Electricity and Magnetism - Exam Notes

General Concepts

• Charge/Point Charge:
  • Charge: A property of particles to measure attractive and repelling forces.
  • Point Charge: Charged objects with sizes much smaller than the distance between them.
• Charge is quantized.
• Charge is conserved.
• Coulomb’s Law: Describes the interaction between electric charges at rest.

Coulomb’s Law

• Coulomb: The amount of charge that flows in one second with a steady current of one ampere.
• Definition: Two stationary electric charges repel or attract one another with a force proportional to the product of the magnitude of the charges and inversely proportional to the square of the distance between them.
• Formula: F = k \frac{q1 q2}{r^2} where k = \frac{1}{4 \pi \epsilon_0} = 8.99 \times 10^9 \frac{N \cdot m^2}{coul^2}
  • F is the force between the charges.
  • q1 and q2 are the magnitudes of the charges.
  • r is the distance between the charges.
  • \epsilon_0 is the permittivity of free space.

Electric Field

• Electric Field: The field produced by any electric charge around it.
• Influence observed when another charge is brought into the area where the field exists.
• The force F experienced by electric charge q gives the electric field lines.
• Formula: E = \frac{F}{q_0}
  • E is the electric field.
  • F is the force experienced by the charge.
  • q_0 is the electric charge.
• Electric Field is also known as electric intensity at that point.

Calculation of Electric Field Strength

• Magnitude of the force acting on q0 as given by Coulomb’s Law: F = \frac{1}{4 \pi \epsilon0} \frac{q q_0}{r^2}
• The strength of the electric field at the site of the test charge:
  E = \frac{F}{q0} = \frac{1}{4 \pi \epsilon0} \frac{q}{r^2}
• The resultant field at a point is obtained by adding vectorially the field contribution due to all charge elements.

Electric Field due to Uniformly Charged Wire

• Consider a section of an infinite line of charge having a charge density \lambda coulomb per meter.

• Magnitude of the field contribution dE at point P due to a charge element dq = \lambda \cdot dx:
  dE = \frac{1}{4 \pi \epsilon_0} \frac{dq}{r^2}
  where r is the distance of point P from the charge element dq.

• The vector dE can be resolved into two components: dEx = -dE \sin{\theta} and dEy = dE \cos{\theta}.

• The minus sign indicates the negative x direction.

• The x and y components of the resultant vector E at point P:

  Ex = \int{-\infty}^{\infty} dEx = \int{-\infty}^{\infty} -\sin{\theta} dE = 0
  Ey = \int{-\infty}^{\infty} dEy = \int{-\infty}^{\infty} \cos{\theta} dE

• Ex must be zero, but Ey components from the right and left halves are equal.

• x = y \tan{\theta}, so dx = y \sec^2{\theta} \cdot d\theta

• r = y/\cos{\theta} = y \sec{\theta}, so r^2 = x^2 + y^2 = y^2 \sec^2{\theta}

• Ey = E = \int0^{\infty} 2 \cos{\theta} dE
  Ey = \int0^{\frac{\pi}{2}} \frac{2 \cos{\theta}}{4 \pi \epsilon0} \frac{\lambda y \sec^2{\theta} d\theta}{y^2 \sec^2{\theta}} = \int0^{\frac{\pi}{2}} \frac{\lambda \cos{\theta}}{2 \pi \epsilon0 y} d\theta Ey = \frac{\lambda}{2 \pi \epsilon0 y} [\sin{\theta}]0^{\frac{\pi}{2}} = \frac{\lambda}{2 \pi \epsilon_0 y}

