Electrostatics

🟢 Lite — Quick Review (1h–1d)

Rapid summary for last-minute revision before your exam.

Electrostatics — Key Facts

Coulomb’s Law: The force between two point charges $q_1$ and $q_2$ separated by distance $r$: $F = \frac{1}{4\pi\varepsilon_0}\frac{q_1 q_2}{r^2} = k\frac{q_1 q_2}{r^2}$. In a medium: $F = \frac{1}{4\pi\varepsilon}\frac{q_1 q_2}{r^2}$ where $\varepsilon = \varepsilon_0 \varepsilon_r$. $k = 1/(4\pi\varepsilon_0) ≈ 8.99 × 10^9$ N·m²/C². Elementary charge $e = 1.602 × 10^{-19}$ C.

Electric Field: Electric field intensity $\vec{E} = \vec{F}/q = \frac{1}{4\pi\varepsilon_0}\frac{Q}{r^2}\hat{r}$ (for a point charge $Q$). For multiple charges: use superposition. Direction away from positive charge, toward negative charge.

Electric Potential: Potential at a point $V = W/q = \frac{1}{4\pi\varepsilon_0}\frac{Q}{r}$. For multiple charges: algebraic sum (scalars). Relationship: $\vec{E} = -\nabla V = -\frac{dV}{dr}\hat{r}$.

⚡ Exam tip: Potential is a scalar (can be positive or negative), electric field is a vector. Always use vector addition for electric field, algebraic addition for potential.

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🟡 Standard — Regular Study (2d–2mo)

Standard content for students with a few days to months.

Electrostatics — CUET Study Guide

Electric Dipole: Dipole moment $\vec{p} = q\vec{d}$ (direction from negative to positive charge). For a dipole in uniform electric field:

• Torque: $\tau = pE\sin\theta = |\vec{p} \times \vec{E}|$
• Potential energy: $U = -\vec{p} \cdot \vec{E} = -pE\cos\theta$ (zero when $\perp$ to field)

Potential at a point due to dipole:

• Axial point (on the axis): $V = \frac{1}{4\pi\varepsilon_0}\frac{p}{r^2}$
• Equatorial point (on perpendicular bisector): $V = -\frac{1}{4\pi\varepsilon_0}\frac{p}{2r^3}$

Electric field at axial point: $E = \frac{1}{4\pi\varepsilon_0}\frac{2p}{r^3}$. At equatorial point: $E = \frac{1}{4\pi\varepsilon_0}\frac{p}{r^3}$ (both directed along the axis).

Gauss’s Law: $\oint \vec{E} \cdot d\vec{A} = \frac{q_{enc}}{\varepsilon_0}$ (for any closed surface). This is always true, but it’s most useful for symmetric charge distributions.

Applications:

1. Spherical shell: Total charge $q$ on surface. $E_{inside} = 0$, $E_{at\ surface} = \frac{1}{4\pi\varepsilon_0}\frac{q}{R^2}$, $E_{outside} = \frac{1}{4\pi\varepsilon_0}\frac{q}{r^2}$. The field outside is the same as if all the charge were at the centre.
2. Infinite uniformly charged plane sheet: $E = \frac{\sigma}{2\varepsilon_0}$ (directed away from positively charged sheet).
3. Uniformly charged solid sphere (radius $R$, total charge $Q$): $E_{inside} = \frac{1}{4\pi\varepsilon_0}\frac{Qr}{R^3}$, $E_{outside} = \frac{1}{4\pi\varepsilon_0}\frac{Q}{r^2}$.

Capacitors: Capacitance $C = Q/V = \varepsilon_0 A/d$ (parallel plate in vacuum). For a medium with dielectric constant $K$: $C = KA\varepsilon_0/d = KC_{vacuum}$. Energy stored: $U = \frac{1}{2}CV^2 = \frac{1}{2}QV^2 = \frac{Q^2}{2C}$. Energy density: $u = \frac{U}{Ad} = \frac{1}{2}\varepsilon_0 E^2$.

Capacitor combinations:

• Series: $1/C_{eq} = 1/C_1 + 1/C_2 + …$ (charge same on all, voltage divides)
• Parallel: $C_{eq} = C_1 + C_2 + …$ (voltage same across all, charge adds)

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🔴 Extended — Deep Study (3mo+)

Comprehensive coverage for students on a longer study timeline.

