Electricity Revision PDF

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PHY-10021/23 Electricity revision sheet Spring Semester 20189-20 Revision Summary I list here most of the common formulae and principles that I will expect you to recall in the exam. Note that this is not meant to be a substitute for revision and going through the lecture material, notes and problems, i.e. you need to be familiar with more than what is in this document, but you could use it as a check list of what you know. The emphasis on the exam will be to apply your knowledge to various physical scenarios. Part A: ELECTRICITY (Physics AND Astrophysics) Electric Fields and Forces • The force between two stationary point charges is given by Coulomb’s law. If the magnitude of the force is denoted by F then: F =

1 |Q1 ||Q2 | r2 4πεo

1 = 8.988 × 109 N m2 C−2 4πεo Physical constants like εo are given in the Maths Handbook. Please note that the force is a vector, and that in the above equation the direction of action of this force is not explicitly stated. It is understood that the force is attractive or repulsive depending on the sign of charges and acts along the line joining the particles. If we wish to include the direction of the force we can express the electric force as ~ = F

1 Q1 Q2 rˆ , 4πεo r 2

where rˆ is a unit vector directed radially outwards from the source of the force. If ~ is directed away from the charge experiencing the the charges repel each other then F force and outward along a line that passes through both charges. If the charges attract ~ is directed from the charge experiencing the force toward the charge each other then F exerting the force along the line joining their centres. ~ and • The electric field due to a single point charge has the vector expression, E, magnitude, E, given by ~ = E

1 Q rˆ , 4πεo r 2

E=

1 |Q| , 4πε0 r 2

respectively, where r ˆ is a unit vector directed radially outward from the charge. The ~ will be directed either radially outward or radially inward deelectric field vector E ~ is a vector quantity. pending on the sign of Q. The electric field, represented by E, −1 −1 ~ The units of E are V m or N C . 1

• Principle of superposition, ~ 2 + F~3 + ..... F~ = F~1 + F ~ =E ~1 + E ~ 2 + E~3 + ..... E N.B. You are expected to be comfortable with adding different vectors in terms of their components and with calculating the magnitude of the vectors. In an exam marks will be lost if you cannot sum different vectors and determine the magnitude of the resultant vector properly. ~ to work out the total field of a A good example of this is to add the components of E system consisting of two oppositely charged particles that are a fixed distance apart, i.e. an electric dipole. ~ is given by, • The electric force on a charge, Q, in an electric field E ~ = QE ~ F You will be expected to be able to use this to calculate the motion of a charged particle in an electric field and its deflection in a given direction. Equating the above expression for the force on the charge , Q, to Newton’s 2nd law, the acceleration of a particle in an electric field is ~ ~a = F~ /m = Q E/m You can then use the equations of linear motion (from PHY-10022) to calculate the velocity or displacement of the particle in a given direction. You should be able to express these as vectors. • Electric Potential Energy For a pair of point charges, Q1 and Q2 , separated by a distance r, with the assumption that the position of one of the charges is fixed (Q1 say) and that U = 0 when r = ∞, the potential energy of the pair is defined as the external work done in order to bring one charge from infinity to a distance r from the other charge with no change in the kinetic energy of the system. Z r ~ E ·d~l = Q2 Q1 = U , (Ur=∞ = 0 J) . W =− F 4πε0 r ∞ where U is the electric potential energy and Q2 Q1 F~E · d~l = dr . 4πε0 r 2 1 Q1 Q2 rˆ , between point charges is directed radially 4πεo r 2 outwards or inwards and dl~ = dr rˆ . So for a pair of point charges a distance r apart, the electric potential energy of the pair is given by ~E = since the electric force, F

U=

1 Q2 Q1 . 4πε0 r

If Q1 and Q2 are initially a distance, ri , apart and then moved to a distance rf the change in the electric potential energy of the pair is   1 Q1 Q2 1 . − ∆U = Uf − Ui = ri 4πε0 rf 2

If there is no change in the kinetic energy of the system it must be the case that Wext = ∆U . Note: If the change in U , i.e. ∆U , is independent of the path taken, the field against ~ Electrostatic electric fields are fields which the work is done is an electrostatic E-field. produced by charge distributions that don’t change with time. ~ and the conservation of total Motion of charges in electrostatic E-fields mechanical energy, E ~ ~ In electrostatic E-fields, the total mechanical energy of charge(s) moving in the E-field is conserved. This means that ∆(K + U ) , where K + U is the total mechanical energy of the charge(s), is zero, i.e. ∆K + ∆U = 0 . This means that Ei = Ef , where E=

1 mv 2 + Uelec , 2

assuming other forms of kinetic energy and potential energy can be ignored. Therefore, electrostatic fields are called conservative fields. • The electric potential The electric potential difference between two positions, e.g. B and A, in an electro~ static E−field can be defined as follows: the electric potential difference between B and A is the work done per unit charge in moving a test charge, Qt , from A to B with no change in its kinetic energy by means of an external force acting against the field. This is expressed as VB − VA =