• Thus,
  E = \frac{\lambda}{2 \pi \epsilon_0 y}

Electric Field at a Point on the Axis of a Charged Circular Ring

• Consider a circular turn of wire of radius a carrying a charge q.
• Calculate E at a point P on the axis of the ring at a distance x from its center.
• The charge contained in this element is dq = \frac{q}{2 \pi a} ds, where 2 \pi a is the circumference of the ring.
• This element produces a field dE at point P.
• The general vector integral E = \int dE becomes a scalar integral E = \int dE \cos{\theta}.
• dE = \frac{1}{4 \pi \epsilon0} \frac{dq}{r^2} = \frac{1}{4 \pi \epsilon0} \frac{q ds}{2 \pi a (a^2 + x^2)}
• \cos{\theta} = \frac{x}{r} = \frac{x}{\sqrt{a^2 + x^2}}
• 
  E = \int dE \cos{\theta} = \int \frac{1}{4 \pi \epsilon0} \frac{q ds}{2 \pi a (a^2 + x^2)} \frac{x}{\sqrt{a^2 + x^2}} E = \frac{q x}{4 \pi \epsilon0 (a^2 + x^2)^{\frac{3}{2}}} \int \frac{ds}{2 \pi a} = \frac{q x}{4 \pi \epsilon_0 (a^2 + x^2)^{\frac{3}{2}}}
• If x >> a, then
  E = \frac{1}{4 \pi \epsilon0} \frac{q x}{x^3} = \frac{q}{4 \pi \epsilon0 x^2}
• This is the same as that produced by a point charge.

Electric Field due to a Charged Disk

• To be derived similarly.

Point Charge in an Electric Field

• A particle of charge q in an electric field E experiences a force F = qE.
• The force will produce an acceleration a = \frac{F}{m} = \frac{qE}{m}.
• If a particle of mass m and charge q is placed at rest in a uniform electric field and released, the resulting motion resembles that of a body falling in the earth’s gravitational field.
  • v = v_0 + at = at = \frac{qEt}{m}
  • y = v_0 t + \frac{at^2}{2}
  • v^2 = v_0^2 + 2ay
  • K = \frac{mv^2}{2}

Electric Field due to an Electric Dipole

• Dipole: Two equal but opposite charges separated by a small distance.

• Let the distance of the point P be y from both the charges.

• E_1 = electric field for +q

• E_2 = electric field for -q

• Total field E = E1 + E2

• Since the charges have the same magnitude and distance from P is the same, E1 = E2 = \frac{1}{4 \pi \epsilon_0} \frac{q}{a^2 + x^2}

• The direction of E1 and E2 components cancel each other.

• The vector sum of E1 and E2 points vertically downwards and has the magnitude

  E = 2E1 \cos{\theta} = 2 \frac{1}{4 \pi \epsilon0} \frac{q}{a^2 + x^2} \frac{a}{\sqrt{a^2 + x^2}} = \frac{1}{4 \pi \epsilon_0} \frac{2aq}{(a^2 + x^2)^{\frac{3}{2}}}

• since x >> a,

  E = \frac{1}{4 \pi \epsilon_0} \frac{2aq}{x^3}

Dipole in an External Electric Field

• An electric dipole is placed in a uniform external electric field E, its dipole moment p making an angle with this field.
• The two forces (F and -F) acting on the charges are equal and opposite:
  F = qE
• The net force on the dipole is clearly zero.
• Since the forces do not act along the same line, there is a net torque on the dipole about an axis passing through the center O of the dipole given by:
  • Torque = magnitude of a force × perpendicular distance between the forces.

Flux and Gauss's Law

• Flux, \varphi: The number of lines of force that cut through a surface.

• The electrical flux is the sum over the surface by an integral:
  \varphi = \int E \cdot dS

• Gauss’s Law: The density of lines of force can be regarded as a measure of E, the electric intensity, electric field strength.

• The magnitude of E at any point equals the number of lines of force crossing unit area surrounding that point, the area being perpendicular to E.

• Now the magnitude of E at all points on the surface is \frac{q}{4 \pi \epsilon0 r^2} where 4 \pi r^2 is the area of the spherical surface and q is the lines of force leave a charge q. \frac{q}{\epsilon0}

• d\Omega = \frac{dS \cos{\theta}}{r^2} is the solid angle subtended by dS at q.

