Electrostatics

Big Picture

Electric charge is a fundamental property of matter. There are two types — positive and negative. Like charges repel; opposite charges attract. The force between charges follows Coulomb's inverse-square law. A charge creates an electric field in the space around it; another charge placed in that field experiences a force. The concept of electric potential (voltage) provides a scalar way to describe the energetic landscape created by charges.

ELI-10: Explain It Like I'm 10

Rub a balloon on your hair and it sticks to the wall. You have moved tiny invisible particles called electrons from your hair to the balloon. The balloon now has extra negative charge. The wall's surface responds by shifting its own charges, creating an attraction. This is static electricity — charges at rest. Positive and negative charges pull toward each other; same charges push apart. Everything electrical — lights, computers, lightning — starts with understanding how charges behave.

10.1 Electric Charge

Core Ideas

• Two types: positive (+) and negative (−). Protons are positive; electrons are negative.
• Quantization: Charge comes in integer multiples of the elementary charge \(e = 1.60 \times 10^{-19}\ \text{C}\).
• Conservation: Total charge in an isolated system is constant.
• Conductors allow charge to flow freely (metals). Insulators do not (rubber, glass).
• Charging by contact: direct transfer of charge. Charging by induction: redistribution of charge without contact.

ELI-10: Explain It Like I'm 10

Everything is made of atoms. Atoms have a positive nucleus surrounded by negative electrons. Normally the positives and negatives balance. Rubbing moves electrons from one material to another, upsetting the balance. The material that gains electrons becomes negative; the one that loses electrons becomes positive. Charge is not created — it is just moved around.

10.2 Coulomb's Law

Core Idea

The force between two point charges is proportional to the product of the charges and inversely proportional to the square of the distance between them.

\[ F = k\frac{|q_1 q_2|}{r^2} \]

Where:

• \(F\) = force magnitude (N)
• \(k = 8.99 \times 10^9\ \text{N·m}^2/\text{C}^2\) (Coulomb's constant)
• \(q_1, q_2\) = charges (coulombs, C)
• \(r\) = separation distance (m)

In vector form: \(\mathbf{F}_{12} = k\frac{q_1 q2}{r^2}\hat{\mathbf{r}}{12}\) (force on \(q_1\) due to \(q_2\)). Like charges → repulsion; opposite charges → attraction.

Superposition: The net force on a charge is the vector sum of forces from all other charges.

Worked Example: Two Charges

Problem: A 2.0 μC charge and a −3.0 μC charge are 0.10 m apart. Find the force between them.

Solution: \(F = (8.99 \times 10^9)\frac{(2.0\times 10^{-6})(3.0\times 10^{-6})}{(0.10)^2} \approx 5.4\ \text{N}\) (attractive).

Worked Example: Superposition with Three Charges

Problem: Three point charges lie on the \(x\)-axis: \(q_1 = +4.0\ \mu\text{C}\) at \(x = 0\), \(q_2 = -2.0\ \mu\text{C}\) at \(x = 0.30\ \text{m}\), and \(q_3 = +3.0\ \mu\text{C}\) at \(x = 0.50\ \text{m}\). Find the net force on \(q_2\).

Solution (step by step):

1. Force on \(q_2\) due to \(q_1\): The distance is \(r_{12} = 0.30\ \text{m}\). The charges are opposite, so the force is attractive — \(q_2\) is pulled toward \(q_1\) (leftward, the \(-x\) direction).

\[ F_{12} = k\frac{|q_1 q2|}{r{12}^2} = (8.99 \times 10^9)\frac{(4.0\times 10^{-6})(2.0\times 10^{-6})}{(0.30)^2} \approx 0.799\ \text{N} \]

Direction: \(-x\) (left). So \(\mathbf{F}_{12} = -0.799\ \hat{\mathbf{i}}\ \text{N}\).

1. Force on \(q_2\) due to \(q_3\): The distance is \(r_{23} = 0.20\ \text{m}\). The charges are opposite, so the force is attractive — \(q_2\) is pulled toward \(q_3\) (rightward, the \(+x\) direction).

\[ F_{23} = k\frac{|q_2 q3|}{r{23}^2} = (8.99 \times 10^9)\frac{(2.0\times 10^{-6})(3.0\times 10^{-6})}{(0.20)^2} \approx 1.349\ \text{N} \]

Direction: \(+x\) (right). So \(\mathbf{F}_{23} = +1.349\ \hat{\mathbf{i}}\ \text{N}\).

