JEE Physics Electric Charges and Fields Notes Set 04

Electric Charge and Coulomb's Law

Charge is a basic property of matter that comes in two kinds, positive and negative. Like charges repel and unlike charges attract. Three facts are used again and again: charge is conserved (it can be moved but not created or destroyed), it is quantised (every charge is an integral multiple of the electron charge, \( q=ne,\ e=1.6\times10^{-19} \) C) and it is additive (the total charge is the algebraic sum).
Coulomb's law gives the force between two point charges at rest: \( F=\frac{1}{4\pi\varepsilon_0}\frac{q_1q_2}{r^2} \), with \( \frac{1}{4\pi\varepsilon_0}=9\times10^9 \) N m²/C². The force acts along the line joining the charges. In a medium of dielectric constant K the force becomes \( F/K \).
With several charges, the force on one of them is the vector sum of the forces due to each of the others (superposition principle). Two like charges q1 and q2 with a fixed total feel the largest force when they are equal.

Electric Field and Field Lines

The electric field at a point is the force per unit positive test charge, \( \vec E=\frac{\vec F}{q_0} \), measured in N/C or V/m. For a point charge \( E=\frac{kq}{r^2} \), directed away from a positive charge and towards a negative one. A charge q placed in a field E feels \( \vec F=q\vec E \) and has acceleration \( \frac{qE}{m} \).
Field lines start on positive charges and end on negative charges. They never intersect, they are closer together where the field is stronger, and they meet a conductor's surface at right angles. In a uniform field they are parallel and equally spaced.
Important fields (k = 1/4πε₀):

1. Ring of charge q, radius R, on its axis at distance x: \( E=\frac{kqx}{(R^2+x^2)^{3/2}} \). It is zero at the centre and has its maximum value \( \frac{2kq}{3\sqrt3R^2} \) at \( x=\frac{R}{\sqrt2} \).
2. Infinite line of charge (density λ): \( E=\frac{\lambda}{2\pi\varepsilon_0 r} \).
3. Infinite non-conducting sheet (density σ): \( E=\frac{\sigma}{2\varepsilon_0} \), independent of distance. Near the surface of a charged conductor: \( E=\frac{\sigma}{\varepsilon_0} \). Between two oppositely charged parallel plates: \( E=\frac{\sigma}{\varepsilon_0} \); outside, zero.
4. Uniformly charged spherical shell: zero inside, \( \frac{kQ}{r^2} \) outside. Uniformly charged solid non-conducting sphere: \( E=\frac{kQr}{R^3}=\frac{\rho r}{3\varepsilon_0} \) inside and \( \frac{kQ}{r^2} \) outside; the field is largest at the surface.

Electric Dipole

A dipole is a pair of equal and opposite charges ±q separated by a small distance 2a. Its dipole moment \( \vec p=q(2\vec a) \) points from the negative to the positive charge (unit C m).
Axial point (on the line of the dipole), distance r from the centre: \( E=\frac{2kp}{r^3} \) along p. Equatorial point: \( E=\frac{kp}{r^3} \) opposite to p. Far from a dipole the field falls as \( 1/r^3 \).
In a uniform field the net force on a dipole is zero but it feels a torque \( \vec\tau=\vec p\times\vec E \), of magnitude \( pE\sin\theta \). The potential energy is \( U=-\vec p\cdot\vec E=-pE\cos\theta \): minimum (stable equilibrium) at \( \theta=0 \), maximum (unstable) at \( \theta=\pi \). Work to rotate from \( \theta_1 \) to \( \theta_2 \) is \( pE(\cos\theta_1-\cos\theta_2) \). For small oscillations the period is \( T=2\pi\sqrt{I/pE} \).
For a system of charges with zero net charge, \( \vec p=\sum q_i\vec r_i \), independent of the origin.

Electric Flux and Gauss's Law

Flux through a surface is \( \phi=\int\vec E\cdot d\vec A \); for a flat surface in a uniform field \( \phi=EA\cos\theta \), where θ is the angle between E and the normal. The SI unit is N m²/C.
Gauss's law: the net flux through any closed surface equals the enclosed charge divided by ε₀, \( \oint\vec E\cdot d\vec A=\frac{q_{enc}}{\varepsilon_0} \). The field E on the surface is due to all charges, inside and outside, but only the enclosed charge fixes the net flux. A closed surface in a uniform field always has zero net flux.

Choose a Gaussian surface that matches the symmetry (sphere, cylinder, pillbox) so that E is constant and normal to it. This gives the results for the line, sheet and sphere listed above.

Conductors in Electrostatics

Inside a conductor in equilibrium the field is zero, the whole conductor is at one potential and any excess charge lies on its outer surface. The field just outside is perpendicular to the surface with magnitude \( \sigma/\varepsilon_0 \). A hollow conductor shields its cavity from outside fields. Charge on a conductor gathers more densely at sharp points, which is why lightning conductors have pointed tips.

Worked Examples

Example 1. Two charges 25 μC and 36 μC are 11 cm apart. Locate the point on the line joining them where the field is zero.
Answer:
Like charges, so the null point lies between them. Let it be x cm from the 25 μC charge.
\( \frac{25}{x^2}=\frac{36}{(11-x)^2}\Rightarrow\frac{5}{x}=\frac{6}{11-x}\Rightarrow x=5 \) cm.
The point is 5 cm from the 25 μC charge.

Example 2. A dipole of moment \( 2\times10^{-8} \) C m is placed at 30° to a field of \( 10^6 \) N/C. Find the torque and the work needed to turn it until it is along the field.
Answer:
Torque \( \tau=pE\sin30°=2\times10^{-8}\times10^6\times0.5=10^{-2} \) N m.
Potential energy changes from \( -pE\cos30° \) to \( -pE \); work done by the field is \( pE(1-\cos30°) \) = \( 2\times10^{-2}\times0.134=2.7\times10^{-3} \) J.
Torque = 0.01 N m; the field itself supplies about 2.7 mJ in aligning the dipole.

Example 3. Find the flux through a closed surface that encloses charges +3 μC and −1 μC, with a further charge of 5 μC lying outside.
Answer:
Only enclosed charge counts: \( q_{enc}=3-1=2 \) μC.
\( \phi=\frac{2\times10^{-6}}{8.85\times10^{-12}}=2.26\times10^5 \) N m²/C.
The outside charge adds equal inflow and outflow, so the answer is 2.26 × 10⁵ N m²/C.