Charge and field
Module 6: Field and particle physics6.1 Fields |
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| 6.1.2 |
Charge and field a) Describe and explain: I) Uniform electric field E = V/d II) The electric field of a charged object, and the force on a charge in an electric field; inverse square law for point charge III) Electrical potential energy and electric potential due to a point charge; 1/r relationship IV) Evidence for discreteness of charge on electron V) The force on a moving charged particle due to a uniform magnetic field VI) Similarities and differences between electric and gravitational fields. b) Make appropriate use of: I) The terms: charge, electric field, electric potential, equipotential surface, electron volt by sketching and interpreting: II) Graphs showing electric potential as area under a graph of electric field versus distance, graphs showing changes in electric potential energy as area under a graph of electric force versus distance between two distance values. III) Graphs showing force as related to the tangent of a graph of electric potential energy versus distance, graphs showing field strength as related to the tangent of a graph of electric potential versus distance IV) Diagrams of electric fields and the corresponding equipotential surfaces. c) Make calculations and estimates involving: I) For radial components [math]F_{\text{electric}} = \frac{kqQ}{r^2}, \qquad E_{\text{electric}} = \frac{F_{\text{electric}}}{q} \\ E_{\text{electric}} = \frac{kQ}{r^2} \left[ k = \frac{1}{4 \pi \varepsilon_0} \right] [/math] II) [math]E_{\text{electric}} = -\frac{dV_{\text{electric}}}{dr}, \quad E_{\text{electric}} = \frac{V}{d} \quad \text{(for a uniform field)}[/math] III) [math]\text{Electrical potential energy} = \frac{kQq}{r}, \quad V_{\text{electric}} = \frac{kQ}{r}[/math] IV) [math]F = qvB[/math] |
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a) Describe and explain:
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I) Uniform Electric Field
- ⇒ Description:
- A uniform electric field is one where the electric field strength E is constant in magnitude and direction.
- Such a field is created, for example, between two parallel conducting plates connected to a voltage source.
- ⇒ Formula:
- [math]E = \frac{V}{d}[/math]
- Where:
- – E: Electric field strength (N/C or V/m),
- – V: Potential difference between the plates (volts),
- – d: Distance between the plates (meters).

- Figure 1 Uniform electric field
- ⇒ Explanation:
- A test charge placed in the field experiences a constant force F due to the uniform E:
- [math]F = qE[/math]
- Where q is the charge on the particle.
- The field lines between the plates are parallel and equally spaced, indicating uniformity.
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II) Electric Field of a Charged Object and Force on a Charge:
- ⇒ Electric Field Due to a Point Charge:
- For a charged object, the electric field at a distance r from the charge Q is given by:
- [math]E = \frac{1}{4 \pi \varepsilon_0} \frac{Q}{r^2}[/math]
- Where:
- – [math]\varepsilon_0[/math]: Permittivity of free space ([math]8.85 × 10^{-12} F/m[/math]).

- Figure 2 Electric field
- ⇒ Force on a Charge in the Field:
- A charge q placed in the field of Q experiences a force given by Coulomb’s law:
- [math]F = qE \\ E = \frac{1}{4 \pi \varepsilon_0} \frac{qQ}{r^2}[/math]
- ⇒ Inverse Square Law:
- The electric field strength and force decrease as [math]\frac{1}{r^2}[/math], demonstrating the inverse square relationship.
- ⇒ Field Lines:
- Field lines originate from positive charges and terminate on negative charges.
- The density of lines indicates the strength of the field.
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III) Electrical Potential Energy and Electric Potential
- ⇒ Electric Potential Energy (U):
- The potential energy of a charge q at a distance r from a charge Q is:
- [math]U = \frac{1}{4 \pi \varepsilon_0} \frac{qQ}{r}[/math]
- ⇒ Electric Potential (V):
- The potential at a distance r from a charge Q is:
- [math]V = \frac{1}{4 \pi \varepsilon_0} \frac{Q}{r}[/math]
- Electric potential is a scalar quantity and represents the energy per unit charge:
- [math]V = \frac{U}{q}[/math]
- 1/r Relationship:
- Both electric potential and electric potential energy follow an inverse relationship with r. This means that as distance increases, the potential and energy decrease.