1. Net force (superposition):

\[ \mathbf{F}{\text{net}} = \mathbf{F}{12} + \mathbf{F}_{23} = (-0.799 + 1.349)\hat{\mathbf{i}} = +0.550\ \hat{\mathbf{i}}\ \text{N} \]

Answer: The net force on \(q_2\) is 0.55 N to the right (\(+x\)). The pull from \(q_3\) (closer and 3.0 μC) outweighs the pull from \(q_1\) (farther away despite 4.0 μC). This illustrates superposition: find each pair-wise force separately with correct direction, then add as vectors.

Limiting Cases of the Inverse-Square Law

Examining what happens in extreme limits builds intuition:

Assumptions Underlying Coulomb's Law

Every equation in physics has a domain of validity. Coulomb's law assumes:

1. Point charges: The charges are treated as mathematical points. For extended bodies, integrate over the charge distribution. This works when the separation is much larger than the size of the charges (\(r \gg\) charge dimensions).

1. Stationary charges (electrostatics): The charges are at rest. If charges move, magnetic fields appear and the full Lorentz force \(\mathbf{F} = q(\mathbf{E} + \mathbf{v} \times \mathbf{B})\) must be used.

1. Vacuum (or uniform medium): The law as written with \(k\) (or \(\epsilon_0\)) applies in vacuum. In a material medium, replace \(k\) with \(k/\kappa\) or \(\epsilon_0\) with \(\kappa\epsilon_0\), where \(\kappa\) is the dielectric constant.

1. Inverse-square holds exactly: Experiment confirms the exponent is \(-2\) to extraordinary precision (deviations smaller than \(10^{-16}\)). Theoretically, this follows from the photon having zero rest mass — if the photon had mass, the exponent would deviate from \(-2\).

ELI-10: Explain It Like I'm 10

Coulomb's law is like gravity but for electric charges. It says the pull (or push) between two charges gets weaker very fast as they move apart — if you double the distance, the force drops to one-quarter. It also says: more charge = stronger force. Unlike gravity, which only pulls, electric forces can pull OR push depending on the charge types.

10.3 Electric Field

Core Idea

An electric field \(\mathbf{E}\) is a vector field that exists in the space around charges. It represents the force per unit charge that a positive test charge would experience at any point.

\[ \mathbf{E} = \frac{\mathbf{F}}{q_0} \]

For a point charge: \(E = k\frac{|q|}{r^2}\) (radially outward if \(q > 0\); radially inward if \(q < 0\)).

• SI unit: N/C (equivalent to V/m).
• Field lines: start on positive charges, end on negative charges. Density of lines indicates field strength.
• For continuous distributions: integrate \(d\mathbf{E}\) from each charge element \(dq\).

Worked Example: Electric Field from Two Point Charges

Problem: Two point charges, \(q_1 = +5.0\ \mu\text{C}\) and \(q_2 = -5.0\ \mu\text{C}\), are placed 0.40 m apart on the \(x\)-axis at \(x = -0.20\ \text{m}\) and \(x = +0.20\ \text{m}\). Find the electric field at the origin \((0, 0)\).

Solution:

1. Field from \(q_1\) at the origin: \(q_1\) is positive, at \(x = -0.20\ \text{m}\), so the vector from \(q_1\) to the origin points right (\(+x\)). A positive charge creates a field pointing away from itself, so \(\mathbf{E}_1\) points right.

\[ E_1 = k\frac{|q_1|}{r_1^2} = (8.99 \times 10^9)\frac{5.0\times 10^{-6}}{(0.20)^2} = 1.124 \times 10^6\ \text{N/C} \] \[ \mathbf{E}_1 = +1.124 \times 10^6\ \hat{\mathbf{i}}\ \text{N/C} \]

1. Field from \(q_2\) at the origin: \(q_2\) is negative, at \(x = +0.20\ \text{m}\), so the vector from \(q_2\) to the origin points left (\(-x\)). However, a negative charge has a field that points toward itself — which means it points from the origin toward \(q_2\), i.e., rightward (\(+x\)).

\[ E_2 = k\frac{|q_2|}{r_2^2} = (8.99 \times 10^9)\frac{5.0\times 10^{-6}}{(0.20)^2} = 1.124 \times 10^6\ \text{N/C} \] \[ \mathbf{E}_2 = +1.124 \times 10^6\ \hat{\mathbf{i}}\ \text{N/C} \]

