Electrostatics — Easy Practice (Set 1)

Quick Recap — Charge & Coulomb's Law

• Charge: quantized ($q = ne$, with $e = 1.6\times10^{-19}$ C), conserved, and additive.
• Coulomb's law: $F = \dfrac{kq_1 q_2}{r^2}$, with $k = \dfrac{1}{4\pi\varepsilon_0} = 9\times10^9$ N m$^2$/C$^2$ — like charges repel, unlike attract.
• Superposition: the net force or field is the vector sum of the individual contributions.
• Electric field: $\vec{E} = \dfrac{\vec{F}}{q}$; for a point charge $E = \dfrac{kq}{r^2}$ (unit N/C).
• Field lines start on positive and end on negative charges, never cross, and are denser where the field is stronger.

Beyond-NCERT JEE Formulae

Section 1 already nails Coulomb's law and the point-charge field, so this sheet gathers the results one level up — the dipole, continuous charge distributions, field energy and the capacitor toolkit. These are the formulae that quietly decide most JEE Main electrostatics questions. Constants used below: $k=\dfrac{1}{4\pi\varepsilon_0}=9\times10^{9}$ SI units and $\varepsilon_0=8.85\times10^{-12}$ SI units.

1. Electric dipole (short dipole, r much greater than its size)

• Axial (end-on) field: $E_{ax}=\dfrac{2kp}{r^{3}}$, directed along $\vec{p}$.
• Equatorial (broadside) field: $E_{eq}=\dfrac{kp}{r^{3}}$, directed opposite to $\vec{p}$; so at equal distance the axial field is exactly twice the equatorial field.
• At a general angle $\theta$ from the axis: $E=\dfrac{kp}{r^{3}}\sqrt{1+3\cos^{2}\theta}$, while the potential is $V=\dfrac{kp\cos\theta}{r^{2}}$ — note $V$ carries NO factor of 2 and falls as $\dfrac{1}{r^{2}}$, whereas $E$ falls as $\dfrac{1}{r^{3}}$.
• In a uniform field: torque $\tau=pE\sin\theta$ (as a vector $\vec{\tau}=\vec{p}\times\vec{E}$), with zero net force; potential energy $U=-pE\cos\theta$; and the work to turn it from $\theta_1$ to $\theta_2$ is $W=pE(\cos\theta_1-\cos\theta_2)$.
• In a NON-uniform field: the two ends feel unequal forces, so besides the torque there is a net translational force $F=p\dfrac{dE}{dx}$ (aligned dipole) dragging it toward the stronger-field region.

When to use: any "short dipole", "two equal and opposite charges $2a$ apart" or "polar molecule in a field" question. [JEE Tip] The axial factor of 2 lives in the FIELD only — never smuggle it into the dipole potential. When a problem compares an axial point with an equatorial point, weigh $\dfrac{2}{r_{ax}^{3}}$ against $\dfrac{1}{r_{eq}^{3}}$, not the bare distances.

2. Fields of continuous charge distributions

• Infinite line charge: $E=\dfrac{2k\lambda}{r}=\dfrac{\lambda}{2\pi\varepsilon_0 r}$, radial and falling as $\dfrac{1}{r}$.
• Infinite non-conducting sheet: $E=\dfrac{\sigma}{2\varepsilon_0}$, uniform and independent of distance.
• Just outside a charged conductor: $E=\dfrac{\sigma}{\varepsilon_0}$ — double the sheet result, because the charge sits on one face only.
• Uniformly charged ring on its axis: $E=\dfrac{kQx}{(x^{2}+R^{2})^{3/2}}$, which is zero at the centre and largest at $x=\dfrac{R}{\sqrt{2}}$.
• Charged conducting sphere or shell (charge Q, radius R): for $r>R$ it acts as a point charge, $E=\dfrac{kQ}{r^{2}}$ with $V=\dfrac{kQ}{r}$; for $r<R$, $E=0$ but $V=\dfrac{kQ}{R}$ stays constant (a shell is NOT at zero potential inside).

