Unit 10: Electric Force, Field, and Potential

Unit 10 covers the physics of electric charge at rest. It starts with charge itself and Coulomb's law for the force between charges, builds the electric field as a way to describe how a charge affects the space around it, then switches to the energy picture with electric potential energy and electric potential. It ends with capacitors, which store separated charge, and with the motion of charges in electric fields.

10.1 Electric Charge and Electric Force

There are two kinds of electric charge, positive and negative. Like charges repel and opposite charges attract. An object is neutral when it carries equal amounts of positive and negative charge, and it is charged when the balance is off. Charging an object does not create charge. It moves electrons from one place to another.

Charge is quantized. The smallest unit of charge that can be isolated is the elementary charge e = 1.60 × 10⁻¹⁹ C. Every observable charge is a whole-number multiple of e. The electron carries −e and the proton carries +e. Conservation of charge says the net charge of an isolated system never changes. In any process, the total charge before equals the total charge after.

Conductors and insulators

In a conductor, some charges are free to move through the material. Metals are the standard example. In an insulator, charges stay essentially fixed where they are. Rubber, glass, and plastic are insulators. Two facts about conductors in electrostatic equilibrium come up constantly. The electric field inside a conductor is zero, and any excess charge sits on the outer surface of the conductor.

Coulomb's law

Coulomb's law gives the magnitude of the electric force between two point charges: F = k|q₁q₂| / r², where r is the separation and k = 8.99 × 10⁹ N·m²/C². The force acts along the line joining the charges, repulsive for like charges and attractive for opposite charges. The two forces form a Newton's third law pair, so they are equal in magnitude even when the charges are very different in size.

Worked example. Two charges of +1.0 μC each are 0.50 m apart. The force magnitude is F = (8.99 × 10⁹)(1.0 × 10⁻⁶)(1.0 × 10⁻⁶) / (0.50)². The numerator is 8.99 × 10⁻³, the denominator is 0.25, and the result is 3.6 × 10⁻² N, or 0.036 N, repulsive. Notice how small the force is. It takes a huge amount of charge to make an everyday-size force, because k is enormous and the coulomb is an enormous unit.

10.2 The Process of Charging

Objects become charged by moving electrons, and there are three processes to know. Charging by friction happens when two materials are rubbed together and electrons transfer from one to the other. One object ends up positive and the other negative, with equal magnitudes, which is conservation of charge in action.

Charging by conduction means charging by direct contact. When a charged conductor touches a neutral one, charge flows until the two are at the same potential, and the total charge is conserved. If the two conductors are identical spheres, they split the total charge evenly. A sphere carrying +6.0 μC touched to an identical neutral sphere leaves each sphere with +3.0 μC.

Charging by induction charges an object without ever touching it with a charged object. Bring a positively charged rod near a neutral metal sphere. The rod attracts the sphere's free electrons toward the near side, leaving the far side positive. While the rod is held in place, ground the sphere. Electrons flow up from the ground onto the sphere, attracted by the rod. Remove the ground connection first, then remove the rod, and the sphere is left with a net negative charge. The order matters. If you remove the rod before breaking the ground connection, the extra electrons flow back out and the sphere ends up neutral.

A related effect is polarization. A charged object brought near a neutral insulator shifts each atom's electron cloud slightly, creating tiny induced dipoles. The near side of the insulator ends up with the opposite sign to the charged object, so the two attract. This is why a charged balloon sticks to a neutral wall. The wall was never given a net charge. Its charges just rearranged.

10.3 Electric Fields

The electric field E at a point is the electric force per unit charge that a charge placed there would feel: E = F / q. Its units are newtons per coulomb (N/C). The field describes how a source charge affects the space around it, whether or not any other charge is actually there to feel it. A single charge sitting alone still produces a field everywhere around it.

For a point charge Q, the field magnitude at distance r is E = k|Q| / r². The direction is defined by the force on a positive test charge: away from a positive source charge and toward a negative one. This definition is the source of the most common sign errors in the unit, so read the trap box below carefully.