1. Total field (superposition):

\[ \mathbf{E}_{\text{total}} = \mathbf{E}_1 + \mathbf{E}_2 = 2.25 \times 10^6\ \hat{\mathbf{i}}\ \text{N/C} \]

Answer: The field at the origin is \(2.25 \times 10^6\ \text{N/C}\) to the right. Notice that for this symmetric dipole configuration, the fields from the two charges add at the midpoint rather than cancel — both point in the same direction because the negative charge's field points toward it (rightward from the origin to \(q_2\)).

ELI-10: Explain It Like I'm 10

Imagine every charge is surrounded by an invisible "force bubble." If you put another charge inside that bubble, it gets pushed or pulled. The electric field describes how strong the bubble is and which way it points at every location. Field lines are like arrows showing the path a tiny positive test charge would take. They point away from positive charges and toward negative charges.

10.4 Gauss's Law

Core Idea

Gauss's law relates the total electric flux through a closed surface to the charge enclosed.

\[ \oint \mathbf{E} \cdot d\mathbf{A} = \frac{q_{\text{enc}}}{\epsilon_0} \]

Where:

• \(\oint \mathbf{E} \cdot d\mathbf{A}\) = electric flux through a closed surface (N·m²/C)
• \(q_{\text{enc}}\) = total charge enclosed (C)
• \(\epsilon_0 = 8.85 \times 10^{-12}\ \text{C}^2/(\text{N·m}^2)\) = permittivity of free space

Practical use: Gauss's law simplifies field calculations when symmetry is present — spherical, cylindrical, or planar. Key results:

• Point charge: \(E = kq/r^2\) (reproduces Coulomb).
• Infinite line of charge: \(E = \lambda/(2\pi\epsilon_0 r)\).
• Infinite sheet of charge: \(E = \sigma/(2\epsilon_0)\).
• Inside a conductor in electrostatic equilibrium: \(E = 0\).

Worked Example: Spherical Symmetry

Problem: A solid insulating sphere of radius \(R = 0.10\ \text{m}\) carries a total charge \(Q = +2.0\ \mu\text{C}\) uniformly distributed throughout its volume. Find the electric field (a) outside the sphere at \(r = 0.30\ \text{m}\) and (b) inside the sphere at \(r = 0.050\ \text{m}\).

Solution (a) — Outside (\(r > R\)):

Choose a spherical Gaussian surface of radius \(r = 0.30\ \text{m}\) concentric with the charged sphere. By symmetry, \(\mathbf{E}\) is radial and has the same magnitude everywhere on the Gaussian surface.

\[ \oint \mathbf{E} \cdot d\mathbf{A} = E \cdot (4\pi r^2) = \frac{Q}{\epsilon_0} \]

\[ E = \frac{1}{4\pi\epsilon_0}\frac{Q}{r^2} = k\frac{Q}{r^2} \]

\[ E = (8.99 \times 10^9)\frac{2.0\times 10^{-6}}{(0.30)^2} \approx 2.00 \times 10^5\ \text{N/C} \]

Direction: radially outward (positive charge). Outside a uniformly charged sphere, the field is identical to that of a point charge \(Q\) at the center. This is the spherical shell theorem for electrostatics.

Solution (b) — Inside (\(r < R\)):

Choose a spherical Gaussian surface of radius \(r = 0.050\ \text{m}\). Only the charge inside this surface contributes. Since charge is uniformly distributed:

\[ q_{\text{enc}} = Q\frac{\text{volume enclosed}}{\text{total volume}} = Q\frac{\frac{4}{3}\pi r^3}{\frac{4}{3}\pi R^3} = Q\left(\frac{r}{R}\right)^3 \]

\[ q_{\text{enc}} = (2.0\times 10^{-6})\left(\frac{0.050}{0.10}\right)^3 = (2.0\times 10^{-6})(0.125) = 2.5 \times 10^{-7}\ \text{C} \]

\[ E \cdot (4\pi r^2) = \frac{q_{\text{enc}}}{\epsilon_0} \quad\Rightarrow\quad E = \frac{1}{4\pi\epsilon0}\frac{q{\text{enc}}}{r^2} = k\frac{Qr}{R^3} \]

\[ E = (8.99 \times 10^9)\frac{(2.0\times 10^{-6})(0.050)}{(0.10)^3} \approx 8.99 \times 10^5\ \text{N/C} \]

Key insight: Inside a uniformly charged insulating sphere, \(E \propto r\) — the field grows linearly from zero at the center to its maximum at the surface. Outside, \(E \propto 1/r^2\), falling off like a point charge. The field is continuous across the boundary \(r = R\).