When to use: rods or wires long compared with the distance (line), broad flat plates (sheet), the field just outside any conductor, and ring-on-axis set-ups. [JEE Tip] Fix the distance law to the geometry first: a point or sphere goes as $\dfrac{1}{r^{2}}$, an infinite line as $\dfrac{1}{r}$, and an infinite sheet stays constant. Forcing a line or sheet into $\dfrac{kq}{r^{2}}$ is the commonest silent blunder.

3. Field energy and the pull on a conductor

• Energy density stored in any electric field: $u=\dfrac{1}{2}\varepsilon_0 E^{2}$, measured in J/m$^3$; the total field energy is $U=\displaystyle\int u\,dV$.
• Electrostatic pressure — the outward force per unit area on a charged conductor's surface: $P=\dfrac{\sigma^{2}}{2\varepsilon_0}=\dfrac{1}{2}\varepsilon_0 E^{2}$.
• Force between capacitor plates: each plate sits in the OTHER plate's field $\dfrac{\sigma}{2\varepsilon_0}$, so $F=\dfrac{Q^{2}}{2\varepsilon_0 A}=\dfrac{\sigma^{2}A}{2\varepsilon_0}$, always attractive.

When to use: "energy stored in the field", "force needed to hold a plate", or "outward pull on a charged shell / soap film" questions. [JEE Tip] That lone factor of $\dfrac{1}{2}$ is the whole game: a plate feels only the other plate's field $\dfrac{\sigma}{2\varepsilon_0}$, not the full gap field $\dfrac{\sigma}{\varepsilon_0}$. Dropping it doubles every force-on-a-plate answer.

4. Capacitor toolkit

• Series: $\dfrac{1}{C_s}=\displaystyle\sum\dfrac{1}{C_i}$ (every capacitor holds the SAME charge). Parallel: $C_p=\displaystyle\sum C_i$ (every capacitor holds the SAME voltage).
• Stored energy: $U=\dfrac{1}{2}CV^{2}=\dfrac{1}{2}QV=\dfrac{Q^{2}}{2C}$. Keep the battery connected and $V$ is fixed (use $\tfrac12 CV^{2}$); once it is disconnected the charge $Q$ is fixed (use $\dfrac{Q^{2}}{2C}$).
• Dielectric slab of thickness t in a gap d (with t less than d): $C=\dfrac{\varepsilon_0 A}{d-t+t/K}$. A conducting slab is the $K\to\infty$ limit, giving $C=\dfrac{\varepsilon_0 A}{d-t}$, and where it sits in the gap makes no difference.
• Combining two charged capacitors (like plates joined): common potential $V=\dfrac{C_1V_1+C_2V_2}{C_1+C_2}$, with heat lost $\Delta U=\dfrac{1}{2}\dfrac{C_1C_2}{C_1+C_2}(V_1-V_2)^{2}$.

When to use: any network reduction, energy-stored question, dielectric insertion, or "two capacitors reconnected" charge-sharing problem. [JEE Tip] The sharing loss $\Delta U$ is ALWAYS positive (energy goes into the wire and the spark) and does not depend on the wire's resistance; it vanishes only when $V_1=V_2$. That sign is a quick sanity check — you can never gain energy on reconnection.

Solved Examples — Beyond-NCERT Formulae

Example 1 — Dipole: axial vs equatorial field

Problem: A short dipole has moment $p=6\times10^{-9}$ C m. Find its field at an axial point and at an equatorial point, each $0.20$ m away, and their ratio. Take $k=9\times10^{9}$ SI units.

Formula: $E_{ax}=\dfrac{2kp}{r^{3}}$ and $E_{eq}=\dfrac{kp}{r^{3}}$.

Working: With $r^{3}=(0.20)^{3}=8\times10^{-3}$ and $kp=(9\times10^{9})(6\times10^{-9})=54$,

$$E_{ax}=\dfrac{2\times54}{8\times10^{-3}}=1.35\times10^{4}\ \text{N/C}$$

$$E_{eq}=\dfrac{54}{8\times10^{-3}}=6.75\times10^{3}\ \text{N/C}$$

Answer: $E_{ax}=1.35\times10^{4}$ N/C and $E_{eq}=6.75\times10^{3}$ N/C, so the ratio is $\dfrac{E_{ax}}{E_{eq}}=2$ exactly — the axial field is always twice the equatorial field at the same distance.