Electric field lines are a visual map of the field. They start on positive charges and end on negative charges (or run to infinity). The field direction at any point is tangent to the line through that point. The line density shows the strength, with crowded lines meaning a stronger field. Field lines never cross, because the field at a point has exactly one direction.

Fields add as vectors. The superposition principle says the total field at a point is the vector sum of the fields from each source charge. Worked example: two +2.0 μC charges are fixed 0.60 m apart. At the midpoint, each charge is 0.30 m away, so each contributes E = (8.99 × 10⁹)(2.0 × 10⁻⁶) / (0.30)² = 2.0 × 10⁵ N/C. The left charge's field points away from it (to the right) and the right charge's field points away from it (to the left). Equal magnitudes in opposite directions cancel, so the net field at the midpoint is zero.

10.4 Electric Potential Energy

Electric potential energy U is the stored energy of a charge configuration. For two point charges, U = kQq / r, with U = 0 defined at infinite separation. The sign carries meaning. Like charges give positive U, since work must be done to push them together against repulsion. Opposite charges give negative U, since they attract and the field would do positive work pulling them together.

More generally, moving a charge q through a potential difference ΔV changes its potential energy by ΔU = qΔV. The work done by the electric field is W = −ΔU. When the field does positive work, the charge's potential energy drops and its kinetic energy rises. If only electric forces act, energy is conserved: ΔK + ΔU = 0, so any loss of potential energy shows up as kinetic energy.

10.5 Electric Potential

Electric potential V is potential energy per unit charge: V = U / q, measured in volts (1 V = 1 J/C). For a point charge Q, V = kQ / r. Potential is a scalar, so it adds by ordinary arithmetic, with the sign of Q included. This makes potential calculations much easier than field calculations. You add numbers instead of adding vectors.

Worked example: the potential 0.10 m from a +2.0 μC charge is V = (8.99 × 10⁹)(2.0 × 10⁻⁶) / 0.10 = 1.8 × 10⁵ V. A −2.0 μC charge at the same distance would give −1.8 × 10⁵ V. Place a +1.0 μC charge there and its potential energy is U = qV = (1.0 × 10⁻⁶)(1.8 × 10⁵) = 0.18 J.

An equipotential surface is a surface on which V is constant. No work is done moving a charge along an equipotential, since ΔV = 0 there. Field lines are always perpendicular to equipotential surfaces, and the field points from higher potential toward lower potential, which is the direction of steepest decrease of V. For parallel plates, the equipotentials are planes parallel to the plates and the field between the plates is uniform, with magnitude E = ΔV / d.

10.6 Capacitors

A capacitor stores separated charge on two conducting plates with an insulating gap between them. Capacitance C is defined by C = Q / ΔV, where Q is the magnitude of charge on one plate and ΔV is the potential difference between the plates. The unit is the farad (F).

For a parallel-plate capacitor, C = ε₀A / d, where A is the plate area, d is the plate separation, and ε₀ = 8.85 × 10⁻¹² C²/N·m² is the permittivity of free space. Larger plates facing each other store more charge, and plates closer together store more charge. The field between ideal parallel plates is uniform: E = ΔV / d = Q / (ε₀A).

Worked example: plates of area 0.020 m² separated by 1.0 mm. C = (8.85 × 10⁻¹²)(0.020) / (1.0 × 10⁻³) = 1.8 × 10⁻¹⁰ F, about 180 pF. Real capacitors are small in farads, so expect microfarads and picofarads in problems.

Combinations of capacitors

Capacitors in parallel all share the same voltage, and their charges add, so C_eq = C₁ + C₂ + ... . Capacitors in series all carry the same charge, and their voltages add, so 1/C_eq = 1/C₁ + 1/C₂ + ... . Example: 2.0 μF and 4.0 μF in parallel give 6.0 μF. In series, 1/C_eq = 1/2.0 + 1/4.0 = 0.75 μF⁻¹, so C_eq = 1.3 μF. The series equivalent is always smaller than the smallest individual capacitor.