ELI-10: Explain It Like I'm 10

Gauss's law is a counting trick. If you draw an imaginary bubble around some charges, the total "electric flow" passing through the bubble's surface tells you exactly how much charge is inside. If the charges are arranged in a nice symmetric pattern (sphere, cylinder, flat sheet), the math simplifies beautifully. Inside a metal conductor at rest, the electric field is always zero — charges quickly rearrange to cancel any field.

10.5 Electric Potential

Core Idea

Electric potential (voltage) is the electric potential energy per unit charge. It provides a scalar description of the electric field.

\[ V = \frac{U}{q_0} \]

• Potential difference: \(\Delta V = V_B - V_A = -\int_A^B \mathbf{E} \cdot d\mathbf{s}\).
• For a point charge: \(V = k\frac{q}{r}\) (reference at infinity, \(V_\infty = 0\)).
• Potential energy of two charges: \(U = k\frac{q_1 q_2}{r}\).
• SI unit of potential: volt (V). \(1\ \text{V} = 1\ \text{J/C}\).

Relation between E and V: \(\mathbf{E} = -\nabla V\). In 1D: \(E_x = -\frac{dV}{dx}\). The electric field points in the direction of steepest decrease in potential.

Worked Example: Potential from Multiple Point Charges

Problem: Three charges — \(q_1 = +2.0\ \mu\text{C}\) at \((0, 0)\), \(q_2 = -3.0\ \mu\text{C}\) at \((0.40\ \text{m}, 0)\), and \(q_3 = +1.0\ \mu\text{C}\) at \((0, 0.30\ \text{m})\) — are arranged in the \(xy\)-plane. Find the electric potential at point \(P = (0.40\ \text{m}, 0.30\ \text{m})\).

Solution:

Electric potential is a scalar — no vectors, just add the contributions with their signs. Compute each distance from the source charge to \(P\):

• \(r_1\): distance from \((0,0)\) to \((0.40, 0.30)\) = \(\sqrt{(0.40)^2 + (0.30)^2} = 0.50\ \text{m}\)
• \(r_2\): distance from \((0.40, 0)\) to \((0.40, 0.30)\) = \(0.30\ \text{m}\)
• \(r_3\): distance from \((0, 0.30)\) to \((0.40, 0.30)\) = \(0.40\ \text{m}\)

\[ V_P = k\left(\frac{q_1}{r_1} + \frac{q_2}{r_2} + \frac{q_3}{r_3}\right) \]

\[ V_P = (8.99 \times 10^9)\left(\frac{+2.0\times 10^{-6}}{0.50} + \frac{-3.0\times 10^{-6}}{0.30} + \frac{+1.0\times 10^{-6}}{0.40}\right) \]

\[ V_P = (8.99 \times 10^9)\left(4.0\times 10^{-6} - 10.0\times 10^{-6} + 2.5\times 10^{-6}\right) \]

\[ V_P = (8.99 \times 10^9)(-3.5 \times 10^{-6}) \approx -3.15 \times 10^4\ \text{V} = -31.5\ \text{kV} \]

Answer: The potential at \(P\) is approximately −31.5 kV. The negative sign means a positive test charge brought from infinity would lose potential energy arriving at \(P\) — the negative \(q_2\) dominates the contribution. Notice how much simpler scalar addition is compared to the vector addition required for electric fields.

The Relationship Between E-Field and Potential — Deep Dive

The connection \(\mathbf{E} = -\nabla V\) is one of the most powerful relationships in electrostatics. Let's unpack it:

1. The negative sign: The electric field points in the direction of decreasing potential. A positive charge accelerates from high \(V\) to low \(V\) (downhill). A negative charge accelerates from low \(V\) to high \(V\) (uphill). This is why \(\mathbf{F} = q\mathbf{E} = -q\nabla V\) — the force always pushes toward lower potential energy, not necessarily lower potential.

2. The gradient: \(\nabla V\) is the vector pointing in the direction of steepest increase in \(V\). Its magnitude is the rate of change of \(V\) with distance in that direction. The steeper the potential "hill," the stronger the field.

