Given q1 = 2 x 10^-6 C, q2 = 3 x 10^-6 C, r = 30 cm = 0.3 m. Using F = k q1 q2 / r^2 = (9 x 10^9)(2 x 10^-6)(3 x 10^-6)/(0.3)^2 = (9 x 10^9 x 6 x 10^-12)/0.09 = 0.054/0.09 = 0.6 N. The force is repulsive since both charges are positive.

F = k q1 q2 / r^2

Given Q = 5 x 10^-6 C, r = 0.3 m. The field due to a point charge is E = kQ/r^2 = (9 x 10^9)(5 x 10^-6)/(0.3)^2 = (45000)/0.09 = 5 x 10^5 N/C. The field is directed radially outward since the charge is positive.

E = kQ/r^2

An electric dipole is a pair of equal and opposite charges (+q and -q) separated by a small distance 2a. The dipole moment p = q x 2a, directed from the negative to the positive charge. On the axial line at distance r from the centre, the field due to +q is k q/(r-a)^2 (away) and due to -q is k q/(r+a)^2 (towards). Net E = kq[1/(r-a)^2 - 1/(r+a)^2] = kq[(4ar)/((r^2-a^2)^2)]. For a short dipole (r much greater than a), E = k(2p)/r^3, directed along the dipole moment.

E(axial) = 2kp/r^3

Gauss's law states that the total electric flux through any closed surface is equal to 1/epsilon0 times the total charge enclosed by that surface. Mathematically, the surface integral of E over the closed surface = q(enclosed)/epsilon0. The closed surface over which the flux is calculated is called a Gaussian surface. The law is useful for finding the electric field of symmetric charge distributions.

flux = q(enclosed)/epsilon0

Consider an infinitely long straight wire with linear charge density lambda. Choose a coaxial cylindrical Gaussian surface of radius r and length L. By symmetry, the field E is radial and uniform over the curved surface. The flux through the curved surface = E x (2 pi r L); the flat ends contribute no flux. Charge enclosed = lambda L. By Gauss's law, E x 2 pi r L = lambda L / epsilon0, so E = lambda/(2 pi epsilon0 r). The field is directed radially and varies as 1/r.

E = lambda/(2 pi epsilon0 r)

Given Q = 4 x 10^-6 C, r = 9 cm = 0.09 m. The potential due to a point charge is V = kQ/r = (9 x 10^9)(4 x 10^-6)/0.09 = (36000)/0.09 = 4 x 10^5 V. Thus the potential at the point is 4 x 10^5 volt.

V = kQ/r

Let a parallel plate capacitor have plate area A, separation d, and charge Q with surface charge density sigma = Q/A. The uniform field between the plates is E = sigma/epsilon0 = Q/(A epsilon0). The potential difference between the plates is V = E d = Q d/(A epsilon0). Hence capacitance C = Q/V = Q/(Q d/(A epsilon0)) = epsilon0 A/d. So the capacitance increases with plate area and decreases with plate separation.

C = epsilon0 A/d

In series: 1/Cs = 1/2 + 1/3 + 1/6 = 3/6 + 2/6 + 1/6 = 6/6 = 1, so Cs = 1 microfarad. In parallel: Cp = 2 + 3 + 6 = 11 microfarad. Thus the series combination gives 1 microfarad and the parallel combination gives 11 microfarad.

series: 1/Cs = sum(1/Ci); parallel: Cp = sum(Ci)

Drift velocity is the small average velocity with which free electrons move in a conductor under an applied electric field, opposite to the field. Consider a conductor of area A with n free electrons per unit volume moving with drift velocity vd. In time t, electrons in a length vd t cross a section, so charge crossing = (n)(A vd t)(e). Current I = charge/time = n A vd e t / t = n A e vd. Hence I = n A e vd.