Energy stored in a capacitor

Separating the charges takes work, and that work is stored as electric potential energy: U = ½QΔV = ½C(ΔV)² = Q² / 2C. All three forms are equivalent, so pick whichever matches the quantities you know. Example: a 4.0 μF capacitor charged to 12 V stores U = ½(4.0 × 10⁻⁶)(12)² = 2.9 × 10⁻⁴ J.

Dielectrics

A dielectric is an insulating material placed between the plates. It polarizes in the plates' field, and the induced field partially cancels the original field, which lets the plates hold more charge at the same voltage. Capacitance increases by the dielectric constant κ: C = κε₀A / d. Two cases matter. If the capacitor stays connected to a battery, the voltage is fixed, so inserting the dielectric increases the charge and the stored energy. If the capacitor is isolated (disconnected, charge fixed), inserting the dielectric lowers the voltage and lowers the stored energy. Name which case you are in before you predict anything.

The same isolated-versus-connected logic applies to changing the geometry. Take an isolated parallel-plate capacitor and double the plate separation. The charge cannot leave, so Q is fixed. Capacitance halves (C = ε₀A/d), the voltage doubles (ΔV = Q/C), and the field between the plates stays the same (E = Q/(ε₀A) has no d in it).

10.7 Conservation of Electric Energy

A charge released in an electric field converts potential energy to kinetic energy according to ΔK = −ΔU = −qΔV. A positive charge speeds up moving toward lower potential. A negative charge speeds up moving toward higher potential. The sign of q decides everything, so write it explicitly.

In a uniform field the force qE is constant, so the acceleration a = qE/m is constant and the kinematics from AP Physics 1 apply directly. A charge entering the field between parallel plates perpendicular to the field follows a parabolic path, exactly like projectile motion with gravity replaced by qE/m.

The electron volt is a convenient energy unit: 1 eV = 1.60 × 10⁻¹⁹ J, the energy an elementary charge gains crossing 1 V. Worked example: an electron starts from rest and accelerates through 200 V. Its kinetic energy is 200 eV = (200)(1.60 × 10⁻¹⁹) = 3.20 × 10⁻¹⁷ J. Setting ½mv² equal to that, with m = 9.11 × 10⁻³¹ kg, gives v = √(2 × 3.20 × 10⁻¹⁷ / 9.11 × 10⁻³¹) = 8.4 × 10⁶ m/s.

Confusions That Cost Points

Practice Questions

Original questions written for this guide in the style of the AP exam. Answers and explanations are on the next page, so complete the questions before checking them.

Answer Key

1. C. F = k|q₁q₂|/r² = (8.99 × 10⁹)(2.0 × 10⁻⁶)(3.0 × 10⁻⁶)/(0.40)² = 0.34 N. Opposite signs attract. D forgot to square the separation (divided by 0.40 instead of 0.16). A got the magnitude right but the direction wrong: opposite charges attract, never repel. B slipped a decimal place.

2. B. The rod attracts electrons to the near side of the sphere. Grounding lets extra electrons flow up from the ground, and removing the ground before the rod traps them there, so the sphere ends negative. A is wrong because protons do not move in a metal; only electrons flow. C confuses induction with polarization of an ungrounded object, which leaves the net charge at zero. D is wrong because the sign of the result follows from the process alone; the rod's exact charge is irrelevant.

3. A. Each charge produces E = kQ/r² = (8.99 × 10⁹)(1.0 × 10⁻⁶)/(0.50)² = 3.6 × 10⁴ N/C at the midpoint, but the two vectors point in opposite directions (each points away from its own source charge), so they cancel to zero. B computed only one charge's contribution and stopped. C added the two magnitudes as scalars, ignoring that fields are vectors. D used the full 1.0 m separation as r instead of the 0.50 m distance from each charge to the midpoint.

4. D. Potential adds as a scalar: V = k(+4.0 nC)/0.10 + k(−4.0 nC)/0.10 = 0. The fields are vectors and both point the same way at P (away from the positive charge and toward the negative charge, i.e., from the positive charge toward the negative one), so they add to a nonzero net field of about 7.2 × 10³ N/C. C assumes zero potential means zero field. A assumes the fields cancel as if the charges had the same sign. B fails to do the scalar addition for V.

