Topic 7: Electric and Magnetic Fields

Topic 7: Content
108) Understand that an electric field (force field) is defined as a region where a charged particle experiences a force
109) Understand that electric field strength is defined as [math] E = \frac{F}{Q} [/math]and be able to use this equation
110) Be able to use the equation  [math]F = \frac{Q_1 Q_2}{4 \pi \varepsilon_0 r^2} [/math], for the force between two charges
111)  Be able to use the equation  [math] E = \frac{Q}{4 \pi \varepsilon_0 r^2}[/math] for the electric field due to a point charge
112) Know and understand the relation between electric field and electric potential
113)  Be able to use the equation [math] E = \frac{V}{d}[/math]for an electric field between parallel plates
114) Be able to use  [math] F = \frac{1}{4 \pi \varepsilon_0 r^2} [/math] for a radial field
115)  Be able to draw and interpret diagrams using field lines and equipotentials to describe radial and uniform electric fields
116)   Understand that capacitance is defined as [math] C= \frac{Q}{V} [/math]and be able to use this equation
117)  Be able to use the equation  [math] W = \frac{1}{2} QV [/math]for the energy stored by a capacitor, be able to derive the equation from the area under a graph of potential difference against charge stored and be able to derive and use the equations [math] W = \frac{1}{2} CV^2 \text{ and } W = \frac{1}{2} \frac{Q^2}{C} [/math]
118) Be able to draw and interpret charge and discharge curves for resistor capacitor circuits and understand the significance of the time constant RC

1. Electric field:

  • A force field can be represented as a vector, which is a mathematical object with both magnitude (size) and direction.
  • The vector representation of a force field is a powerful tool for analyzing and visualizing the forces that act on objects within the field.
  • When the electric field is applied, the pieces of thread line up along the field lines, in the same way that iron filings follow magnetic field lines.
  • Figure 1 Electric field
  • – Electric field lines are continuous and unbroken.
    – They form a vector field, with both magnitude and direction.
    – They can be used to visualize and analyze electric fields around various charge configurations

2. Electric field strength:

  • Two-point charges exerting a force on each other.
  • A charge produces an electric field around it, which exerts a force on another charged object.
  • This idea is similar to a magnetic field close to a magnet, or a gravitational field around a planet.
  • Electric field strength (E) is the magnitude of the electric field at a given point in space.
  • It is a vector quantity, typically measured in units of Newtons per Coulomb (N/C) or Volts per meter (V/m).
  • [math]  E = \frac{F}{Q} [/math]
  • – E is the electric field strength (N/C or V/m)
  • – F is the force exerted on a test charge (N)
    – Q is the magnitude of the test charge (C)
  • The direction of the electric field is defined as the direction of the force on a positive charge.
  • Electric field is a vector quantity because it has both magnitude and direction.
  • Figure 2 Electric field strength
  • Physical Interpretation:
    – The electric field strength (E) represents the intensity of the electric field at a given point.
    – It is a vector quantity, with both magnitude and direction.
    – The direction of E is the same as the direction of the force on a positive test charge.
    – E is a radial vector, decreasing in magnitude with increasing distance from the source charge.
  • Application:
  • Electric field strength is crucial in understanding various phenomena, such as:
    – Electric forces and torques
    – Electric potential and potential difference
    – Capacitance and capacitors
    – Electric currents and resistance
    – Electromagnetic induction and waves

  • Example:

    A small charge of +2µC is placed in the electric field in figure 3. What force does it experience?

  • Figure 3 Electric field for 400 N/C
  • Given data:
  • Electric field [math] = E = 400 N/C [/math]
  • Charge [math] = Q = +2µC = 2 * 10^{-6} C [/math]
    Applied force = F=?
    Formula:
  • [math] E = \frac{F}{Q} \\ F = EQ [/math]
  • Solution:
  • [math] F = EQ  \\ F = (400)(2 *10^{-6}) \\ F = 8 *10^{-4} \, \text{N (downwards)} [/math]

