What Students Should Master in This Unit
Electrostatics explains how stationary electric charges interact, how charge creates electric fields, and how electric potential stores energy. This unit is a bridge between mechanics, circuits, and modern technology.
Use conservation of charge, quantization, conductors, insulators, polarization, and charging methods.
Apply Coulomb's law, electric field definitions, vector direction rules, and superposition.
Analyze electric potential, equipotentials, work, parallel plates, capacitors, and Gauss's law ideas.
Jump to a Topic
1. Electric Charge Basics
Electric charge is a property of matter that causes electric forces. There are two types of charge: positive and negative. Like charges repel, and opposite charges attract.
Core Principles
- Conservation of charge: charge is transferred, not created or destroyed in ordinary interactions.
- Quantization: charge comes in whole-number multiples of e.
- Neutral object: equal total positive and negative charge.
- Charged object: imbalance between protons and electrons.
- In solids, electrons usually move more easily than protons because protons are bound inside nuclei.
2. Charging Methods
Objects become charged when electrons are transferred or rearranged. The total charge of the system is still conserved.
| Method | What Happens | Typical Result |
|---|---|---|
| Friction | Rubbing transfers electrons between materials. | Objects usually become oppositely charged. |
| Contact | A charged object touches a neutral conductor. | Charge is shared; both can end with same sign. |
| Induction | A nearby charged object rearranges charge without touching. | With grounding, a neutral conductor can gain net charge. |
| Polarization | Charges shift slightly inside a neutral object. | Neutral objects can be attracted to charged objects. |
| Grounding | Charge flows to or from Earth. | Object can be neutralized or charged by induction. |
3. Conductors, Insulators, and Charge Distribution
Materials respond differently to electric charge depending on how freely electrons can move.
Conductor Facts Students Should Know
- Excess charge on a conductor moves to the outer surface.
- The electric field inside a conductor in electrostatic equilibrium is zero.
- Electric field lines meet conductor surfaces at right angles.
- Charge density is larger near sharp points, so electric fields can be strongest there.
- Shielding occurs because conductors rearrange charge to cancel internal electric fields.
4. Coulomb's Law
Coulomb's law gives the magnitude of the electric force between two point charges. The force acts along the line connecting the charges.
Direction Rules
- Like charges repel.
- Opposite charges attract.
- Coulomb force is a vector, so net force requires vector addition.
- Use absolute value for magnitude, then decide direction from charge signs.
5. Electric Fields
An electric field describes how a source charge affects the space around it. A test charge placed in the field experiences an electric force.
Field Direction
- Electric field points in the direction a positive test charge would be pushed.
- Field points away from positive source charges.
- Field points toward negative source charges.
- Field line density shows field strength: closer lines mean stronger field.
6. Electrostatics Diagrams
Use these diagrams to connect charge signs, force direction, electric field lines, and equipotential ideas before doing calculations.
7. Superposition of Forces and Fields
When more than one charge acts on a point, calculate each force or field separately, then add the vectors.
Problem Strategy
- Draw a clean charge diagram.
- Mark distances from each source charge to the point of interest.
- Find each field or force magnitude.
- Choose direction using charge sign and geometry.
- Add vectors using signs, components, or symmetry.
8. Electric Potential Energy
Electric potential energy belongs to a system of charges. It depends on charge signs, distance, and the reference point chosen.
Sign Meaning
- Like charges have positive potential energy when separated by finite distance.
- Opposite charges have negative potential energy because energy must be added to separate them to infinity.
- Potential energy is scalar, so add values algebraically, not as vectors.
9. Electric Potential and Voltage
Electric potential describes electric potential energy per unit charge. Potential is a scalar, while electric field is a vector.
Voltage Vocabulary
- Electric potential: voltage at a location.
- Potential difference: voltage difference between two locations.
- High potential: more electric potential energy per positive charge.
- Electron behavior: electrons tend to accelerate opposite the electric field because their charge is negative.
10. Equipotential Lines and Surfaces
An equipotential line connects points with the same electric potential. Moving a charge along an equipotential requires no electric work because ΔV = 0.
How to Read Equipotential Maps
- Closer equipotential lines mean a stronger electric field.
- Electric field points from higher potential to lower potential for positive test charge direction.
- Equipotential lines never cross because one point cannot have two different voltages.
11. Parallel Plates and Uniform Fields
Parallel plates are useful because the field between large closely spaced plates is approximately uniform, except near the edges.
Motion in Uniform Electric Fields
- A positive charge accelerates in the direction of the field.
- A negative charge accelerates opposite the field.
- If the charge enters sideways, motion can look like projectile motion with electric acceleration replacing gravity.
12. Capacitors
A capacitor stores separated charge and electric potential energy. Capacitors are important in circuits, sensors, cameras, power supplies, and many electronic devices.