I = n A e vd

In parallel: 1/Rp = 1/2 + 1/3 + 1/6 = 3/6 + 2/6 + 1/6 = 6/6 = 1, so Rp = 1 ohm. In series: Rs = 2 + 3 + 6 = 11 ohm. Thus the parallel combination gives 1 ohm (less than the smallest resistance) and the series combination gives 11 ohm.

parallel: 1/Rp = sum(1/Ri); series: Rs = sum(Ri)

A Wheatstone bridge has four resistances P, Q, R and S arranged in a quadrilateral, with a galvanometer between the junctions of P-Q and R-S, and a cell across the other diagonal. When the bridge is balanced, no current flows through the galvanometer, so the junction connected by the galvanometer are at the same potential. Applying Kirchhoff's laws, the potential drops give P/Q = R/S. This is the balance condition; if three resistances are known, the fourth can be found.

P/Q = R/S

Given emf E = 2 V, internal resistance r = 0.5 ohm, external resistance R = 4.5 ohm. Current I = E/(R + r) = 2/(4.5 + 0.5) = 2/5 = 0.4 A. Terminal potential difference V = I R = 0.4 x 4.5 = 1.8 V (equivalently V = E - I r = 2 - 0.4 x 0.5 = 1.8 V).

I = E/(R + r); V = E - I r

The force on a current-carrying conductor is F = B I L sin theta. Here B = 0.2 T, I = 4 A, L = 0.5 m and theta = 90 deg (so sin theta = 1). F = 0.2 x 4 x 0.5 x 1 = 0.4 N. The direction of the force is given by Fleming's left-hand rule.

F = B I L sin theta

Consider a circular loop of radius R carrying current I. By the Biot-Savart law, each element I dl is perpendicular to the line joining it to the centre (theta = 90 deg), so dB = (mu0/4 pi)(I dl)/R^2. All elements produce fields in the same direction (perpendicular to the plane of the loop) at the centre, so they simply add. Integrating dl around the loop gives the circumference 2 pi R: B = (mu0/4 pi)(I/R^2)(2 pi R) = mu0 I/(2R). For N turns, B = mu0 N I/(2R).

B = mu0 I/(2R)

Ampere's circuital law states that the line integral of the magnetic field B around any closed loop is equal to mu0 times the total current enclosed by the loop: the integral of B.dl = mu0 I(enclosed). Applying it to a long solenoid with n turns per unit length carrying current I, the magnetic field inside is uniform and given by B = mu0 n I, directed along the axis; the field outside an ideal long solenoid is nearly zero.

B = mu0 n I

When a charged particle of charge q and mass m enters a magnetic field B perpendicularly with speed v, the magnetic force qvB provides the centripetal force m v^2/r. Equating: q v B = m v^2/r, so the radius r = m v/(q B). The time period T = 2 pi r/v = 2 pi m/(q B), which is independent of the speed and radius. This principle is used in the cyclotron.

r = mv/(qB); T = 2 pi m/(qB)

A moving coil galvanometer works on the principle that a current-carrying coil placed in a magnetic field experiences a torque. The deflection of the coil is directly proportional to the current passing through it (I is proportional to the deflection). To convert a galvanometer into an ammeter, a small resistance (shunt) is connected in parallel with it, so most of the current bypasses the galvanometer. To convert it into a voltmeter, a high resistance is connected in series with it, so that only a small current flows and it can measure potential difference.

The three elements of the earth's magnetism are: (1) Magnetic declination - the angle between the geographic meridian and the magnetic meridian at a place. (2) Magnetic dip or inclination - the angle made by the earth's total magnetic field with the horizontal at a place. (3) Horizontal component of the earth's field (BH) - the component of the earth's total magnetic field in the horizontal direction, given by BH = B cos(dip). These three completely specify the earth's field at a place.

BH = B cos(dip)

Consider a conducting rod of length l moving with velocity v perpendicular to a uniform magnetic field B. In time dt the rod sweeps an area dA = l (v dt), so the change in flux is d(flux) = B dA = B l v dt. By Faraday's law, the induced emf e = d(flux)/dt = B l v. Alternatively, the free electrons in the rod experience a force qvB that pushes them to one end, setting up an emf e = B v l across the rod.

e = B l v

Given N = 500, initial flux = 0.01 Wb, final flux = 0.05 Wb, so change in flux = 0.05 - 0.01 = 0.04 Wb, and time = 0.1 s. The induced emf (magnitude) = N (change in flux)/time = 500 x 0.04/0.1 = 500 x 0.4 = 200 V.

e = N (change in flux)/time