3. Coulomb’s law:

  • Coulomb’s Law states that “the magnitude of the electrostatic force (F) between two-point charges [math] (Q_1 \, \text{and} \, Q_2) [/math] in a vacuum is directly proportional to the product of the charges and inversely proportional to the square of the distance (r) between them”.
  • Mathematically, this is expressed as:
  • [math] F \propto Q_1 Q_2 \hspace{1cm} (1) \\
    F \propto \frac{1}{r^2} \hspace{1cm} (2) \\
    \text{Combining these two equations:} \\
    F \propto \frac{Q_1 Q_2}{r^2} \\
    F = k \frac{Q_1 Q_2}{r^2} \hspace{1cm} (3)
    [/math]
  • where:
    – F is the electrostatic force
    – k is Coulomb’s constant (approximately [math] 8.99 \times 10^9 Nm^2 .C^{-2} [/math])
  • Or [math] k = \frac{1}{4 \pi \epsilon_0} [/math]. So, the equation 3 becomes
  • [math] F = \frac{1}{4 \pi \epsilon_0} \frac{Q_1 Q_2}{r^2} \hspace{1cm} (4) [/math]
  • Figure 4 Coulomb’s forces applied on charges
  • –  and are the magnitudes of the two-point charges
    – r is the distance between the centers of the two charges
  • The direction of the force is along the line joining the two charges, and the force is:
    – Attractive if the charges are opposite (i.e., one positive and one negative)
    – Repulsive if the charges are like (i.e., both positive or both negative)
  • Coulomb’s Law applies to point charges in a vacuum, but it can also be used to approximate the force between charged objects in other environments, like air or water.

  • ⇒ Examples

  •  Calculate the force of attraction between two-point charges A and B separated by a distance of 0.2m. The charge at A is +2µC and the charge at B is -1µC .
  • Solution:
  • First charge [math] = Q_1 = +2μC = +2 *10^{-6} C [/math]
    Second charge [math] = Q_2 = -1μC= -1 *10^{-6}C [/math]
    Distance between these charges [math] = r = 0.2 [/math]
    Force of attraction = F =?
  • Formula
  • [math] F = \frac{1}{4 \pi \epsilon_0} \frac{Q_1 Q_2}{r^2}  [/math]
  • Solution
  • [math] F = \frac{1}{4(3.14)(8.854 * 10^{-12})} \frac{(+2 * 10^{-6})(-1 * 10^{-6})}{(0.2)^2} \\ F = -0.45 \, \text{N} [/math]
  • The significance of the minus sign is to remind us that the force is attractive, but it is not really necessary to include it.
  • Figure 5 The electric field lines close to a positively charged sphere
  • We can produce a formula for the electric field close to a sphere as follows.
  • We know from Coulomb’s law that the force between a sphere, carrying charge Q, and a small charge q at a distance r from the centre is
  • [math] F = \frac{Q q}{4 \pi \varepsilon_0 r^2} [/math]
  • We also know that
  • [math] F = Eq [/math]
  • It follows that the electric field close to the sphere is given by the formula
  • [math] E = \frac{Q}{4 \pi \varepsilon_0 r^2} [/math]
  • So, the strength of the electric field obeys an inverse square law.
    As you can see from Figure 5, the electric field is a vector quantity.
    So, when we consider the field close to two or more-point charges, we must take account of the direction of the electric field.

4. Know and understand the relation between electric field and electric potential.

  • Electric Field (E):
    – Measures the force per unit charge exerted on a charged particle at a given point in space
    – Units: Newtons per Coulomb (N/C) or Volts per meter (V/m)
    – Direction: Points in the direction of the force on a positive test charge
  • Electric Potential (V):
    – Measures the potential energy per unit charge at a given point in space
    – Units: Volts (V)
    – Direction: Scalar quantity, no direction
  • The relationship between electric field and electric potential is:
  • [math] \Delta V = -\int \mathbf{E} \cdot d\mathbf{l}[/math]
    Where:
    – ΔV is the change in electric potential
    – E is the electric field
    – dl is the displacement vector
  • In other words:
    – Electric field is the negative gradient of electric potential
    – Electric potential is the antiderivative of electric field
  • Key points:
    – Electric field is a vector, while electric potential is a scalar
    – Electric field points in the direction of the force, while electric potential indicates the potential energy
    – Electric field and electric potential are related but distinct concepts
  • Some important relationships:
    [math] E = -dV/dl [/math](electric field is the negative derivative of electric potential)
    [math]\Delta V = -\int \mathbf{E} \cdot d\mathbf{l}[/math]
    (electric potential is the antiderivative of electric field)
  • Understanding the relationship between electric field and electric potential is crucial for analyzing and solving problems in electricity and electromagnetism.