Capacitor Ideas
- Larger plate area increases capacitance.
- Smaller plate separation increases capacitance.
- A dielectric material can increase capacitance and reduce the field for a given free charge.
- Capacitors can hold charge even after a power source is removed, so real capacitors require safety care.
13. Gauss's Law
Gauss's law connects electric flux through a closed surface to the charge enclosed inside that surface. In Grade 11-12 courses, it is often used conceptually and in AP-style symmetry problems.
When Gauss's Law Is Useful
- Spherical symmetry, such as charge on a conducting sphere.
- Cylindrical symmetry, such as long charged wires.
- Planar symmetry, such as very large charged sheets.
- Conceptual flux questions where enclosed charge matters more than surface shape.
14. Simulation Labs for This Unit
These PhET simulations help students visualize charge transfer, electric force, field lines, potential maps, and charged-particle motion.
Place positive and negative charges, view electric field arrows, map electric potential, and compare field strength at different points.
Lab idea: map field strength along a straight line from a source charge.Investigate how charge size and separation distance change the electric force between two charges.
Lab idea: double distance and verify the inverse-square relationship.Explore charge transfer, attraction, repulsion, polarization, and how rubbing can move electrons.
Lab idea: explain why a charged balloon sticks to a neutral wall.Model charge buildup and discharge, including the connection between static electricity and sparks.
Lab idea: describe charge movement before and during a discharge.Use electric fields and forces to guide a charged particle through a goal while avoiding obstacles.
Lab idea: place charges strategically and justify each force direction.15. Electrostatics Lab Skills
Electrostatics labs require careful control of materials, humidity, grounding, and distance measurements. Small changes can strongly affect observations.
Common Labs
- Charging by friction using rods, fur, wool, plastic, or rubber.
- Electroscope observations for charge detection.
- Charging by induction and grounding demonstration.
- Coulomb force investigation using a simulation or torsion balance model.
- Electric field mapping around point charges or parallel plates.
- Equipotential mapping using conductive paper or a simulation.
- Capacitor charge and discharge observation with teacher-approved equipment.
Good Data Habits
- Keep distance measurements center-to-center for point-charge models.
- Record charge signs before calculating direction.
- Separate force magnitude calculations from direction reasoning.
- Use vector diagrams for multi-charge arrangements.
- Repeat trials because static charge can leak away through air or surfaces.
16. Worked Examples
An object has charge -4.8 × 10-9 C. How many excess electrons does it have?
N = |q|/e = (4.8 × 10-9)/(1.60 × 10-19) = 3.0 × 1010 electrons.
q1 = +2.0 µC and q2 = -3.0 µC are 0.50 m apart. Find the force magnitude.
F = k|q1q2|/r2 = (8.99 × 109)(6.0 × 10-12)/(0.50)2 = 0.216 N.
The force is attractive because the charges have opposite signs.
Find E at a point 0.20 m from a +5.0 µC charge.
E = k|Q|/r2 = (8.99 × 109)(5.0 × 10-6)/(0.20)2 = 1.12 × 106 N/C.
Direction is away from the positive charge.
A -2.0 µC charge is in a 3.0 × 104 N/C field to the right. Find force.
|F| = |q|E = (2.0 × 10-6)(3.0 × 104) = 0.060 N.
Because q is negative, the force is to the left.
+3.0 µC and -3.0 µC are 0.40 m apart. Find E at the midpoint.
Each charge is 0.20 m away, so E = (8.99 × 109)(3.0 × 10-6)/(0.20)2 = 6.74 × 105 N/C.
Both fields point from + toward -, so Enet = 1.35 × 106 N/C toward the negative charge.
Find V at 0.30 m from a +4.0 µC point charge.
V = kQ/r = (8.99 × 109)(4.0 × 10-6)/(0.30) = 1.20 × 105 V.
Find U for +2.0 µC and -5.0 µC separated by 0.25 m.
U = kq1q2/r = (8.99 × 109)(-1.0 × 10-11)/(0.25) = -0.360 J.
Two plates have potential difference 120 V and separation 0.020 m. Find E.
E = ΔV/d = 120/0.020 = 6.0 × 103 V/m.
An electron is in an electric field of 4.0 × 103 N/C. Find acceleration magnitude.
F = |q|E = (1.60 × 10-19)(4.0 × 103) = 6.4 × 10-16 N.
a = F/m = (6.4 × 10-16)/(9.11 × 10-31) = 7.0 × 1014 m/s2.
A capacitor stores 30 µC at 12 V. Find capacitance.
C = Q/V = (30 × 10-6)/12 = 2.5 × 10-6 F = 2.5 µF.
A 4.0 µF capacitor is charged to 9.0 V. Find stored energy.
