#### 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 E
_{k}and lose E_{p} - A negative charge moves in the opposite direction of the electric field: they lose E
_{k}and gain E_{p} - 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

- E =
- Vm
^{-1}- E =
^{V}/_{d} - V is potential difference
- d is the distance between plates

- E =

- NC
- 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

- V
_{AB}=^{Q}/_{4πε0}(^{1}/_{b}–^{1}/_{a}) - V
_{AB}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 × 10
^{6}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}

- E =
- Substitute the given information:
- 2 × 10
^{6}=^{1}/_{4πε0}×^{Q}/_{0.35²} - ∴ Q = 2.7 × 10
^{-5}C

- 2 × 10

- 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}

- V =
- substitute the given information:
- V =
^{1}/_{4πε0}×^{2.6 × 10-5}/_{0.35} - ∴ V = 7.0 × 10
^{5}V

- 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}

- C =
- 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
- Q
_{T}= Q_{1}+ Q_{2}+ Q_{3}

- Q
- Apply Q = CV and V is constant in parallel:
- Q
_{T}= V(C_{1}+ C_{2}+ C_{3}) ^{QT}/_{V}= C_{1}+ C_{2}+ C_{3}

- Q
- Hence
- C
_{T}= C_{1}+ C_{2}+ C_{3}

- C

**Capacitors in Series**

- The total p.d. in a circuit is the sum of the individual p.d.
- VT = V
_{1}+ V_{2}+ V_{3}

- VT = V
- Apply Q = CV and Q is constant in series:
- V
_{T}= Q(^{1}/_{C1}+^{1}/_{C2}+^{1}/_{C3}) ^{VT}/_{Q}=^{1}/_{C1}+^{1}/_{C2}+^{1}/_{C3}

- V
- 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 = 4πε
_{0}r

- V =

**Examples**

- An isolated metal sphere of radius 63 cm is charged to a potential of 1.2 × 10
^{6}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

- C = 4π × 8.85 × 10

- 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 × 10
^{6})² = CV²

- 25% × C × (1.2 × 10
- Cancel C and calculate V:
- V = 6.0 × 10
^{5}V

- V = 6.0 × 10

- 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