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
- NC-1
- 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/r²
- F = Qq/4πε0r²
Electric Field of a Point Charge
- Electric field strength is force per unit positive charge
- Dividing force by charge, q: E = Q/4πε0r²
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/b – 1/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
Examples
- 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- the charge that can be stored on the sphere
- Maximum field strength is given, therefore the fields strength formula should be used:
- E = Q/4πε0r²
- Substitute the given information:
- 2 × 106 = 1/4πε0 × Q/0.35²
- ∴ Q = 2.7 × 10-5 C
- Maximum field strength is given, therefore the fields strength formula should be used:
- 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
- using the potential equation:
- the charge that can be stored on the sphere
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
Equipotential
- 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
Capacitance
Capacitors
- 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
Examples
- 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
- 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
- Using the equation derived above:
- 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
- Using the equation for energy:
- Calculate the capacitance of the sphere
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