.
  • (b) Charges are enclosed, but because the net charge enclosed is zero, the net flux through the closed surface is also zero.
  • (c) The shape and size of the surfaces that enclose a charge does not matter because all surfaces enclosing the same charge have the same flux.

Gauss's Law - Generalization

• Gauss's Law generalizes this result to the case of any number of charges and any location of the charges in the space inside the closed surface.
• Gauss's Law: The flux (\Phi) of the electric field (\vec{E}) through any closed surface S, also called a Gaussian surface, is equal to the net charge enclosed (q{enclosed}) divided by the permittivity of free space (\epsilon0):
  • \Phi = \oint E \cdot dA = \frac{q{enclosed}}{\epsilon0}
• This equation holds for charges of either sign: since the area vector of a closed surface is defined to point outward, if the enclosed charge is negative, then the flux through either S or S' is negative.

Gauss's Law - Integration - Continuous Distribution

• Note: The Gaussian surface is a mathematical construct that may be of any shape, provided that it is closed. However, since our goal is to integrate the flux over it, we tend to choose shapes that are highly symmetrical.
• For discrete point charges, then we simply add them to find their net sum:
  • \Phi = \frac{\sum (qi){enc}}{\epsilon0} = \frac{q1+q2 + q5 +q6}{\epsilon0}
• If the charge is described by a continuous distribution, then we need to integrate appropriately to find the total charge that resides inside the enclosed volume:
  • \Phi = \frac{1}{\epsilon0} \int{Venc} \rho dV
  • \Phi = \frac{1}{\epsilon_0} \int dA

Applying Gauss's Law - How to Solve

• Problem-Solving Strategy:
  • (1) Identify the spatial symmetry of the charge distribution. This is an important first step that allows us to choose the appropriate Gaussian surface. As examples, an isolated point charge has spherical symmetry, and an infinite line of charge has cylindrical symmetry.
  • (2) Choose a Gaussian surface with the same symmetry as the charge distribution and identify its consequences. With this choice, E \cdot \hat{n} may be easily determined over the Gaussian surface.
  • (3) Evaluate the integral \Phi = \oint E \cdot \hat{n} dA over the Gaussian surface; i.e., calculate the flux through the surface. The symmetry of the Gaussian surface allows us to factor E \cdot \hat{n} outside the integral.
  • (4) Determine the amount of charge enclosed by the Gaussian surface. This is an evaluation of the right-hand side of the equation representing Gauss's Law; it is often necessary to perform an integration to obtain the net enclosed charge.
  • (5) Evaluate the electric field of the charge distribution. The field may now be found using the results of steps 3 and 4.

Applying Gauss's Law - Geometries

• Common Gauss's Law Geometries
  • Point Charge
  • Line Charge
  • Sheet of Charge
  • Conducting Cylinder
  • Conducting Sphere
  • Sphere with Uniform Charge Density
  • Cylinder with Uniform Charge Density
  • Charge on Surface of Conductor
  • Charged Conducting Plates

Applying Gauss's Law - Spherical Symmetry

• Spherical Symmetry: Occurs when the charge density function depends only on the radial distance from the center.
  • (a) Uniform charge density \rho_0 throughout the sphere
  • (b) Upper half of the sphere has a different charge density from the lower half
  • (c) Charges are in spherical shells of different charge densities, which means that charge density is only a function of the radial distance from the center; i.e., \rho = \rho(r).

Applying Gauss's Law - Spherical Symmetry 2

• Spherical Symmetry: Computing Electric Flux and the Electric Field
  • The electric field E_p at any point P of the spherical Gaussian surface for a spherically symmetrical charge distribution is parallel to the area element vector at that point, and the electric flux is thus simply the product of the magnitude of electric field and the value of the area. In addition, the magnitude of the electric field must be the same everywhere on a spherical surface concentric with the distribution.
  • \vec{Ep}(r) = Ep(r) \hat{r}
  • \Phi = \oint E \cdot \hat{n} dA = \oint Ep dA = Ep(4 \pi r^2) = \frac{q{enc}}{\epsilon0}
  • Ep(r) = \frac{1}{4 \pi \epsilon0} \frac{q_{enc}}{r^2}
  • \vec{Ep}(r) = \frac{1}{4 \pi \epsilon0} \frac{q_{enc}}{r^2} \hat{r}

Applying Gauss's Law - Spherical Symmetry 3

• Spherical Symmetry: Computing Enclosed Charge
  • A spherically symmetrical charge distribution and the Gaussian surface used for finding the field (a) inside and (b) outside the distribution.
  • q{enc} = q{within r < R}
  • q{enc}= q{total}
  • E{in}(r) = \frac{1}{4 \pi \epsilon0} \frac{q_{within r < R}}{r^2} \hat{r}
  • E{out}(r) = \frac{1}{4 \pi \epsilon0} \frac{q_{total}}{r^2} \hat{r}

Applying Gauss's Law - Cylindrical Symmetry

• Cylindrical Symmetry: Occurs when the charge density does not depend on the polar angle of the cross-section or along the axis.
  • (a) Uniform charge density \rho_0 throughout the cylinder.
  • (b) Upper half of the cylinder has a different charge density from the lower half.
  • (c) Left half of the cylinder has a different charge density from the right half.
  • (d) Charges are constant in different cylindrical rings, so the charge density depends only on the radial distance from the central axis; i.e., \rho = \rho(r).