• Hence, d\varphi, the flux of E across dS is given by:

  d\varphi = E \cdot dS = E dS \cos{\theta} = \frac{q}{4 \pi \epsilon0 r^2} dS \cos{\theta} = \frac{q}{4 \pi \epsilon0} d\Omega

• Now the total solid angle subtended by S at q is 4 \pi.

• Hence the total flux of E over S is given by:

  \varphi = \oint E \cdot dS = \frac{q}{4 \pi \epsilon0} \oint d\Omega = \frac{q}{4 \pi \epsilon0} 4 \pi = \frac{q}{\epsilon_0}

• General theorem: The outward flux of E over any closed surface is equal to the algebraic sum of the enclosed charges divided by \epsilon_0.

Gauss’s Theorem

• If a surface encloses equal and opposite charges, the flux, \varphi is zero.
• Charges outside the surface make no contribution to the electric flux.
• The total flux does not depend on the exact location of the charges inside the closed surface.

Application of Gauss’ Law

• (i) Deduction of Coulomb’s law:

  • Coulomb’s law can be deduced from Gauss’ law.

  • E and dS at any point on the Gaussian surface are directed radially outward.

  • The angle between them is zero, and the quantity E \cdot dS = E dS \cos{\theta} = EdS.

  • Gauss’ law then becomes:

    \oint E \cdot dS = \frac{q}{\epsilon0} \implies \oint EdS = \frac{q}{\epsilon0}

  • Since E is constant for all points on the surface of the sphere, it can be taken outside the integral, and consequently:

    E \oint dS = \frac{q}{\epsilon_0}

  • But \oint dS is simply the area 4 \pi r^2 of the sphere. Therefore:

    E (4 \pi r^2) = \frac{q}{\epsilon0} \implies E = \frac{1}{4 \pi \epsilon0} \frac{q}{r^2}

  • This is the magnitude of electric field strength E at any point a distance r from an isolated point charge q.

  • Again E = \frac{F}{q0} or F = Eq0.

• (ii) Electric field due to a charged sphere

• (iii) Electric field due to a line of charge

Potential Difference

• The work done in moving a test charge q_0 from a point A to a point B with constant speed is known as potential difference then

  VA - VB = \frac{W{AB}}{q0}

  1 \text{ Volt} = \frac{1 \text{ joule}}{coul}

• The potential difference between two points is 1 volt if one joule of work is done in moving one coulomb of charge from one point to the other.

• Potential & Electric field strength:

  • The electric force on the charge is qE and points downward.

  • To move the charge, the force on q_0 must be countered by an external force F of the same magnitude but directly upward.

  • The work done W{AB} = \intA^B F \cdot d\mathbf{l} = \intA^B q0 E \cdot d\mathbf{l} = q_0Ed

  • VA - VB = \frac{W{AB}}{q0} = \frac{q0Ed}{q0} = Ed

  • To find the total work for all infinitesimal segments

    VA - VB = - \int_A^B E \cdot d\mathbf{l}

Potential due to a Point Charge

• In Fig A and B are two points near an isolated point charge q.

• For simplicity assume that A, B, and q lie on a straight line.

• The potential difference between points A and B is given by

  VB - VA = - \intA^B E \cdot d\mathbf{l} = - \intA^B E dl \cos{\theta}

  Where E point to the right whereas dl points to the left, so that the angle between them is 180o therefore,

  VB - VA = \int_A^B Edl

  E = \frac{1}{4 \pi \epsilon_0} \frac{q}{r^2}, dl = dr

  VB - VA = \frac{q}{4 \pi \epsilon0} \int{rA}^{rB} \frac{dr}{r^2} = \frac{q}{4 \pi \epsilon0} [-\frac{1}{r}]{rA}^{rB} = \frac{q}{4 \pi \epsilon0} (\frac{1}{rA} - \frac{1}{r_B})

  V = \frac{q}{4 \pi \epsilon_0 r}

Potential due to a Dipole

• Two equal charges, q of opposite sign, separated by a small distance 2a, constitute an electric dipole.

• The electric dipole moment p has the magnitude 2aq and points from the negative charge to the positive charge.

• The expression for the electrical potential, V due to a dipole at a long distance.

