Coulomb's law states that the force of attraction or repulsion between two point charges is directly proportional to the product of the magnitudes of the charges and inversely proportional to the square of the distance between them, acting along the line joining them. Mathematically, F = k q1 q2 / r^2, where k = 1/(4 pi epsilon0) = 9 x 10^9 N m^2 C^-2 in free space.

F = k q1 q2 / r^2

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

The electric field intensity at a point is the force experienced by a unit positive test charge placed at that point. It is a vector quantity given by E = F/q0, where q0 is the small positive test charge. Its SI unit is newton per coulomb (N/C) or equivalently volt per metre (V/m).

E = F/q0

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

When a dipole of moment p is placed in a uniform electric field E at an angle theta, the two charges experience equal and opposite forces qE, forming a couple. The magnitude of the torque = force x perpendicular distance = qE x (2a sin theta) = (q x 2a) E sin theta = pE sin theta. In vector form, torque = p x E. The torque tends to align the dipole with the field.

torque = pE sin theta = p x E

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)

The electric potential at a point is the work done in bringing a unit positive charge from infinity to that point against the electric field. It is a scalar quantity and its SI unit is the volt (V), where 1 volt = 1 joule per coulomb. The potential due to a point charge Q at distance r is V = kQ/r = Q/(4 pi epsilon0 r).

V = kQ/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

The electric field is the negative gradient of the electric potential: E = -dV/dr. This means the field points in the direction in which the potential decreases most rapidly. The negative sign indicates that the electric field is directed from a region of higher potential to a region of lower potential, that is, potential decreases along the direction of the field.

E = -dV/dr

The capacitance of a conductor is the ratio of the charge given to it to the resulting rise in its potential: C = Q/V. It is a measure of the ability of the conductor to store charge. Its SI unit is the farad (F), where 1 farad = 1 coulomb per volt. Capacitance depends on the size and shape of the conductor and the surrounding medium.

C = Q/V

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)

The energy stored in a charged capacitor is the work done in charging it. It can be written in three equivalent forms: U = (1/2) C V^2 = (1/2) Q V = Q^2/(2C), where C is the capacitance, Q the charge and V the potential difference. This energy is stored in the electric field between the plates.

U = (1/2) C V^2 = (1/2) QV = Q^2/(2C)

When a dielectric of dielectric constant K is fully inserted between the plates, the capacitance increases K times: C = K C0, where C0 is the capacitance with air. This is because the dielectric gets polarised and sets up an internal field opposite to the applied field, reducing the net field and hence the potential difference for the same charge; since C = Q/V, a smaller V means a larger C.

C = K C0

Ohm's law states that, at constant temperature, the current flowing through a conductor is directly proportional to the potential difference across its ends, so V = IR, where R is a constant called the resistance. Resistance is the opposition offered by a conductor to the flow of current and is defined as R = V/I. Its SI unit is the ohm.

V = IR

Resistivity (specific resistance) is the resistance of a conductor of unit length and unit area of cross-section; it depends on the material and temperature but not on its dimensions. The resistance of a wire is R = rho L/A, so it is directly proportional to its length L and inversely proportional to its area of cross-section A. The SI unit of resistivity is the ohm metre.

R = rho L/A

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)