Applying Gauss's Law - Cylindrical Symmetry 2

• Cylindrical Symmetry: Computing Electric Flux and the Electric Field
  • The electric field in a cylindrically symmetrical charge distribution depends only on the radial distance from the axis. Therefore, the electric field E_p at any point P of the cylindrical Gaussian surface for a cylindrically symmetrical charge distribution is either parallel (on the lateral surface) or perpendicular (on the top and bottom surfaces) to the area element vector at that point.
  • \vec{Ep}(r) = Ep(r) \hat{r}
  • \Phi = \oint E \cdot \hat{n} dA = \int{Side} E \cdot \hat{n} dA + \int{Top} E \cdot \hat{n} dA + \int_{Bottom} E \cdot \hat{n} dA
  • \Phi = \int{Side} Ep(r) (\hat{r} \cdot \hat{r}) dA + \int{Top} Ep(r) (\hat{r} \cdot \hat{z}) dA + \int{Bottom} Ep(r) (\hat{r} \cdot - \hat{z}) dA
  • \Phi = Ep (r) \int{Side} dA + 0 + 0 = E_p (2 \pi r L)
  • Ep(r) = \frac{q{enc}}{2 \pi \epsilon_0 L r}
  • \vec{Ep}(r) = \frac{q{enc}}{2 \pi \epsilon_0 L r} \hat{r}

Applying Gauss's Law - Cylindrical Symmetry 3

• Cylindrical Symmetry: Computing Enclosed Charge:
  • Let R be the radius of the cylinder having a cylindrically symmetrical charge distribution, and let the field point P be at a distance r from the axis.
    • r < R: q{enc} = q{within r < R}
    • r > R: q{enc} = q{total}
  • Using the linear charge density of enclosed charge, \lambda{enc} = \frac{q{enc}}{L}
  • Ep(r) = \frac{q{enc}}{2 \pi \epsilon_0 L r} \hat{r}
  • E{in}(r) = \frac{\lambda{within r < R}}{2 \pi \epsilon_0 r} \hat{r}
  • E{out}(r) = \frac{\lambda{total}}{2 \pi \epsilon_0 r} \hat{r}

Applying Gauss's Law - Planar Symmetry

• Planar Symmetry: Occurs when charges are uniformly spread over a large flat surface; all points in a plane parallel to the plane of charge are identical with respect to the charges.
• At any point P, the components of the electric field parallel to a plane of charges cancel out the two charges located symmetrically from the field point P.
• Therefore, the field at any point is pointed vertically from the plane of charges.
• Therefore, if the charge distribution is in the xy-plane, the electric field at P(x, y, z) depends only on the distance z from the plane and has a direction either toward the plane or away from the plane:
  • \vec{E} = E(z) \hat{z}
    • where z is the distance from the plane and \hat{z} is the unit vector normal to the plane. Note that in this system E(-z) = E(z), although of course they point in opposite directions.

Applying Gauss's Law - Planar Symmetry 2

• Planar Symmetry: Computing Electric Flux and the Electric Field
  • For a thin charged sheet, we use a Gaussian box (straddling the xy-plane) for finding the electric field at the field point P. The area vector to each face of the box is from inside the box to outside. On two faces of the box (top and bottom), the electric fields are parallel to the area vectors, and on the other four faces (sides), the electric fields are perpendicular to the area vectors.
  • \Phi = \oint E \cdot \hat{n} dA = \int{Side I} E \cdot \hat{n} dA + \int{Side II} E \cdot \hat{n} dA + …
  • \Phi = \int{Side I} Ep (+\hat{z}) \cdot (+\hat{z}) dA + \int{Side II} Ep (-\hat{z}) \cdot (-\hat{z}) dA + 4(0)
  • \Phi = Ep \int{Side I} dA + Ep \int{Side II} dA
  • \Phi = Ep A + Ep A = 2E_p A
  • Ep = \frac{q{enc}}{2 \epsilon_0 A}
  • Ep = \frac{\sigma0}{2 \epsilon_0}
  • \vec{Ep}(r) = \frac{\sigma0}{2 \epsilon_0} (± \hat{z})

Conductors in Electrostatic Equilibrium

• Polarization of a metallic sphere by an external point charge +q. The near side of the metal has an opposite surface charge compared to the far side of the metal. The sphere is said to be polarized. When the external charge is removed, the polarization of the metal also disappears.

Conductors in Electrostatic Equilibrium 2

• In the presence of an external charge +q, the charges in a metal redistribute and cause an induced electric field due to the polarization. The electric field at any point has three contributions, from +q and the induced charges - \sigmaA and \sigmaB.
• The redistribution of charges is such that the sum of the three contributions at any point P inside the conductor is
  • EP = Eq + EA + EB = 0
• Due to Gauss's Law, there is no net charge enclosed by a Gaussian surface that is solely within the volume of the conductor at equilibrium; i.e., q{enc}=0 and thus the electric field E{net}=0 (at all points inside a conductor).

Conductors in Electrostatic Equilibrium 3

• Inside a Cavity:
  • A charge inside a cavity in a metal. The distribution of charges at the outer surface does not depend on how the charges are distributed at the inner surface, since the electric field inside the body of the metal is zero. That magnitude of the charge on the outer surface does depend on the magnitude of the charge inside, however.