Electrostatics — Comprehensive Physics Notes

Continuous Charge Distributions:

Linear charge density $\lambda = Q/L$ (C/m). Surface charge density $\sigma = Q/A$ (C/m²). Volume charge density $\rho = Q/V$ (C/m³).

Electric field of an infinite line charge (linear density $\lambda$): Using a cylindrical Gaussian surface of radius $r$ and length $L$: $\oint \vec{E} \cdot d\vec{A} = E(2\pi r L) = \lambda L/\varepsilon_0$. So $E = \frac{\lambda}{2\pi\varepsilon_0 r}$ (directed radially outward for positive $\lambda$).

Electric field of a uniformly charged ring (axis): For a ring of radius $R$ with total charge $Q$, on the axis at distance $x$: $E = \frac{1}{4\pi\varepsilon_0}\frac{Qx}{(x^2+R^2)^{3/2}}$. Maximum at $x = R/\sqrt{2}$ where $E = \frac{1}{4\pi\varepsilon_0}\frac{Q}{R^2} \cdot \frac{2\sqrt{3}}{9}$.

Method of Images: Used for finding potential when a conductor is placed near point charges. The conductor’s surface is an equipotential. Replace the conductor with an image charge such that the boundary conditions are satisfied.

Example: A point charge $+q$ at distance $d$ from an infinite grounded conducting plane. The induced charge on the plane is equivalent to an image charge $-q$ at distance $d$ behind the plane. Force on $+q$: attractive, $F = \frac{1}{4\pi\varepsilon_0}\frac{q^2}{(2d)^2}$.

Dielectrics — Detailed: When a dielectric is placed in an electric field, polar molecules align with the field (orientational polarisation) and the electron cloud distorts (electronic polarisation). The effective field inside is reduced: $E_{eff} = E_{applied}/K$. Capacitance increases by factor $K$ (dielectric constant). For vacuum $K = 1$, air $K ≈ 1.0006$ (≈1), water $K ≈ 80$, glass $K ≈ 4-8$.

Energy with dielectric: $U = \frac{1}{2}CV^2$ still holds, but $C = KC_0$.

Dielectric breakdown: When $E$ exceeds a critical value, the dielectric becomes conductive. For air: $E_{break} ≈ 3 × 10^6$ V/m. This is why lightning strikes occur — the electric field in storm clouds exceeds this value.

Van de Graaff Generator: A large hollow sphere supported by an insulating column. Charge is sprayed onto a moving belt, carried inside the sphere, and collected by a brush. The sphere reaches a potential limited by corona discharge: $V = kQ/r$. Maximum potential: $V_{max} = E_{break} × r$ (where $E_{break} ≈ 3 × 10^6$ V/m for air at STP). For a sphere of radius $r = 1$ m, $V_{max} ≈ 3$ GV. Practically limited to ~10 MV due to air breakdown.

Electric Pressure: The force per unit area (pressure) on a charged conductor surface: $P = \frac{\sigma^2}{2\varepsilon_0} = \frac{\varepsilon_0 E^2}{2}$. This pressure pushes outward — it’s what makes balloon surfaces taut.

Capacitor Dielectric Insertion: When a dielectric is inserted between the plates of a capacitor connected to a battery: voltage stays constant ($V$ = battery voltage), charge increases ($Q = CV$ increases because $C$ increases), energy increases (from $U = \frac{1}{2}CV^2$). The dielectric is attracted into the capacitor (work done by the field).

When a dielectric is inserted into an isolated (uncharged) capacitor: charge $Q$ stays constant, capacitance increases, voltage decreases ($V = Q/C$ decreases), energy decreases (the energy decrease appears as work done pulling the dielectric in).

JEE Pattern Analysis: CUET/JEE questions frequently test: (1) Gauss’s law applications (spherical shell, infinite plane, solid sphere), (2) Electric potential from dipole and point charges, (3) Capacitor combinations (series and parallel), (4) Energy stored in capacitors with and without dielectrics, (5) Force between capacitor plates $F = \frac{Q^2}{2\varepsilon_0 A}$. JEE 2023: “Two capacitors $C_1 = 2 \mu F$ and $C_2 = 4 \mu F$ are connected in series across a 100 V supply. Find the voltage across $C_1$.” Answer: Charge on both is same in series: $Q = C_{eq} V = (4/3) \mu F × 100 V = 133.3 \mu C$. $V_1 = Q/C_1 = 133.3/2 = 66.7$ V.

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