5. Electric Field between Parallel Plates:

  • Two parallel plates, separated by a distance d, with a potential difference V between them.
  • The electric field E is uniform and constant between the plates.
  • The direction of E is from the positive plate to the negative plate.
  • The magnitude of E is proportional to the potential difference V and inversely proportional to the distance d.
  • Mathematical Expression:
  • [math] E = \frac{V}{d} [/math]
  • Where:
    – E is the electric field (in N/C or V/m)
    – V is the potential difference (in V)
    – d is the distance between the plates (in m)
  • Key Points:
  • The electric field is uniform and constant between the plates.
  • The electric field lines are straight and parallel to each other.
  • The electric field strength increases with increasing potential difference and decreases with increasing distance.
  • Applications:
    – Capacitors
    – Electrometers
    – Electrostatic precipitation
    – Particle accelerators
  • Important Note:
    – The electric field between parallel plates is a classical example of a uniform electric field.
    – This configuration is often used to study electric fields, capacitors, and electrostatics.

6. Electric field and potential:

  • Electric field the relationship between the electric field E and the electric potential difference V is
  • [math] E = \frac{V}{d} [/math]
  • Another way of writing
  • [math] E = \frac{∆V}{∆d} [/math]
  • Or
  • [math] ∆V = E∆r [/math]
  • Where ∆V is the difference in potential between two points in the field that are a distance r apart.
  • In a uniform field E is constant and so ∆V/∆r is constant.
  • ∆r becomes smaller and smaller, then
  • [math] E = -\frac{dV}{dr} \qquad (1)\\
    dV = -E \, dr [/math]
  • Integrate this equation then
  • [math] \int dV = -\int E \, dr \\
    V = -\int \left( k\frac{ Q}{4 \pi \varepsilon_0} \frac{1}{r^2} \right) dr \\
    V = -k \frac{ Q}{4 \pi \varepsilon_0} \int \left( \frac{1}{r^2} \right) dr \\
    V = -k \frac{ Q}{4 \pi \varepsilon_0} \left(-\frac{1}{r}\right) \\
    V = k \frac{Q}{4 \pi \varepsilon_0} \frac{1}{r} [/math]
  • The minus sign (equation 1) means that the direction of the electric field is in the opposite sense to the potential difference.
  • The electric field strength is equal to the (negative) potential gradient at any point in the field.
  • Potential (V) varies with r in a field where [math] E ∝ r^{-2}[/math].
  • [math] V \propto \frac{1}{r} \\
    V = k \frac{Q}{r} [/math]
  • [math] k = \frac{1}{4 \pi \varepsilon_0} [/math] so,
  • [math] V = \frac{1}{4 \pi \varepsilon_0} \frac{Q}{r} [/math]
  • Which is shown in figure 6

  • Figure 6 Equipotential-around a charge sphere

7. Be able to draw and interpret diagrams using field lines and equipotentials to describe radial and uniform electric fields:

  • Radial Electric Field:
    – Field lines emerge from a positive charge (Q) and terminate on a negative charge (-Q).
    – Field lines are straight and radial, extending outward from the positive charge.
    – Equipotentials are circular and centered on the positive charge.
    – Closer equipotentials indicate a stronger electric field.

  • Figure 7 Radial electric field lines
  • Uniform Electric Field:
    – Field lines are parallel and evenly spaced.
    – Equipotentials are parallel planes perpendicular to the field lines.
    – The distance between equipotentials indicates the strength of the electric field.
    – Arrows on field lines indicate the direction of the electric field (figure 8).
  • Figure 8 Uniform electric field
  • Some key points to keep in mind:
    – Field lines never intersect or cross each other.
    – Equipotentials never intersect or cross each other.
    – Field lines are always perpendicular to equipotentials.
  • When interpreting diagrams:
    – Identify the type of electric field (radial or uniform).
    – Note the direction of the electric field.
    – Observe the spacing and shape of field lines and equipotentials.
    – Use the diagram to determine the strength and direction of the electric field at a given point.

