Electrostatics

Electric charge is a fundamental property of matter. It is conserved (total charge in an isolated system remains constant), quantized (q = ne, where n is an integer and e = 1.6 x $10^{-19}$ C; [e] = [A T], SI unit: coulomb, C), and additive (algebraic sum of individual charges).

Coulomb's law gives the force between two point charges: F = kq1q2/$r^{2}$, where k = 1/(4 $\pi \; \epsilon_{0}$) = 9 x $10^{9}$ N $\frac{m^{2}}{C^{2}}$, q1 and q2 are the charges (C), and r is the separation (m).
Dimensional formula: [F] = [M L $T^{-2}$], SI unit: newton (N).
The permittivity of free space $\epsilon_{0}$ = 8.85 x $10^{-12} \; C^{2}$/(N $m^{2}$); [$\epsilon_{0}$] = [$M^{-1} \; L^{-3} \; T^{4} \; A^{2}$]. The superposition principle states that the net force on a charge is the vector sum of individual Coulomb forces.

The electric field at a point is E = F/$q_{0}$ = kQ/$r^{2}$, where $q_{0}$ is a small positive test charge; [E] = [M L $T^{-3} \; A^{-1}$], SI unit: N/C = V/m. Field lines originate from positive charges, terminate on negative charges, never cross, and their density represents field strength.
For a uniformly charged ring on the axis, E = kQx/($R^{2}$ + $x^{2}$)^($\frac{3}{2}$), which is zero at the center and maximum at x = R/$\sqrt{2}$.

An electric dipole consists of two equal and opposite charges separated by distance 2l.
Dipole moment p = q .
2l (direction from -q to +q); [p] = [A T L], SI unit: C m.
The torque on a dipole in a uniform field is $\tau$ = pE sin($\theta$); [$\tau$] = [M $L^{2} \; T^{-2}$], SI unit: N m.
The field on the axial line is $E_{axial}$ = 2kp/$r^{3}$ (for r >> l), and on the equatorial line $E_{eq}$ = kp/$r^{3}$ — note the factor of 2 difference.

Gauss's law: Electric flux $\Phi$ = integral(E .
dA) = $\frac{q_{enc}}{\epsilon_{0}}$; [$\Phi$] = [M $L^{3} \; T^{-3} \; A^{-1}$], SI unit: V m.
Applications: infinite line charge E = $\lambda$/(2 $\pi \; \epsilon_{0}$ r), infinite plane sheet E = $\sigma$/(2 $\epsilon_{0}$), conducting sphere E = kQ/$r^{2}$ (outside), E = 0 (inside), and insulating sphere E = kQr/$R^{3}$ (inside, linearly increasing).

Electric potential: V = kQ/r; [V] = [M $L^{2} \; T^{-3} \; A^{-1}$], SI unit: volt (V). The relation E = -dV/dr connects field and potential. Equipotential surfaces are perpendicular to field lines, and no work is done moving a charge along them. Potential energy U = kq1q2/r; [U] = [M $L^{2} \; T^{-2}$], SI unit: joule (J). Sign convention: U > 0 for like charges (repulsion), U < 0 for unlike charges (attraction).

Capacitance: C = Q/V; [C] = [$M^{-1} \; L^{-2} \; T^{4} \; A^{2}$], SI unit: farad (F).
For a parallel plate capacitor, C = $\epsilon_{0}$ A/d (vacuum) or C = K $\epsilon_{0}$ A/d (with dielectric constant K).
Series: 1/$C_{eq}$ = 1/C₁ + 1/C₂.
Parallel: $C_{eq}$ = C₁ + C₂.
Energy stored: U = ($\frac{1}{2}$)$CV^{2}$ = $Q^{2}$/(2C) = ($\frac{1}{2}$)QV; [U] = [M $L^{2} \; T^{-2}$], SI unit: J.

The key testable concept is Gauss's law applications and capacitor combinations with dielectrics, which together account for the majority of NEET electrostatics questions.

Solved Numericals

N₁. Two charges +3 uC and -3 uC are placed 20 cm apart. Find the electric field at a point on the axial line 30 cm from the center of the dipole.

Given: q = 3 uC = 3 x $10^{-6}$ C, 2l = 20 cm = 0.20 m so l = 0.10 m, r = 30 cm = 0.30 m. Dipole moment: p = q x 2l = 3 x $10^{-6}$ C x 0.20 m = 6 x $10^{-7}$ C m.