• Let the point P, where the potential will be calculated by the quantities r and \theta in Fig.

• From symmetry, it is clear that the potential will not change as point P rotates about the y-axis, r and \theta being fixed.

• Now the resultant potential at the point P due to charge +q and -q.

  VP = \frac{q}{4 \pi \epsilon0} [\frac{1}{r1} - \frac{1}{r2}]

  r1^2 = r2^2 + (2a)^2 + 2r2(2a) \cos{\theta'} or r1^2 - r2^2 = 2a (2a + 2r2 \cos{\theta'})

  r1 - r2 = \frac{2a (2a + 2r2 \cos{\theta'})}{r1 + r_2}

  VP = \frac{q}{4 \pi \epsilon0} [\frac{r2 - r1}{r1 r2}] = \frac{q}{4 \pi \epsilon0} [\frac{2a (2a + 2r2 \cos{\theta'})}{(r1 + r2)r1 r2}]

• For an ideal dipole (2a << r), so r1 \approx r, r2 \approx r and \theta' \approx \theta

  VP = \frac{q}{4 \pi \epsilon0} \frac{2a \cos{\theta}}{r^2} = \frac{p \cos{\theta}}{4 \pi \epsilon_0 r^2}

• Some special cases:

  • (i) When the point P lies on the axial line of the dipole on the side of the positive charge, \theta = 0 \implies \cos{\theta} = 1.

    VP = \frac{p}{4 \pi \epsilon0 r^2}

  • (ii) When the point P lies on the axial line of the dipole on the side of the negative charge, \theta = 180 \implies \cos{\theta} = -1.

    VP = - \frac{p}{4 \pi \epsilon0 r^2}

  • (iii) When the point P lies on the equatorial line of the dipole on the side of the positive charge, \theta = 90 \implies \cos{\theta} = 0.

    So V = 0

Capacitor and Capacitance

• Capacitor: Two conductors of arbitrary shape, completely isolated from each other and their surroundings.

• Capacitor is a passive two terminal electrical component used to store electrical energy temporarily in an electrical field.

• When the capacitor is charged by connecting the plates to the opposite terminals of a battery, equal and opposite charges appear on the two plates of the capacitor.

• Capacitors are widely used in electronic circuits for blocking direct current while allowing alternating current to pass.

• Capacitance: The charge q of a capacitor is found to be directly proportional to the potential difference between the plates.

  q \propto V \implies q = CV \implies C = \frac{q}{V}

• Capacitance is defined as the ratio of the electric charge q on each conductor to the potential difference V between them.

• The SI unit of capacitance is the farad (F), which is equal to one coulomb per volt (1 C/V).

• Typical capacitance values range from about 1 pF (10^{-12} F) to about 1 mF (10^{-3} F).

• The value of capacitance depends on:

  • The geometry of each plate
  • The spatial relationship between the plates
  • The medium in which the plates are immersed.

Capacitance of a Parallel-Plate Capacitor

• A parallel-plate capacitor formed of two parallel conducting plates of area A and separated by a distance d.

• After connecting a battery, the charges appear +q and -q and the electric field strength E between the plates will be uniform.

• Let the capacitance be C and imagine a Gaussian surface of height h closed by plane caps of area A.

• Thus, the only part of the Gaussian which contributes to the electric flux is the Gaussian surface that lies between the plates.

  • Here E is constant and according to Gauss’ law

    \oint E \cdot dS = \frac{q}{\epsilon0} \implies EA = \frac{q}{\epsilon0} \implies E = \frac{q}{\epsilon_0 A}

  • The potential difference V between the plates can be obtained from

    V = \int -E \cdot dl = \int{-}^{+} E dl = \int{-}^{+} E dl = Ed

  • Since E is constant and can be taken outside the integral and \int dl is simply the plate separation d.

  • Substituting the values of E,

• The capacitance of the parallel-plate capacitor is, therefore

  V = \frac{qd}{\epsilon0 A} \implies C = \frac{q}{V} = \frac{\epsilon0 A}{d}

Capacitance of a Spherical Capacitor

• Fig. shows a central cross-section of a capacitor that consists of two concentric spherical shells of radii a and b.