8. Capacitance:

  • The ability of any object to store charge is called capacitance.
  • Capacitance is given the symbol C, and the S1 unit is the farad (F).
  • The capacitance of a capacitor depends on the area of the metal plates, the distance between the plates and the electrical properties of the material separating the plates.
  • The amount of charge, Q, that can be stored on a capacitor depends on the size of the capacitance, C, and the potential difference, V, across the capacitor causing the separation of the charge:
  • [math] Q ∝ V \\ Q = CV [/math]
  • The capacitance of a capacitor can then be defined by
  • [math] C = \frac{Q}{V} [/math]
  • So, one farad is equal to one coulomb per volt ([math] C.V^{-1}[/math]). Actually, 1 F is quite a large capacitance, and useful ‘real-life’ capacitors have capacitances measured in microfarads (μF), nanofarads (nF) or picofarads (pF).

    9. The energy stored by a capacitor:

    • When a capacitor is charged up, the pd from the electricity supply (or the energy per unit charge) causes electrons to flow off one plate, through the external circuit and onto the other plate.
    • This separation of charge is kept steady provided that the pd. is continuously applied, and that there is no leakage of charge.
    • Once the pd. is removed, and a complete discharging circuit is connected to the capacitor, the electrical energy stored by the separated charge can be released as the electrons flow back off the negatively charged plate and back onto the positively charged plate.
    • Figure 9 (a) Graph of Q versus V for a capacitor (b) The energy stored on a capacitor is equivalent to the area under a Q against V graph.
    • If the pd applied to the plates is increased, more charge and therefore energy is stored on the plates. A graph of potential difference against charge for a capacitor is shown in Figure 9 (a).
    • By the definition of potential difference that
    • [math] V = \frac{W}{Q} [/math]
    • where W is the amount of work done per unit charge, Q.
    • In the context of a capacitor, the potential difference V is the amount of work done in moving unit charge off one plate and onto another plate.
    • At any potential difference V, the work done moving an amount of charge is therefore
    • [math] W = V_{\text{av}} \Delta Q \\
      W = \frac{1}{2} V_{\text{max}} \Delta Q [/math]
    • This is represented by the shaded area in the graph in Figure 9 (a).
    • The total energy E (W) stored on the capacitor, charging the capacitor from empty up to a charge of Q at a potential difference V, is calculated by adding up all the similar shaped areas from Q = 0 up to a charge Q.
    • In other words, this is the whole area under the graph up to Q. Because the shape is a triangle (Figure 9 (b)),
    • [math] W = \frac{1}{2} QV[/math]
    • But
    • [math] Q = CV [/math]
    • SO
    • [math] W = \frac{1}{2} (CV) V \\
      W = \frac{1}{2} C V^2 \qquad (1) \\
      Q = C V \\
      V = \frac{Q}{C} [/math]
    • Put in equation 1
    • [math] W = \frac{1}{2} C \left( \frac{Q}{C} \right)^2 [/math]
    • and
    • [math]E = \frac{1}{2} \frac{Q^2}{C} [/math]

    10. Time constant:

    • The time constant RC is the product of the resistance (R) and capacitance (C) in a circuit.
    • It represents the time it takes for a capacitor to charge or discharge by approximately 63.2% of its final value.
    • The unit of [math] \tau [/math] is seconds (s).
    • Mathematical Equation:
    • [math] \tau = R * C [/math]
    • Where:
      – [math] \tau [/math] (tau) is the time constant
      – R is the resistance
      – C is the capacitance
    • Discharging a capacitor:
    • Consider the circuit shown in Figure 10.