UB − UA . Qt

~ We can apply this to any charge Q assuming there is no change in the E-field that Q moves in. We can define an electric potential function, V , if we define the zero of the electric potential. For a point charge, Q, the electric potential, V , associated with the field produced by Q at a point a distance r from it is given by V =

1 Q . 4πεo r

The electric potential is a scalar. Thus it has a numerical value, but no direction. ~ , which is a vector. Compare this to E The value of the electric potential at a point in space due to the collection of charges, Q1 , Q2 , . . . QN , is the algebraic sum of the electric potentials due to the individual charges. This can be written as Vtotal =

QN Q2 Q1 . + ... + + 4πεo rN 4πεo r1 4πεo r2

~ will be the vector sum from the individual charges. In contrast the total E-field

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• The electric potential difference between two points, e.g. between B and A, in an ~ is electrostatic electric field, E, ∆V = VB − VA = −

Z

B

~ ~l . E·d

A

Since the electric field is electrostatic one should choose d~l based on the direction of ~ since this is easiest path of integration to choose. E For a point charge Q the potential difference between two locations in the field is   Z B Z B Z B Q 1 Q Q 1 ~ ~l = − rˆ· ˆr dr = − dr = . − ∆V = VB −VA = − E·d 2 2 4πεo rB rA A A 4πεo r A 4πεo r ~ = Er r, ˆ Er = Q/(4πεo r 2 ) . It can be We note that for a point charge we can write E seen that for a point charge Er and V are related by Er = −

dV . dr

Note: that you will be expected to know how to calculate the potential difference ~ ~ over a path joining the two points. between two points in an E-field, by integrating E You should know how to do this for the various fields calculated using Gauss’ law. Choosing d~l = ndr ˆ , where n ˆ is a unit vector in the same or opposite direction of E~ (as required), is the simplest choice of dl~ to take. ~ The potential difference in a uniform electrostatic E-field ~ r , where E ~ You should also know that for a uniform electric field ∆V = −E·∆~ is the electric field and ∆~r is the change in position. This can be simplified to ~ par∆V = −E cos(θ )∆r , where E cos(θ) is the component of the electric field E rf − ~ri | . allel or anti-parallel to ∆~ r = r~f − ~ri , with ∆r = |~ In the case of an electric field line, the potential difference between the final (f) and initial point (i) is ∆V

= Vf − Vi = −E|~ rf − ~ri |

∆V

= Vf − Vi = E|~ rf − ~ri |

~ , (in the direction of E) ~ . (in the opposite direction to E)

If ∆~r is not parallel or anti-parallel to E~ we write ∆V = ∓Ed where d is the magnitude of the displacement along the field line, i.e. d = |~rf − r~i || cos(θ)|, the − sign indicates displacement in the direction of the field, and the + sign indicates displacement against the field. (This explanation was not mentioned in lectures, but it is valid. The total mechanical energy in terms of the electric potential Assuming other forms of potential energy need not be considered, the change in the kinetic energy of a charged particle, Q, moving in an electrostatic electric field can be written as ∆K = −∆U = −Q(∆V ) , where ∆V = Vf −Vi , and Q(∆V ) is the electric potential energy of the charged particle.

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• Electric flux and Gauss’ law The electric flux through a flat open surface is ~ A ~ = (E cos θ) A , ΦE = E· ~ is the area vector perpendicular to the flat open surface and θ is the angle where A ~ and A ~. between E Gauss’ law, where the electric flux (ΦE ) out of a Gaussian surface is given by I ~ dA ~ = Qenclosed . ΦE = E· εo In words, this means that the electric flux through a closed surface is equal to the net charge enclosed by the surface divided by εo . The unit of electric flux is the volt-metre (V · m). Gauss’ waw is one of the fundamental laws of electromagnetism that you will need to remember.You will need to be able to explain Gauss’ law in your own words. Be careful to ensure that your explanation and understanding of Gauss’ law matches that given in the definition above. ~ due to various You are expected to be able to apply Gauss’ law to calculate the E-fields distributions of charge, i.e. spherical, cylindrical charge distributions, or sheets or lengths of charge. These are derived in detail in the lecture notes. See the example derivations in section 6, and in the context of capacitors those in section 7. For example, in the case of an infinitely long line of charge the magnitude of the electric field at a perpendicular distance r from the line of charge is E=

λ , 2πεo r

where λ = Q/l is the charge per unit length. Remember that in general the total charge enclosed by a Gaussian surface is the charge density integrated over the volume enclosed by the surface. It is Z Q= ρ dV . You should also understand the relationship between the charge enclosed and the resultant electric fields for conductors, i.e. E = 0 in the interior of a conductor since there is no excess charge inside a conductor – it can only reside on the surface and be in equilibrium.