• draw a sphere of radius r concentric with the two shells. Applying Guass’ law to the surface

  \oint E \cdot dS = \frac{q}{\epsilon0} \implies E (4 \pi r^2) = \frac{q}{\epsilon0} \implies E = \frac{q}{4 \pi \epsilon_0 r^2}

  Potential difference between the two concentric spheres

  V = - \inta^b E \cdot dr = - \inta^b \frac{q}{4 \pi \epsilon0 r^2} dr = \frac{q}{4 \pi \epsilon0} (\frac{1}{a} - \frac{1}{b}) = \frac{q (b-a)}{4 \pi \epsilon_0 ab}

  C = \frac{q}{V} = \frac{4 \pi \epsilon_0 ab}{b-a}

Capacitors in Series and Parallel

• Equivalent capacitance is meant the capacitance of a single capacitor that can be substituted for the combination.

• One can easily solve unknown quantities in the ckt.

• Capacitors connected in series: Fig shows three capacitors connected in series to a battery, maintaining a potential difference of V volts across the left and right terminals of the series combination.

  • The potential differences produced across the individual capacitors are V1, V2, V3 and Capacitances are C1, C2, C3 respectively.
  • The equivalent capacitance is the reciprocal of the sum of the reciprocals of the individual capacitances.
  • The left-hand plate of every capacitor of the series connection carries A charge q of one sign while the right-hand plate of every capacitor Carries a charge of equal magnitude.
  • V = V1 + V2 + V_3
  • V1 = \frac{q}{C1}, V2 = \frac{q}{C2}, V3 = \frac{q}{C3}

• Equivalent capacitance:
  \frac{q}{C{eq}} = \frac{q}{C1} + \frac{q}{C2} + \frac{q}{C3}

  \frac{1}{C{eq}} = \frac{1}{C1} + \frac{1}{C2} + \frac{1}{C3}

• Capacitors connected in parallel: Fig. shows three capacitors connected in parallel to a battery V.

  • The terminals of the battery are wired directly to the plates of the three capacitors.
  • The connection is said to be parallel when the potential difference is the same.
  • The total charge q that is delivered by the battery to the combination is shared among the capacitors.

• For each capacitor,

  q1 = C1V, q2 = C2V, q3 = C3V

• The total charge on the parallel combination is then

  q = q1 + q2 + q3 = C1V + C2V + C3V = (C1 + C2 + C_3)V

  C{eq} = \frac{q}{V} = C1 + C2 + C3

• The equivalent capacitance of a parallel combination is simply the sum of the capacitance of the individual capacitors.
  \sum{j=1}^n C{j}

Energy stored in a Charged Capacitor

• Suppose a charge q' has been transferred from one plate to the other.

• The potential difference V' between the plates at that instant will be \frac{q'}{C} where C is the capacitance.

• If an extra amount of charge dq' is to be transferred then, the increase in work done will be

  dW = V' dq' = \frac{q'}{C} dq'

• The work required to charge the capacitor plates up to a final value q is

  W = \int0^q dW = \int0^q \frac{q'}{C} = \frac{q^2}{2C}

• The work is stored as the potential energy U in the capacitor:

  U = \frac{q^2}{2C} = \frac{1}{2}CV^2

Capacitor with Dielectric

• A dielectric material is an electrical insulator that can be polarized by an applied electric field.
• The effect of the presence of a dielectric in the space between the plates of a capacitor was first investigated by M. Faraday.
• He showed that when both the capacitor charged to the same potential difference, the charge on the capacitor with dielectric was greater than that on the other.
• The dielectric constant is the ratio of the permittivity of a substance to the permittivity of free space.
• If C is the capacitance of the capacitor with dielectric and C0 is the capacitance of no dielectric material then the ratio \frac{C}{C0} is called the dielectric constant k of the dielectric material.
• Materials with high dielectric constants are useful in the manufacture of high-value capacitors.
• For example, dry air has a low dielectric constant, but it makes an excellent dielectric material for capacitors used in high-power radio- frequency transmitters.