Exact formula (axial): E = 2kpr/($r^{2}$ - $l^{2}$)² E = 2 x 9 x $10^{9}$ N $\frac{m^{2}}{C^{2}}$ x 6 x $10^{-7}$ C m x 0.30 m / (0.$30^{2}$ - 0.$10^{2}$)² $m^{4}$ E = 2 x 9 x $10^{9}$ x 6 x $10^{-7}$ x 0.30 / (0.09 - 0.01)² E = 3.24 x $10^{3}$ / 6.4 x $10^{-3}$ = 5.06 x $10^{5}$ / 6.4 x $10^{-3}$ Numerator: 2 x 9 x $10^{9}$ x 6 x $10^{-7}$ x 0.30 = 3240 N $m^{2}$/(C $m^{2}$) ... Let us recalculate step by step. Numerator = 2 x (9 x $10^{9}$) x (6 x $10^{-7}$) x 0.30 = 2 x 9 x 6 x 0.30 x 10^(9-7) = 32.4 x $10^{2}$ = 3240. Denominator = (0.09 - 0.01)² = (0.08)² = 6.4 x $10^{-3} \; m^{4}$. E = 3240 / 6.4 x $10^{-3}$ = 5.0625 x $10^{5}$ N/C.

Approximate formula (r >> l): $E_{approx}$ = 2kp/$r^{3}$ = 2 x 9 x $10^{9}$ x 6 x $10^{-7}$ / (0.30)³ = 10800 / 0.027 = 4.0 x $10^{5}$ N/C.

The approximate formula gives a lower value because it neglects the $l^{2}$ term. The exact value is 5.06 x $10^{5}$ N/C. At r = 0.30 m with l = 0.10 m, r is only 3 times l, so the approximation has noticeable error (~21%).

N₂. A parallel plate capacitor of capacitance 100 pF is charged to 200 V. Find the energy stored. If a dielectric of K = 5 is inserted with battery disconnected, find new capacitance, voltage, and energy.

Initial: C = 100 pF = 100 x $10^{-12}$ F = $10^{-10}$ F, V = 200 V. Energy: U = ($\frac{1}{2}$)$CV^{2}$ = ($\frac{1}{2}$) x $10^{-10}$ F x (200 V)² = ($\frac{1}{2}$) x $10^{-10}$ x 4 x $10^{4}$ = 2 x $10^{-6}$ J = 2 uJ. Charge stored: Q = CV = $10^{-10}$ x 200 = 2 x $10^{-8}$ C = 20 nC.

Battery disconnected (Q remains constant): New capacitance: C' = KC = 5 x 100 pF = 500 pF = 5 x $10^{-10}$ F. New voltage: V' = Q/C' = 2 x $10^{-8}$ C / 5 x $10^{-10}$ F = 40 V. New energy: U' = ($\frac{1}{2}$)C'V'² = ($\frac{1}{2}$) x 5 x $10^{-10}$ x (40)² = ($\frac{1}{2}$) x 5 x $10^{-10}$ x 1600 = 4 x $10^{-7}$ J = 0.4 uJ.

Energy decreases from 2 uJ to 0.4 uJ (by factor K = 5). The energy is spent in pulling the dielectric into the capacitor.

N₃. Using Gauss's law, derive the electric field at distance r from an infinitely long straight wire with linear charge density $\lambda$ = 5 uC/m. Calculate E at r = 10 cm.

Choose a cylindrical Gaussian surface of radius r and length L, coaxial with the wire. By symmetry, E is radial and constant over the curved surface, and zero through the flat ends. Flux: $\Phi$ = E x 2 $\pi$ r L (curved surface only). Charge enclosed: $q_{enc}$ = $\lambda$ L. By Gauss's law: E x 2 $\pi$ r L = $\lambda$ L / $\epsilon_{0}$. Therefore: E = $\lambda$ / (2 $\pi \; \epsilon_{0}$ r); [E] = [M L $T^{-3} \; A^{-1}$], SI unit: N/C.

Calculation: $\lambda$ = 5 uC/m = 5 x $10^{-6}$ C/m, r = 10 cm = 0.10 m. E = 5 x $10^{-6}$ / (2 $\pi$ x 8.85 x $10^{-12}$ x 0.10) = 5 x $10^{-6}$ / (5.563 x $10^{-12}$) = 8.99 x $10^{5}$ N/C ~ 9.0 x $10^{5}$ N/C.

Dimensional check: [C/m] / [$C^{2}$/(N $m^{2}$) x m] = [C/m] x [N $\frac{m^{2}}{C^{2}}$] / [m] = [N/C]. Verified.