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IV. Evidence for the Discreteness of Charge on the Electron
- ⇒ Experiment:
- Millikan’s Oil Drop Experiment provided direct evidence of quantized charge.
- ⇒ Description:
- – Fine oil droplets were sprayed into a chamber and allowed to fall under gravity.
- – A uniform electric field was applied to balance the weight of the droplets.
- – The force on the droplets was calculated using: [math]F = qE[/math]
- – The charges on the droplets were measured and found to be multiples of a fundamental charge([math]e = 1.6 × 10^{-19} C[/math]).
- ⇒ Conclusion:
- The smallest measurable charge was identified as the charge of a single electron.
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V. Force on a Moving Charged Particle in a Uniform Magnetic Field
- ⇒ Lorentz Force:
- A charged particle of charge q, moving with velocity v in a magnetic field B, experiences a force given by:
- [math]F = qvB \, sinθ[/math]
- Where:
- – θ: Angle between v and B.
- The force is maximum when [math]θ = 90^0[/math] (particle moving perpendicular to the field).
- The force is zero when [math]θ = 0^0[/math](particle moving parallel to the field).
- The force causes the particle to move in a circular or helical path, with radius:
- [math]r = \frac{mv}{qB}[/math]
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VI. Similarities and Differences Between Electric and Gravitational Fields
| Aspect | Electric Field | Gravitational Field |
|---|---|---|
| Source | Electric Charges | Masses |
| Force Law | [math]F = \frac{1}{4 \pi \varepsilon_0} \frac{q_1 q_2}{r^2}[/math] | [math]F = G \frac{m_1 m_2}{r^2}[/math] |
| Field Strength | [math]E = \frac{F}{q} = \frac{1}{4 \pi \varepsilon_0} \frac{Q}{r^2}[/math] | [math]g = \frac{F}{m} = \frac{G M}{r^2}[/math] |
| Potential | [math]V = \frac{1}{4 \pi \varepsilon_0} \frac{Q}{r}[/math] | [math]\phi = -\frac{G M}{r}[/math] |
| Attractive / Repulsive | Can be both attractive and repulsive | Always attractive |
| Field Lines | Begin at positive charges, end at negative charges | Always point toward the mass |
- ⇒ Similarity:
- Both fields follow the inverse square law.
- ⇒ Difference:
- Electric forces can be attractive or repulsive, whereas gravitational forces are always attractive.
b) Make appropriate use of:
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I. Definitions of Key Terms
- ⇒ Charge:
- Definition:
- – Charge (q) is a fundamental property of matter that causes it to experience a force in an electric field.
- Unit: Coulomb (C).
- Types: Positive (+q) and negative (−q).
- ⇒ Electric Field (E):
- Definition:
- – The region around a charged object where other charges experience a force.
- Formula:
- [math]E = \frac{F}{q} \\
E = \frac{1}{4 \pi \varepsilon_0} \frac{Q}{r^2}[/math] - – E: Electric field strength (N/C or V/m)
- – F: Force experienced by a test charge (q)
- – Q: Source charge,
- – r: Distance from the source charge.
- Direction:
- – Outward from positive charges, inward to negative charges.
- ⇒ Electric Potential (V):
- Definition:
- – The work done per unit charge in bringing a test charge from infinity to a point in an electric field.
- Formula:
- [math]V = \frac{1}{4 \pi \varepsilon_0} \frac{Q}{r^2}[/math]
- – V: Electric potential (volts),
- – Q: Source charge,
- – r: Distance from the source charge.
- ⇒ Equipotential Surface:
- Definition:
- A surface where the electric potential is the same at every point.
- No work is done when moving a charge along an equipotential surface
- Equipotential surfaces are always perpendicular to electric field lines.
- ⇒ Electron-volt (eV):
- Definition:
- The energy gained or lost by an electron moving through a potential difference of 1 volt.