Charging and Discharging a Capacitor

• It is important to study what happens while a capacitor is charging and discharging.
• It is the ability to control and predict the rate at which a capacitor charges and discharges that makes capacitors really useful in electronic timing circuits.
• The rate at which a capacitor can be charged or discharged depends on:
  • the capacitance of the capacitor and
  • the resistance of the circuit through which it is being charged or is discharging.
• The Capacitor is Charging
• Ohm's law
• Ohm's law states that the potential difference between the ends of a conductor varies directly as the current flowing in it, provided the temperature does not change and the physical state of the conductor remains the same.

V = IR

• Where I is the current through the conductor in units of amperes, V is the voltage measured across the conductor in units of volts and R is the resistance of the conductor in units of ohms.
• The ohm is defined as the resistance of a conductor in which a potential difference of 1 volt is developed when a current of 1 ampere flows through it.
• The specific resistance or resistivity is the resistance offered by a conductor of unit length and unit cross section. Its unit is ohm-metre or mho.

Kirchhoff's Circuit Laws

• (1st Law): The algebraic sum of currents in a network of conductors meeting at a point is zero.
• Or The current entering any junction is equal to the current leaving that junction.
• Kirchhoff's voltage law (KVL, 2nd Law) Kirchoffs Voltage Law or KVL, states that “in any closed loop network, the total voltage around the loop is equal to the sum of all the voltage drops within the same loop” which is also equal to zero.

Application of Kirchhoff’s Law to Wheatstone’s Bridge

• When the galvanometer shows, “zero deflection” the points B and D are the same potential and ig=0.

  \frac{P}{Q} = \frac{R}{S}

  S = R\frac{Q}{P}

• Find the value of the unknown resistance S in terms of known resistance P, Q, and R.

Alternating Currents

• An alternating current is a periodic function of time or one which passess through a cycle of changes at regular intervals.

AC vs DC

• AC is the type of electric current which varies instantaneously with time, whereas DC remains constant with time.
• Flow of electrons in AC is bi-directional, whereas DC is uni-directional.
• Frequency of AC is between 50 Hz to 60 Hz (country dependent), whereas DC is zero.
• Power factor for AC lies between 0 and 1, whereas DC is always 1.

Circuit Containing Pure Resistance Only

• Let an AC source of emf E be connected to a pure resistance R.

• The instantaneous emf from the source is given by E = E0 sin ωt

• Ohm's law:

  I = \frac{E}{R} = \frac{E0}{R} \sin{\omega t} \implies I = I0 \sin{\omega t}

Circuit Containing Pure Inductor Only

• Inductors oppose changes in current through them, by dropping a voltage directly proportional to the rate of change of current.

• Let an AC source be connected across a pure inductive element. If the alternating current I = I0 sin ωt flows through it. Then

  E = L \frac{dI}{dt} = L \omega I_0 \cos{\omega t}

Circuit Containing Pure Capacitance Only

• Let an AC source be connected across a pure inductive element. If E = E0sinωt then, q = EC = CE0 sin ωt

Flow of A.C through inductor, capacitor, and resistance in series

• Consider the RLC circuit. We can determine it in vector form the emf of the circuit is

  I = \frac{V}{R + j(\omega L - \omega^{-1} C)}

Special Cases of RLC Circuits

• When \omega L > \frac{1}{\omega C}, the circuit behaves as an inductive circuit.
• When \omega L < \frac{1}{\omega C}, the circuit behaves as a capacitive circuit.
• When \omega L = \frac{1}{\omega C}, the circuit behaves as a purely resistive circuit.

The Parallel Resonance Circuit

• A parallel circuit containing a resistance, R, an inductance, L, and a capacitance, C, will produce a parallel resonance circuit when the resultant current through the parallel combination is in phase with the supply voltage.
• At resonance there will be a large circulating current between the inductor and the capacitor due to the energy of the oscillations, and then parallel circuits produce current resonance.