3. In one dimension: \(E_x = -dV/dx\). If \(V(x) = kq/x\) (a point charge), then \(E_x = -d(kq/x)/dx = kq/x^2\), recovering Coulomb's result. The minus sign correctly orients the field: for \(q > 0\), \(V\) decreases as \(x\) increases, so \(E_x > 0\) (field points outward).

4. Equipotential spacing: In a map of equipotential contours, closely spaced contours mean a large \(|\nabla V|\) and therefore a strong \(\mathbf{E}\). Widely spaced contours mean a weak field.

5. Path independence: The potential difference \(\Delta V = -\int_A^B \mathbf{E} \cdot d\mathbf{s}\) is independent of path — a consequence of the electrostatic field being conservative (\(\nabla \times \mathbf{E} = 0\)). This is why voltage is well-defined: it does not matter which route you take from \(A\) to \(B\), the work per unit charge is the same.

6. Units connection: \(1\ \text{N/C} = 1\ \text{V/m}\). This equivalence is not obvious but follows from dimensional analysis: \(\text{N/C} = (\text{N·m})/(\text{C·m}) = \text{J}/(\text{C·m}) = \text{V/m}\).

ELI-10: Explain It Like I'm 10

Voltage is like height on a hill. A positive charge naturally "rolls downhill" — from high voltage to low voltage. A negative charge rolls "uphill." The steeper the hill (the faster the voltage changes with position), the stronger the electric field. A battery creates a voltage difference that pushes charges through a circuit, like a pump creating a height difference that pushes water through pipes.

10.6 Equipotential Surfaces

Core Idea

An equipotential surface is a surface on which the electric potential is constant. No work is done moving a charge along an equipotential.

Key properties:

• Electric field lines are always perpendicular to equipotential surfaces.
• Conductors in electrostatic equilibrium are equipotential bodies.
• Equipotential surfaces are closer together where the field is stronger.

ELI-10: Explain It Like I'm 10

Contour lines on a topographic map connect points of equal height. Walking along a contour line, you do not go uphill or downhill — it is flat. Equipotential surfaces are contour lines for voltage. A charge can move along an equipotential surface without gaining or losing energy, just like walking along a contour. The electric field always points straight downhill — perpendicular to the contour lines.

10.7 Conductors in Electrostatic Equilibrium

Conductors (metals) contain free electrons that move in response to electric fields. When a conductor reaches electrostatic equilibrium — all charges are at rest, no currents flow — a set of powerful, universal properties emerge. These are among the most tested ideas in Physics 2.

Property 1: Zero Electric Field Inside

\(\mathbf{E} = 0\) everywhere inside the conducting material. If there were a non-zero field, free electrons would experience a force \(\mathbf{F} = -e\mathbf{E}\) and would move, contradicting the assumption of equilibrium. The charges rearrange themselves until the internal field is exactly canceled.

Consequence: Any net charge on an isolated conductor resides entirely on its surface. If charge were inside, Gauss's law with a surface just inside the conductor would require \(E \neq 0\), which is impossible in equilibrium.

Property 2: Field Perpendicular to the Surface

Just outside a conductor, the electric field is perpendicular to the surface. If the field had a tangential component, surface charges would move along the surface, again violating equilibrium. The magnitude just outside is:

\[ E = \frac{\sigma}{\epsilon_0} \]

where \(\sigma\) is the local surface charge density.

Property 3: The Conductor Is an Equipotential

Since \(\Delta V = -\int \mathbf{E} \cdot d\mathbf{s}\) and \(\mathbf{E} = 0\) inside, there is no potential difference between any two points inside or on the surface. The entire conductor — interior and surface — sits at a single potential.

Property 4: Charge Concentrates at Sharp Points

Surface charge density \(\sigma\) is highest where the radius of curvature is smallest — at sharp points and edges. A sphere has uniform \(\sigma\); a teardrop-shaped conductor concentrates charge at the pointed end. The field just outside is correspondingly stronger near sharp features, which is why lightning rods work: the strong field near the tip ionizes the air, providing a conductive path for charge to dissipate safely.