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Electric currents, Ohm’s law, Resistor, Kirchhoff’s Rules for Electric Circuits, Charging/Discharging a capacitor • Definition of electric current:

dQ dt i.e. current is the rate of flow of charge. It is defined in terms of the net positve charge flowing perpendicularly through the cross-sectional area A of a conductor. The current is positive. The average current is ∆Q/∆t . I=

• Microscopic definition of current. I = An|q|vd Where:- n, Number of conducting charge carriers per unit volume. |q| the magnitude of the charge carrier (which is the electron in metal conductors). vd magnitude of the drift velocity. I Current. A Cross-sectional area of the conductor. • Current density.

I A i.e. the current per unit cross sectional area of the conductor (units A m−2 ). Thus: J=

J = n|q|vd • Ohm’s law. At a given temperature, if it can be said that J~ in a conductor is directly proportional ~ , we can write to E ~ = ρJ~ , E where ρ is the resitivity of the conducting material in the circuit. The ratio E/J = ρ doesn’t depend on the field strength, E. This is Ohm’s law. You should also recall the alternative form of Ohm’s Law in terms of the electrical conductivity, σ , (when σ doesn’t depend on the electric field strength E) i.e.:J = σE , where E is the strength of the electric field applied across a length of conductor. For a uniform/homogeneous conductor (i.e. the conductor has the same properties at every point) of length L and cross-sectional area A we can write the resistance R of the conducting material as R=ρ

L . A

The resistance is defined as the ratio of the electric potential difference between the two ends of the conductor and the current I flowing through it, i.e. V /I = R .

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Only for constant resistivity, ρ , can it be said that V ∝ I and can it be written V = IR , where R is a constant. • Kirchhoff’s Junction rule: The algebraic sum of the currents into/out of a node (or junction) is zero. This arises through conservation of charge. • Kirchhoff’s Loop Rule: The algebraic sum of the voltages (electric potential differences) across all components around a closed loop of a circuit is zero. This arises through conservation of energy. You are expected to be able to apply these rules to an electric circuit containing series/parallel combinations of e.m.f., resistance and capacitance. • Electrical Power: The energy delivered per unit time by a source of emf, E , having internal resistance r and “generating” a current I is Pemf = IVter , where Vter = E − Ir is the potential difference between the terminals of the emf. • The electrical power dissipated by a circuit element having resistance R, a potential difference V across it, and current I flowing through it is: P =

dW V2 . = V I = I 2R = R dt

• Series resistance. Req = R1 + R2 + R3 + ............ • Parallel resistance.

1 1 1 1 + ......... + + = R3 R1 R2 Req

• Capacitance is defined by, Q , V where V is the potential difference between the positive and negative “plates” of the capacitor, i.e. V = Vp − Vn . The unit of capacitance is the farad (F). You should be able to use Gauss’ law to derive the capacitance of various capacitors, i.e. the capacitance of a parallel plate capacitor, the capacitance of a capacitor consisting of a pair of concentric spherical conducting shells, and the capacitance per unit length of a pair of co-axial cylindrical conducting shells, as per the lecture notes. You should also be able to apply the formulae relevant to these examples. C=

• Capacitance of a parallel plate capacitor is given by, C= With a dielectric material κ, C=

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εo A d κεo A d

• Series capacitance.

1 1 1 1 + ...... + + = C2 C1 C2 Ceq

• Parallel capacitance. Ceq = C1 + C2 + C3 + .......... • Energy stored in a capacitor. UE =

Q2 1 = CV 2 2 2C

• Charging capacitors in a RC circuit: q(t) = Qf (1 − e−t/τ ) v(t) = Vf (1 − e−t/τ ) dq = I0 e−t/τ dt where τ = RC is the time constant of the circuit. Qf is the final charge stored in the capacitor Vf is the final voltage (potential difference) across the capacitor I0 is the initial current (I0 = E /R, where E is the ideal e.m.f). In the case of a charging RC circuit, τ tells us how long it takes the for the charge on the capacitor to reach a value that is approximately 0.6321 times the final charge, i.e. q(τ ) ≈ 0.6321 Qf . i(t) =

• Capacitor discharge q(t) = Q0 e−t/τ , i(t) = −

E dq = I0 e−t/τ , I0 = . dt R

In the case of a discharging RC circuit, τ tells us how long it takes the for the charge on the capacitor to reach a value that is approximately 0.3679 times the initial charge, i.e. q(τ ) ≈ 0.3679 Q0 . I do not expect you to remember all the above relations for discharging/charging of a capacitor in an electric circuit. However, you are expected to be able derive them from first principles, e.g. starting with Kirchoff’s loop rule to form a 1st order differential equation, then solving for the charge q as a function of time. You would also be expected to manipulate them, e.g. to calculate the charge or current at a certain time t, or to determine the rate of growth/decay (via the time constant τ = RC).

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