5. A. The electron gains KE = qΔV = (1.60 × 10⁻¹⁹ C)(200 V) = 3.20 × 10⁻¹⁷ J. From ½mv² = 3.20 × 10⁻¹⁷ J, v = √(2 × 3.20 × 10⁻¹⁷ / 9.11 × 10⁻³¹) = 8.4 × 10⁶ m/s. B is the speed for 100 V, from halving the potential difference. C comes from an extra factor of 2 in the denominator, a slip in handling the one-half in ½mv². D confuses potential with potential energy: the electron moves toward higher V, which for a negative charge means lower potential energy and therefore a gain, not a loss, of kinetic energy.

6. B. Series capacitors add reciprocally: 1/C_eq = 1/2.0 + 1/4.0 = 0.75 μF⁻¹, so C_eq = 1/0.75 = 1.3 μF. A applied the parallel rule to a series combination. C averaged the two capacitances, which is not a combination rule at all. D computed the reciprocal sum 0.75 and forgot to invert it back, which is the most common arithmetic slip on this problem type.

7. C. Isolated means Q is fixed. C = ε₀A/d, so doubling d halves C. With Q fixed, ΔV = Q/C doubles. A is wrong because the voltage cannot stay the same when C changes with Q fixed. B has the capacitance change backwards and treats the capacitor as battery-connected. D assumes capacitance depends on the charge, which it does not; C is set by geometry alone.

8. D. U = ½C(ΔV)² = ½(4.0 × 10⁻⁶)(12)² = 2.9 × 10⁻⁴ J. A dropped the one-half, computing B(ΔV)². B used ΔV instead of (ΔV)². C slipped a decimal place, a factor of 10 too large.

One-Page Recall Check

Say each answer out loud before you look back, and mark the ones you cannot finish. Anything you cannot say out loud yet belongs in your flashcard deck. In Rycal, add those items to the Electric Force, Field, and Potential deck and let spaced review bring them back over the next few days.

• State the two kinds of charge, what a neutral object is, and the value of the elementary charge.
• State conservation of charge and explain what moves when an object is charged.
• Explain the difference between a conductor and an insulator, and state where excess charge sits on a conductor.
• Write Coulomb's law, define every symbol, and compute the force between two 1.0 μC charges 0.50 m apart.
• Describe charging by conduction and charging by induction, including the role of grounding and the order of steps.
• Explain polarization and why a charged object attracts a neutral insulator.
• Define the electric field, give its units, and state the field of a point charge with its direction rule.
• List the field line rules and explain what superposition means for fields.
• Explain why the force on a negative charge points opposite the field direction.
• Write the potential energy of two point charges, and relate ΔU to qΔV and to the work done by the field.
• Define electric potential, give its units, and explain why it adds as a scalar.
• Describe equipotential surfaces and their geometric relationship to field lines.
• Explain why V = 0 at a point does not imply E = 0 there, with an example.
• Define capacitance, write the parallel-plate formula, and compute C for A = 0.020 m² and d = 1.0 mm.
• State the series and parallel combination rules for capacitors and say how they differ from the resistor rules.
• Write the three forms of the stored energy and compute the energy in a 4.0 μF capacitor at 12 V.
• Explain what a dielectric does, and contrast the isolated and battery-connected cases.
• Describe the motion of a charge released in a uniform field, and define the electron volt.

Key terms for this unit

Electric charge, Elementary charge, Conservation of charge, Conductor, Insulator, Charging by friction, Charging by conduction, Charging by induction, Grounding, Polarization, Coulomb's law, Electric field, Electric field lines, Superposition, Electric potential energy, Electric potential, Volt, Equipotential surface, Capacitance, Farad, Parallel-plate capacitor, Permittivity of free space, Dielectric, Dielectric constant, Series combination, Parallel combination, Electron volt, Test charge.

About this guide. Written for Rycal and aligned to the College Board AP Physics 2 course framework, Unit 10. All questions and explanations are original Rycal writing. Rycal is independent and is not affiliated with or endorsed by the College Board.