- Conversion:
- [math]1eV= 1.6 × 10^{-19} J[/math]
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II. Graphs and Their Interpretations
- ⇒ Graph of Electric Potential vs. Distance:
- Description:
- – Electric potential (V) decreases with increasing distance (r) from a positive charge
- – For a point charge:
- [math]V(r) = \frac{1}{4 \pi \varepsilon_0} \frac{Q}{r}[/math]
- Graph:
- – A curve that asymptotically approaches zero as r→∞.

- Figure 3 Graph between electric potential and distance
- Interpretation:
- – The area under the graph of E r represents the electric potential difference.
- ⇒ Electric Potential Energy Change (ΔU) vs. Distance:
- Formula:
- [math]\Delta U = q \Delta V \\
\Delta U = \frac{qQ}{4 \pi \varepsilon_0} \left( \frac{1}{r_1} – \frac{1}{r_2} \right)[/math] - Graph:
- – A decreasing curve, as potential energy decreases with distance.

- Figure 4 Potential energy curve
- Interpretation:
- – The area under the graph of F r between two points represents the change in electric potential energy (ΔU).
- ⇒ Force and Tangent of Potential Energy Graph:
- Relation:
- – Force is the negative gradient of potential energy (U) vs. distance:
- [math]F = – \frac{dU}{dr}[/math]
- Graph:
- – A steep negative slope indicates a strong force.
- ⇒ Field Strength and Tangent of Potential Graph:
- Relation:
- – Electric field strength is the negative gradient of the electric potential (V) vs. distance:
- [math]E = – \frac{dV}{dr}[/math]
- Graph:
- – A steep slope of V r indicates a strong electric field.
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III. Equipotential Surfaces and Electric Field Diagrams
- ⇒ Representation of Electric Field Lines:
- Positive Point Charge:
- – Field lines radiate outward
- – Equipotential surfaces are concentric spheres centered on the charge.
- Negative Point Charge:
- – Field lines converge inward.
- – Equipotential surfaces are concentric spheres.
- ⇒ Uniform Electric Field:
- Field Lines:
- – Parallel and equally spaced.
- Equipotential Surfaces:
- – Planes perpendicular to the field lines.
- ⇒ Parallel Plates:
- Field Lines:
- – Straight lines from the positive plate to the negative plate.
- Equipotential Surfaces:
- – Planes parallel to the plates.
- ⇒ Dipole:
- Field Lines:
- – Originate from the positive charge and terminate on the negative charge.
- Equipotential Surfaces:
- – Symmetric curves around the dipole.
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IV. Practical Uses and Applications
- Electric Field Strength (E):
- – Determining the force on a test charge:
- [math]F = qE[/math]
- Electric Potential (V):
- – Calculating work done in moving a charge:
- [math]W = qΔV[/math]
- Energy Conversions:
- – Conversion of potential energy to kinetic energy:
- [math]\frac{1}{2} mv^2 = q∆V[/math]
- Graphical Analysis:
- Helps in visualizing and calculating quantities like work, potential difference, and field strength.
c) Make calculations and estimates involving:
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I) Radial Electric Force
- ⇒ Formula:
- [math]F_{\text{electric}} = k \frac{qQ}{r^2}[/math]
- – [math]F_{\text{electric}} [/math]: Force between two point charges (N),
- – q: Test charge (C),
- – Q: Source charge (C),
- – r: Distance between charges (m)
- – k: Coulomb’s constant, given by:
- [math]k = \frac{1}{4 \pi \varepsilon_0} \\
k = 8.99 \times 10^9 \, \text{(Nm}^2\text{/C}^2)[/math] - ⇒ Explanation:
- The force is proportional to the product of the charges and inversely proportional to the square of the distance between them.
- Direction:
- – If q and Q are of the same sign, the force is repulsive.
- – If q and Q are of opposite signs, the force is attractive.