      Figure 10 A capacitor discharge circuit

    • When switch S is closed, the capacitor C immediately charges to a maximum value given by Q = CV.
    • As switch S is opened, the capacitor starts to discharge through the resistor R and the ammeter.
    • At any time t, the p.d. V across the capacitor, the charge stored on it and the current (I), flowing through the circuit and the ammeter are all related to each other by two equations.
    • Appling Kirchhoff’s second circuit law around the capacitor-resistor loop.
    • [math] V_R + V_C = 0 [/math]
    • Where [math]V_C \text{ and } V_R [/math] are the p.d. across the capacitor and the resistor, respectively. Substituting for both using Q = CV and V = IR gives
    • [math] V_C = IR \\
      V_R = \frac{Q}{C} \\
      \frac{Q}{C} + IR = 0 [/math]
    • But
    • [math]I = \frac{∆Q}{∆t} [/math]
    • So,
    • [math] \frac{Q}{C} + IR = 0 \\
      \frac{Q}{C} = -IR \\
      \frac{Q}{C} = -\frac{\Delta Q}{\Delta t} R [/math]
    • Rearranging
    • [math]\Delta Q = – \frac{Q}{RC} \Delta t [/math]
    • When integrating a differential equation, such as the one here, we let ∆t→0, and then change to calculus notation (∆ to d) and group like terms on each side.
    • [math] \frac{dQ}{Q} = – \frac{1}{RC} dt [/math]
    • Showing a constant ratio, and
    • [math] \int_{Q_0}^{Q} \frac{dQ}{Q} = – \int_{0}^{t} \frac{1}{RC} dt \\
      \left[ \ln Q \right]_{Q_0}^{Q} = – \left[ \frac{t}{RC} \right]_{0}^{t} \\
      \ln Q – \ln Q_0 = – \frac{t}{RC} [/math]
    • Then using the rules of logs
    • [math]\ln \frac{Q}{Q_0} = – \frac{t}{RC} [/math]
    • So, finally by exponential
    • [math]Q = Q_0 e^{-\frac{t}{RC}} [/math]

    • ⇒ Charging a capacitor

    • The charging time must be considered, though, if the charging procedure is a component of a circuit that needs a greater resistance.
    • Kirchhoff’s second circuit law tells us that the sum of the e.m.f. in the circuit equals the sum of p.d. in the circuit, so
    • [math]V_R + V_C = V_s [/math]
    • Where [math]V_s [/math] is the e.m.f. of the source, [math] V_R [/math] is the p.d. across the resistor and [math]V_C [/math] is the p.d. across the capacitor. Using Ohm’s law and the basic capacitor equation, this becomes

    • [math]V_s = IR + \frac{Q}{C} [/math]

    • During a small time, interval Δt when the capacitor is charging, [math]V_s [/math] and C do not change, [math] \frac{∆V_s}{∆t} = 0 [/math], unlike I and Q, which do change. So in the small time interval
    • [math] 0 = \frac{\Delta I}{\Delta t} R + \frac{1}{C} \frac{\Delta Q}{\Delta t} [/math]
    • But [math] \frac{\Delta Q}{\Delta t} = I, [/math] so,
    • [math]0 = \frac{\Delta I}{\Delta t} R + \frac{I}{C} [/math]
    • So,
    • [math] \frac{\Delta I}{\Delta t} R = -\frac{I}{C} \\
      \frac{\Delta I}{\Delta t} = -\frac{I}{RC} [/math]
      [math]\text{Rearranging and replacing:} [/math]
      [math] \frac{dI}{dt} = -\frac{I}{RC} \\
      \frac{dI}{I} = -\frac{1}{RC} dt  [/math]
    • Integrating in a similar way to before gives
    • [math] I = I_0 e^{-\frac{t}{RC}} [/math]
    • At [math] t = 0, \, V_C = 0, \, \text{and} \, I = I_0 = \frac{V_s}{R} [/math] , and so at any time (t),
    • [math]\begin{gather*}
      \text{We can write:} \\
      I = \frac{V_s}{R} e^{-\frac{t}{RC}} \\
      \text{or} \\
      IR = V_s e^{-\frac{t}{RC}} \\
      V_R = V_s e^{-\frac{t}{RC}} \\
      \text{According to Kirchhoff’s second circuit law:} \\
      V_C = V_R – V_s \\
      \text{Substituting for } V_R \text{ gives:} \\
      V_C = V_s – V_s e^{-\frac{t}{RC}} \\
      \text{Factorizing this gives:} \end{gather*} [/math]
      [math]V_C = V_s \left(1 – e^{-\frac{t}{RC}}\right) [/math]
      [math]\begin{gather*}\text{Then using the capacitor equation:} \\
      Q = C V
      \end{gather*} [/math]
    • So,[math] V = \frac{Q}{C} [/math] ,then we get
    • [math] Q = C V_s \left(1 – e^{-\frac{t}{RC}}\right) [/math]
    • Figure 11 Charging and discharging curves (a) shows charging curve (b) shows discharging curve
    • But  is the maximum charge that can be stored on the capacitor when it is fully charged. That is equal to , the initial charge on the capacitor when it is about to discharge. So finally
    • [math] Q = Q_0 \left(1 – e^{-\frac{t}{RC}}\right) [/math]
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