JEE Physics Electric Charges and Fields Notes Set 06

Charge and Coulomb's Law

Charge is quantised (\( q=ne \), \( e=1.6\times10^{-19} \) C), conserved in every isolated process, additive as a scalar, and invariant with speed. Like charges repel and unlike charges attract. A body is charged by friction, by conduction, or by induction; only electrons move, so a body gains negative charge by gaining electrons.
Coulomb's law for two point charges in vacuum: \( F=\frac{1}{4\pi\varepsilon_0}\frac{q_1q_2}{r^2}=k\frac{q_1q_2}{r^2} \), \( k=9\times10^9 \) N m\( ^2 \)/C\( ^2 \). The force acts along the line joining the charges and the two forces form an action-reaction pair. In a medium of dielectric constant K the force becomes \( F/K \).
With more than one charge, the force on any one charge is the vector sum of the forces due to the others taken one at a time (superposition); the presence of a third charge does not change the force between two.

Equilibrium of Point Charges

To find a point where the field of two like charges \( q_1 \) and \( q_2 \) vanishes, look between them, at distances in the ratio \( \sqrt{q_1}:\sqrt{q_2} \) from the charges. For unlike charges the point lies outside the pair, on the side of the smaller charge.
For three collinear charges to be in equilibrium together, the outer two must be of the same sign and the middle one of opposite sign. The middle charge must be smaller than the larger outer charge in magnitude, and the equilibrium is unstable. Write the net force on any one of the three as zero to get the unknown magnitude.

Electric Field

Field at a point is the force per unit positive test charge, \( \vec E=\frac{\vec F}{q_0} \), unit N/C or V/m. A charge q in a field feels \( \vec F=q\vec E \), along the field for q positive and opposite for q negative. Fields from several sources add as vectors.
Point charge: \( E=\frac{kq}{r^2} \), radially outward for positive q.
Infinitely long line charge (density \( \lambda \)) at distance r: \( E=\frac{\lambda}{2\pi\varepsilon_0r} \), perpendicular to the line.
Infinite sheet (density \( \sigma \)): \( E=\frac{\sigma}{2\varepsilon_0} \) on both sides, independent of distance. Two parallel sheets of equal and opposite density give \( E=\frac{\sigma}{\varepsilon_0} \) between them and zero outside.
Ring of radius R and charge Q, on its axis at distance x: \( E=\frac{kQx}{(R^2+x^2)^{3/2}} \). It is zero at the centre and has a maximum \( \frac{kQ\cdot2}{3\sqrt3R^2} \) at \( x=\frac{R}{\sqrt2} \).
Uniformly charged thin spherical shell: \( E=0 \) inside, \( \frac{kQ}{r^2} \) outside, and \( \frac{\sigma}{2\varepsilon_0} \) (the average of the two sides) just at the surface. Uniformly charged solid non-conducting sphere: \( E=\frac{kQr}{R^3} \) inside and \( \frac{kQ}{r^2} \) outside.
For a finite rod or an arc, take an element \( dq=\lambda\,dl \), write its field, resolve into components and use symmetry to cancel what can be cancelled before integrating. A uniformly charged semicircular wire of radius R has field \( \frac{2k\lambda}{R} \) at the centre, along the symmetry axis.

Electric Dipole

Two equal and opposite charges \( \pm q \) separated by a small distance 2a (or l) form a dipole with moment \( \vec p=q\cdot\vec{l} \), directed from the negative to the positive charge. Unit C m.
Far from the dipole (r much larger than the separation): on the axis \( E=\frac{2kp}{r^3} \) along \( \vec p \); on the equatorial line \( E=\frac{kp}{r^3} \) opposite to \( \vec p \). So the axial field is twice the equatorial field at the same distance. At a general point, at angle \( \theta \) from \( \vec p \), the radial and transverse components are \( \frac{2kp\cos\theta}{r^3} \) and \( \frac{kp\sin\theta}{r^3} \), giving \( E=\frac{kp}{r^3}\sqrt{1+3\cos^2\theta} \).
In a uniform field the net force on a dipole is zero but there is a torque \( \vec\tau=\vec p\times\vec E \), \( \tau=pE\sin\theta \), that tries to align p with E. Potential energy \( U=-\vec p\cdot\vec E=-pE\cos\theta \): minimum (stable) at \( \theta=0 \), maximum (unstable) at \( \theta=180^\circ \). Work to turn it from \( \theta_1 \) to \( \theta_2 \) is \( pE(\cos\theta_1-\cos\theta_2) \).
In a non-uniform field there is also a net force, and the dipole is pulled towards the stronger-field region when aligned with the field. Small oscillations about the stable position have period \( T=2\pi\sqrt{\frac{I}{pE}} \).

Electric Lines of Force

A line of force is the path along which a free positive charge would start moving; the tangent gives the field direction and the density of lines shows the field strength. Lines begin on positive charges and end on negative charges (or run to infinity), never cross, never form closed loops in electrostatics, and meet a conductor's surface at right angles. Their number from a charge q is proportional to q. Lines are not necessarily the path a charge follows, unless it starts from rest in a straight-line field.