Property 5: Faraday Cage

A hollow conductor with no charges inside the cavity has \(\mathbf{E} = 0\) in the cavity, regardless of what happens outside. External fields cannot penetrate the conducting shell — charges on the outer surface rearrange to cancel the field in the interior. This is the principle behind shielding sensitive electronics. Conversely, if a charge is placed inside the cavity, an equal and opposite charge is induced on the inner cavity wall, and a compensating charge appears on the outer surface.

Worked Example: Charge on a Conducting Sphere

Problem: A solid conducting sphere of radius \(R = 0.15\ \text{m}\) carries a net charge of \(Q = +6.0\ \mu\text{C}\). Find (a) the surface charge density, (b) the electric field just outside the sphere, and (c) the potential of the sphere.

Solution (a): All charge resides on the surface.

\[ \sigma = \frac{Q}{4\pi R^2} = \frac{6.0\times 10^{-6}}{4\pi(0.15)^2} \approx 2.12 \times 10^{-5}\ \text{C/m}^2 \]

Solution (b): Just outside a conductor, \(E = \sigma/\epsilon_0\). Equivalently, using Gauss's law with a spherical Gaussian surface of radius \(R^+\):

\[ E = k\frac{Q}{R^2} = (8.99\times 10^9)\frac{6.0\times 10^{-6}}{(0.15)^2} \approx 2.40 \times 10^6\ \text{N/C} \]

Direction: radially outward.

Solution (c): Since the conductor is an equipotential, its potential equals the potential at its surface (and every interior point). Taking \(V_\infty = 0\):

\[ V = k\frac{Q}{R} = (8.99\times 10^9)\frac{6.0\times 10^{-6}}{0.15} \approx 3.60 \times 10^5\ \text{V} = 360\ \text{kV} \]

ELI-10: Explain It Like I'm 10

A metal object is full of free electrons — tiny negative charges that can zip around. When you put charge on a metal ball, all the extra charges rush to the surface and spread out until they cannot push each other anymore. Inside, everything cancels perfectly — zero electric field, like a calm lake. The surface becomes a "voltage plateau" — every point on it has the same electric height. And if you make a hollow metal box (a Faraday cage), the inside is completely shielded from outside electric fields — that is why your phone loses signal in an elevator.

Topic Summary

• Electric charge is quantized (\(e = 1.60 \times 10^{-19}\ \text{C}\)) and conserved. Two types: ±.
• Coulomb's law: \(F = k|q_1 q_2|/r^2\), with superposition. Like charges repel; opposites attract. The law assumes point charges, static conditions, and vacuum; it diverges at \(r \to 0\) (point-charge model breaks) and vanishes as \(r \to \infty\).
• Electric field \(\mathbf{E} = \mathbf{F}/q_0\). Point charge: \(E = k|q|/r^2\). Field lines show direction. Vector superposition for multiple sources.
• Gauss's law: flux through closed surface ∝ enclosed charge. Simplifies calculations with symmetry. For a uniformly charged sphere: \(E \propto r\) inside, \(E \propto 1/r^2\) outside.
• Electric potential \(V = U/q\). Voltage = J/C. Scalar superposition — much simpler than vector addition for fields. Relation: \(\mathbf{E} = -\nabla V\); in 1D, \(E_x = -dV/dx\).
• Equipotential surfaces are perpendicular to field lines. Conductors in equilibrium have uniform potential.
• Conductors in equilibrium: \(\mathbf{E} = 0\) inside, charge on surface, field perpendicular to surface, charge concentrates at sharp points, and Faraday cage shielding.

Essential Equations

Concept Check

1. Two identical metal spheres, one charged and one neutral, are touched together and separated. What is the final charge on each? Why?
2. Electric field lines never cross. Why?
3. A proton and an electron are placed in a uniform electric field. Compare the magnitude and direction of the forces on each.
4. Why is the electric field inside a conductor zero in electrostatic equilibrium?
5. If the electric potential is zero at a point, must the electric field also be zero there? Give a counterexample.
6. Three equal positive charges sit at the vertices of an equilateral triangle. Where (if anywhere) is the electric field zero? Where is the potential zero?
7. A point charge \(+Q\) sits at the center of a hollow conducting sphere with net charge zero. Describe the charge distribution on the sphere and the electric field in each region (inside the cavity, within the metal, outside).

Open Educational References

• OpenStax, College Physics, Chapters 18–19: Electric Charge, Electric Potential
• OpenStax, University Physics, Volume 2, Chapters 5–7: Electric Charge, Gauss's Law, Electric Potential