4 Ijesha Close, Ilupeju, Lagos
+2348 097 685 118

Electric Fields and Capacitance

Electric Fields
Concept of Electric Fields
  • Electric fields can be described as a field of force; it can move charged particles by exerting a force on them
  • A positive charge moves in the direction of the electric field: they gain Ek and lose Ep
  • A negative charge moves in the opposite direction of the electric field: they lose Ek and gain Ep
  • The electric field of a charge is the space around the charge in which an electric force due to that charge is experienced
  • The direction of the field lines show the direction of the field — always from the positive charge to the negative
  • A higher density of the lines shows a stronger region of field
Diagrammatic Representation
  • Parallel Plates
  • Points
Electric Field Strength
  • It is the force per unit positive charge acting at a point; it is a vector
  • Its units include:
    • NC-1
      • E = F/q
      • E is the electric field strength
      • F is the force
      • q is the charge
    • Vm-1
      • E = V/d
      • V is potential difference
      • d is the distance between plates
  • The higher the voltage, the stronger the electric field
  • The greater the distance between the plates, the weaker the electric field
Coulomb’s Law
  • Any two points charges exert an electrical force on each other that is proportional to the product of the charges and inversely proportional to the square of separation
  • F ∝ Qq/
  • F = Qq/4πε0
Electric Field of a Point Charge
  • Electric field strength is force per unit positive charge
  • Dividing force by charge, q: E = Q/4πε0
Electric Potential
  • Electric potential at a point is the work done in bringing a unit positive charge form infinity to that point
  • W = VQ and W = Fd
  • V = Fd/Q
  • V = Q/4πε0r
  • The potential difference between two points A and B from an isolated charge Q is defined as the work done in taking a unit positive charge from B to A

  • VAB = Q/4πε0(1/b1/a)
  • VAB is equal to the gain in electrical potential energy if Q is positive, and loss if Q is negative
  • In general
    • If a +ve is moved in the direction of the electric field, its electric potential energy will decrease
    • If a -ve charge is moved in the direction of the electric field, its electric potential energy will increase
    • If a charge is accelerated in the field, its electric potential energy will be converted to kinetic
      ∴ Vq = ½mv²
  1. The maximum field strength at the surface of a sphere before electrical breakdown (sparking) occurs is 2.0 × 106 Vm-1. The sphere has a radius r of 0.35 m.
    Calculate the maximum values of

    1. the charge that can be stored on the sphere
      • Maximum field strength is given, therefore the fields strength formula should be used:
        • E = Q/4πε0
      • Substitute the given information:
        • 2 × 106 = 1/4πε0 × Q/0.35²
        • ∴ Q = 2.7 × 10-5 C
    2. the potential at the surface of the sphere
      • using the potential equation:
        • V = Q/4πε0r
      • substitute the given information:
        • V = 1/4πε0 × 2.6 × 10-5/0.35
        • ∴ V = 7.0 × 105 V
Potential Due to a Conducting Sphere
  • A charge +Q on an isolated conducting sphere is uniformly distributed over its surface
  • The charge remains on the surface and at all points inside the sphere, the field strength is 0
  • As there is no field inside the sphere, the potential difference from any point inside the sphere to the surface is zero.
  • Therefore, the potential at any point inside a charged hollow sphere is the same as its surface
  • An equipotential surface is a surface where the electric potential is constant
  • Equipotential lines are drawn such that the potential is constant between intervals
  • As the potential is constant, the potential gradient = 0, hence E along the surface = 0
  • Hence no work is done when a charge is moved along this surface
  • Electric field lines must meet equipotential surfaces at right angles
  • Spacing will be closer when the field is stronger
Similarity and Difference between Electric and Gravitational Potential
  • Similarities:
    • Ratio of work done to mass/charge
    • Work done moving a unit mass/charge from infinity
    • Both have zero potential at infinity
  • Differences:
    • Gravitational forces are always attractive
    • Electric forces can be attractive or repulsive
    • For gravitational, work gets out as masses come together
    • For electric, work done is on charges if they have the same sign, and work done gets out if opposite charges come together

  • Capacitors are used to store energy
  • A dielectric is an electrical insulator
  • How Capacitors Store Energy:
    • On a capacitor, there is a separation of charge with +ve on one plate and -ve on the other
    • To separate the charges, work must be done, hence energy is released when charges come together
Capacitance and Farad
  • Capacitance is the ratio of the charge stored by a capacitor to the potential difference across it
  • Farad is the unit of capacitance. It represents 1 coulomb per volt
    • C = Q/V
  • The capacitance of a capacitor is directly proportional to the area of the plates and inversely proportional to the distance between the plates
Dielectric Breakdown
  • An electric field can cause air to become conducting by:
    • The electric field causes forces in opposite directions on the electrons and the nucleus of atoms in air
    • This results in the field causing electrons to be stripped off the atom
    • Results in a spark — air now contains oppositely charged particles which can carry charge
Capacitors in Parallel

  • By conservation of energy and hence charge (W = QV), the total charge in a circuit is the sum of individual charges
    • QT = Q1 + Q2 + Q3
  • Apply Q = CV and V is constant in parallel:
    • QT = V(C1 + C2 + C3)
    • QT/V = C1 + C2 + C3
  • Hence
    • CT = C1 + C2 + C3
Capacitors in Series

  • The total p.d. in a circuit is the sum of the individual p.d.
    • VT = V1 + V2 + V3
  • Apply Q = CV and Q is constant in series:
    • VT = Q(1/C1 + 1/C2 + 1/C3)
    • VT/Q = 1/C1 + 1/C2 + 1/C3
  • Hence
    • 1/CT = 1/C1 + 1/C2 + 1/C3
Capacitance of a Body
  • Any isolated body can have a capacitance
  • Considering a sphere of radius r, carrying a charge Q, the potential at the surface is
    • V = Q/4πε0r
    • C = Q/V = Q ÷ Q/4πε0r
    • C = 4πε0r
  1. An isolated metal sphere of radius 63 cm is charged to a potential of 1.2 × 106 V. At this potential, there is an electrical discharge in which it loses 75% of its energy
    1. Calculate the capacitance of the sphere
      • Using the equation derived above:
        • C = 4π × 8.85 × 10-12 × 63 × 10-2
        • ∴ C = 7.0 × 10-11 Farad
    2. Calculate the potential of the sphere after the discharge has taken place
      • Using the equation for energy:
        • W = CV²
      • After the discharge, the sphere contains 25% of the energy before, so equating the energy before and after:
        • 25% × C × (1.2 × 106)² = CV²
      • Cancel C and calculate V:
        • V = 6.0 × 105 V
Energy Stored in a Capacitor

  • Area under a potential-charge graph is equal to the work done
    • W = ½QV = ½CV²
  • The half comes in because:
    • When the first charge flows onto the capacitor plates, there is no potential difference opposing the flow
    • As more charges flow, the potential difference increases, so more work is done
    • The average potential difference is equal to half the maximum potential difference