U = 1/2 CV2 = 0.5(4.0 × 10-6)(9.0)2 = 1.62 × 10-4 J.
A closed surface has no net charge inside it. What is the net electric flux through the surface?
By Gauss's law, ΦE,total = qenclosed/ε0 = 0. Field lines may enter and leave, but net flux is zero.
17. Practice Problems
Try each problem first. Then open the answer check and compare units, powers of ten, vector directions, and signs.
1. How many excess electrons are on an object with charge -8.0 × 10-9 C?
Answer
N = |q|/e = 5.0 × 1010 electrons.
2. A neutral object loses electrons. Does it become positive or negative?
Answer
Positive, because losing negative charge leaves excess positive charge.
3. Two +1.0 µC charges are 0.30 m apart. Find electric force magnitude.
Answer
F = (8.99 × 109)(1.0 × 10-12)/(0.30)2 = 0.100 N, repulsive.
4. If distance between two charges triples, what happens to the force?
Answer
Force becomes 1/9 as large.
5. Find E at 0.50 m from a +2.0 µC charge.
Answer
E = (8.99 × 109)(2.0 × 10-6)/(0.50)2 = 7.19 × 104 N/C away from the charge.
6. A +4.0 µC charge sits in a 2.5 × 103 N/C field. Find force.
Answer
F = qE = (4.0 × 10-6)(2.5 × 103) = 1.0 × 10-2 N in the field direction.
7. A -3.0 µC charge experiences a 0.060 N force to the left. If the field is horizontal, what is E?
Answer
|E| = F/|q| = 0.060/(3.0 × 10-6) = 2.0 × 104 N/C. Direction is right because the charge is negative.
8. What is the field direction near a negative point charge?
Answer
Toward the negative charge.
9. At the midpoint between two equal positive charges, what is the net electric field?
Answer
Zero, because the fields have equal magnitude and opposite directions.
10. At the midpoint between +Q and -Q, what direction is the net field?
Answer
From the positive charge toward the negative charge.
11. Find V at 0.40 m from a -3.0 µC charge.
Answer
V = kQ/r = (8.99 × 109)(-3.0 × 10-6)/0.40 = -6.74 × 104 V.
12. A +2.0 µC charge moves through a potential difference of 50 V. Find ΔU.
Answer
ΔU = qΔV = (2.0 × 10-6)(50) = 1.0 × 10-4 J.
13. What is the work done moving a charge along an equipotential line?
Answer
Zero, because ΔV = 0.
14. Plate voltage is 240 V and separation is 0.030 m. Find field strength.
Answer
E = ΔV/d = 240/0.030 = 8.0 × 103 V/m.
15. A proton is released in a uniform electric field. Does it accelerate with or against the field?
Answer
With the field because a proton is positive.
16. An electron is released in a uniform electric field. Does it accelerate with or against the field?
Answer
Against the field because an electron is negative.
17. A capacitor has Q = 48 µC and V = 16 V. Find C.
Answer
C = Q/V = 48 µC/16 V = 3.0 µF.
18. A 10 µF capacitor is connected to 12 V. Find stored charge.
Answer
Q = CV = (10 × 10-6)(12) = 1.2 × 10-4 C = 120 µC.
19. A 2.0 µF capacitor is charged to 20 V. Find stored energy.
Answer
U = 1/2 CV2 = 0.5(2.0 × 10-6)(20)2 = 4.0 × 10-4 J.
20. What is the electric field inside a conductor in electrostatic equilibrium?
Answer
Zero.
21. A closed Gaussian surface encloses +6.0 µC. Write the net flux expression.
Answer
ΦE,total = qenclosed/ε0 = (6.0 × 10-6)/ε0.
22. In the Charges and Fields simulation, what happens to field arrow size as you move farther from a point charge?
Answer
The arrows get smaller because field strength decreases with distance squared.
18. What to Know Before Moving On
- Charge is conserved and quantized in multiples of e = 1.60 × 10-19 C.
- Like charges repel; opposite charges attract.
- Objects can be charged by friction, contact, induction, or polarization.
- Coulomb's law is Fe = k|q1q2|/r2.
- Electric field is force per unit positive test charge: E = F/q.
- A point charge creates field E = k|Q|/r2.
- Electric field points away from positive charges and toward negative charges.
- For multiple charges, add force or field vectors by superposition.
- Electric potential energy for two point charges is U = kq1q2/r.
- Electric potential is V = U/q and point-charge potential is V = kQ/r.
- Equipotential lines are perpendicular to electric field lines.
- Between parallel plates, E = ΔV/d.
- Capacitance is C = Q/ΔV and capacitor energy is U = 1/2 C(ΔV)2.
- Gauss's law relates net electric flux to enclosed charge.