Electric Flux and Gauss's Law

Flux through a small area \( d\vec A \) is \( d\phi=\vec E\cdot d\vec A \); for a uniform field and flat surface \( \phi=EA\cos\theta \), where \( \theta \) is the angle between E and the normal. It is a scalar, unit N m\( ^2 \)/C. For a closed surface the normal points outward, so lines leaving count positive.
Gauss's law: \( \oint\vec E\cdot d\vec A=\frac{q_{enclosed}}{\varepsilon_0} \). The flux depends only on the enclosed charge, but E on the surface is due to all charges, inside and outside. The law is useful for finding E when the charge distribution has spherical, cylindrical or planar symmetry, with a Gaussian surface chosen so that E is constant over it and perpendicular or parallel to it.
Solid angle idea: the flux through a surface from a point charge is \( \frac{q}{4\pi\varepsilon_0}\Omega \), where \( \Omega \) is the solid angle subtended at the charge. A charge at the centre of a cube sends \( \frac{q}{6\varepsilon_0} \) through each face; a charge at a corner of the cube sends \( \frac{q}{24\varepsilon_0} \) through each of the three faces not touching the corner and zero through the three that do.

Conductors in Electrostatic Equilibrium

Inside a conductor E = 0 and the whole conductor is at one potential, so its surface is an equipotential. Any excess charge lies on the outer surface. Field just outside the surface is perpendicular to it with magnitude \( E=\frac{\sigma}{\varepsilon_0} \), and the outward pressure on the surface charge is \( \frac{\sigma^2}{2\varepsilon_0} \) per unit area.
A cavity with no charge inside has no field and no charge on its inner wall (electrostatic shielding). If a charge +q is placed inside the cavity, the inner wall gets \( -q \) and the outer surface gets \( +q \) added, whatever the shape.
Grounding a conductor makes its potential zero and draws in or sends out whatever charge is needed. Two conductors joined by a wire reach a common potential and charge flows until that happens. On a conductor of irregular shape, charge density is greater where the radius of curvature is smaller, so the field is strongest at sharp points.

Worked Examples

Worked Example 1. Charges q and 4q are held a distance L apart. Where must a third charge be placed, and what must it be, so that all three are in equilibrium?
Answer:
Between the like charges, at distance x from q: \( \frac{kq}{x^2}=\frac{4kq}{(L-x)^2} \) gives \( L-x=2x \), so \( x=\frac L3 \) from q.
For q to be in equilibrium, the force from the third charge \( q_3 \) must balance the force from 4q: \( \frac{kqq_3}{(L/3)^2}=-\frac{4kq^2}{L^2} \) in magnitude and opposite in direction, so \( q_3=-\frac{4q}{9} \). The same value keeps 4q in equilibrium.

Worked Example 2. A ring of radius 12 cm carries a charge of 6 nC spread uniformly. Find the field on the axis at 5 cm from the centre, and the distance at which the axial field is greatest.
Answer:
\( E=\frac{kQx}{(R^2+x^2)^{3/2}}=\frac{9\times10^9\times6\times10^{-9}\times0.05}{(0.0144+0.0025)^{3/2}}=\frac{2.7}{2.197\times10^{-3}}\approx1.23\times10^3 \) N/C.
Setting \( \frac{dE}{dx}=0 \) gives \( x=\frac{R}{\sqrt2}=\frac{12}{\sqrt2}=8.5 \) cm.

Worked Example 3. A dipole of moment \( 3\times10^{-8} \) C m is placed in a uniform field \( 2\times10^5 \) N/C at 30 degrees to the field. Find the torque, and the work needed to turn it through a further 30 degrees away from the field direction.
Answer:
\( \tau=pE\sin30^\circ=3\times10^{-8}\times2\times10^5\times0.5=3\times10^{-3} \) N m.
\( W=pE(\cos30^\circ-\cos60^\circ)=6\times10^{-3}\times(0.866-0.5)=2.2\times10^{-3} \) J.

Worked Example 4. A solid non-conducting sphere of radius 10 cm holds 5 microcoulomb uniformly through its volume. Find the field at 5 cm, 10 cm and 20 cm from the centre.
Answer:
Inside: \( E=\frac{kQr}{R^3}=\frac{9\times10^9\times5\times10^{-6}\times0.05}{(0.1)^3}=2.25\times10^6 \) N/C.
On the surface: \( E=\frac{kQ}{R^2}=4.5\times10^6 \) N/C.
At 20 cm: \( E=\frac{kQ}{r^2}=\frac{9\times10^9\times5\times10^{-6}}{0.04}=1.125\times10^6 \) N/C.

Worked Example 5. A point charge q sits at one corner of a cube. Find the total flux through the cube and through each face.
Answer:
Place seven more identical cubes around the corner to build a large cube with q at its centre; total flux is \( \frac{q}{\varepsilon_0} \), so each small cube accounts for \( \frac{q}{8\varepsilon_0} \), all through its three outer faces.
Those three faces are equivalent, so each carries \( \frac{q}{24\varepsilon_0} \). The three faces meeting at the corner lie in planes containing q and carry no flux. Total through the cube: \( 3\times\frac{q}{24\varepsilon_0}=\frac{q}{8\varepsilon_0} \).