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II) Radial Electric Field Strength:
- ⇒ Relationship to Force:
- [math]E_{electric} = \frac{F_{electric}}{q}[/math]
- Substituting
- [math]F_{\text{electric}} = k \frac{qQ}{r^2} \\
E_{\text{electric}} = \left( k \frac{qQ}{r^2} \right) \frac{1}{q} \\
E_{\text{electric}} = k \frac{Q}{r^2}[/math] - – [math]E_{\text{electric}}[/math]: Electric field strength (N/C or V/m),
- – Q: Source charge,
- – r: Distance from the source charge.
- ⇒ Interpretation:
- – The electric field strength is independent of the test charge (q).
- – It depends only on the source charge (Q) and the distance (r).
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III) Electric Field from Electric Potential
- ⇒ Radial Field:
- [math]E_{\text{electric}} = -\frac{dV_{\text{electric}}}{dr}[/math]
- – [math]V_{\text{electric}}[/math]: Electric potential due to a point charge,
- [math]V_{\text{electric}} = k \frac{Q}{r} [/math]
- – Differentiating [math]V_{\text{electric}}[/math] with respect to r:
- [math]E_{\text{electric}} = -\frac{d}{dr} \left( k \frac{Q}{r} \right) \\
E_{\text{electric}} = k \frac{Q}{r^2}[/math] - This confirms that the electric field is the negative gradient of the electric potential.
- ⇒ Uniform Field:
- [math]E_{electric} = \frac{V_{electric}}{d}[/math]
- – d: Distance between two points in the uniform electric field.
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IV) Electrical Potential Energy
- ⇒ Formula:
- [math]U_{electric} = k \frac{Qq}{r}[/math]
- – [math]U_{electric}[/math]: Electrical potential energy (J),
- – Q: Source charge,
- – q: Test charge,
- – r: Distance between charges.
- ⇒ Interpretation:
- Electrical potential energy decreases as the charges are moved farther apart (r increases).
- If Q and q have the same sign,
- [math]U_{electric} > 0[/math]; for opposite signs, [math]U_{electric} < 0[/math].
- V) Magnetic Force on a Moving Charge
- ⇒ Formula:
- [math]F = qvB[/math]
- – F: Magnetic force (N),
- – q: Charge of the particle (C),
- – v: Velocity of the particle perpendicular to the magnetic field (m/s)
- – B: Magnetic flux density (T, teslas).
- Interpretation:
- – The force is maximum when the charge moves perpendicular to the magnetic field ([math]θ = 90^0).[/math]).
- – The force is zero when the charge moves parallel to the field ([math]θ = 0^0[/math]).
- ⇒ Direction (Right-Hand Rule):
- Thumb: Direction of velocity (v),
- Fingers: Direction of magnetic field (B),
- Palm: Direction of force for positive charges (opposite for negative charges).
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VI) Examples and Applications
- ⇒ Example 1: Electric Force Calculation
- Two charges, Q=2.0 and q=1.0 , are separated by r=0.1 m. Find the force between them.
- [math]F_{\text{electric}} = k \frac{qQ}{r^2} \\
F_{\text{electric}} = \frac{(8.99 \times 10^9) \times \left( (2 \times 10^{-6}) (1 \times 10^{-6}) \right)}{(0.1)^2} \\
F_{\text{electric}} = 1.798 \, \text{N}[/math] - ⇒ Example 2: Electric Potential Energy
- Using the same charges as above, calculate the potential energy at r=1m.
- [math]U_{\text{electric}} = k \frac{Qq}{r} \\
U_{\text{electric}} = \frac{(8.99 \times 10^9) \times \left( (2 \times 10^{-6}) (1 \times 10^{-6}) \right)}{0.1} \\
U_{\text{electric}} = 0.1798 \, \text{J}[/math] - ⇒ Example 3: Magnetic Force
- A proton (q= [math]1.6 × 10^{-19} C[/math]) moves with a velocity [math]v = 1 × 10^6 m/s[/math] perpendicular to a magnetic field B=0.01 T. Find the force on the proton.
- [math]F = q v B \\
F = (1.6 \times 10^{-19}) \times (1 \times 10^6) \times (0.01) \\
F = 1.6 \times 10^{-15} \, \text{N}[/math]