Example 2 — Dipole: torque, energy and work to rotate

Problem: A dipole of moment $p=3\times10^{-8}$ C m sits in a uniform field $E=5\times10^{4}$ N/C. Find the torque when it makes $30^\circ$ with the field, and the work an external agent does to turn it from alignment ($0^\circ$) to $90^\circ$.

Formula: $\tau=pE\sin\theta$ and $U=-pE\cos\theta$, so $W=U(\theta_2)-U(\theta_1)=pE(\cos\theta_1-\cos\theta_2)$.

Working: First $pE=(3\times10^{-8})(5\times10^{4})=1.5\times10^{-3}$. Then

$$\tau=pE\sin 30^\circ=(1.5\times10^{-3})(0.5)=7.5\times10^{-4}\ \text{N m}$$

$$W=pE(\cos 0^\circ-\cos 90^\circ)=(1.5\times10^{-3})(1-0)=1.5\times10^{-3}\ \text{J}$$

Answer: torque $=7.5\times10^{-4}$ N m and work $=1.5\times10^{-3}$ J $=1.5$ mJ. The work equals $pE$ here because the dipole is turned all the way to perpendicular, where $\cos\theta=0$.

Example 3 — Field of an infinite line charge

Problem: A very long straight wire carries a uniform linear charge density $\lambda=4\times10^{-6}$ C/m. Find the field at $0.20$ m from it, and again at $0.40$ m. Take $k=9\times10^{9}$ SI units.

Formula: for an infinite line, $E=\dfrac{2k\lambda}{r}$ — an inverse-first-power law, not an inverse square.

Working: With $2k\lambda=2(9\times10^{9})(4\times10^{-6})=7.2\times10^{4}$,

$$E(0.20)=\dfrac{7.2\times10^{4}}{0.20}=3.6\times10^{5}\ \text{N/C}$$

$$E(0.40)=\dfrac{7.2\times10^{4}}{0.40}=1.8\times10^{5}\ \text{N/C}$$

Answer: $3.6\times10^{5}$ N/C and $1.8\times10^{5}$ N/C. Doubling the distance HALVES the field rather than quartering it — the signature of a line charge as opposed to a point charge.

Example 4 — Field of an infinite sheet and the force on a charge

Problem: A large non-conducting sheet carries a uniform surface density $\sigma=8.85\times10^{-6}$ C/m$^2$. Find the field it produces and the force it exerts on a $2\ \mu$C charge placed near it. Take $\varepsilon_0=8.85\times10^{-12}$ SI units.

Formula: for an infinite sheet, $E=\dfrac{\sigma}{2\varepsilon_0}$, the same at every distance; then $F=qE$.

Working:

$$E=\dfrac{8.85\times10^{-6}}{2(8.85\times10^{-12})}=5\times10^{5}\ \text{N/C}$$

$$F=qE=(2\times10^{-6})(5\times10^{5})=1.0\ \text{N}$$

Answer: $E=5\times10^{5}$ N/C and $F=1.0$ N. Because the field is uniform, the force is the same however far the charge sits from the sheet. Had this been a conducting surface, $E=\dfrac{\sigma}{\varepsilon_0}$ would double both answers.

Example 5 — Energy density and electrostatic pressure

Problem: (a) At a point the electric field is $E=2\times10^{6}$ N/C; find the energy density there and the energy stored in $1$ cm$^3$ around it. (b) A charged conductor has surface density $\sigma=8.85\times10^{-6}$ C/m$^2$; find the outward electrostatic pressure on its surface. Take $\varepsilon_0=8.85\times10^{-12}$ SI units.

Formula: energy density $u=\tfrac12\varepsilon_0 E^{2}$; electrostatic pressure $P=\dfrac{\sigma^{2}}{2\varepsilon_0}=\tfrac12\varepsilon_0 E_{s}^{2}$, where $E_s$ is the field just outside.

Working (a):

$$u=\tfrac12(8.85\times10^{-12})(2\times10^{6})^{2}=17.7\ \text{J/m}^3$$

In a volume of $1$ cm$^3$, i.e. $1\times10^{-6}$ m$^3$, the stored energy is $U=uV=17.7\times10^{-6}$ J $=17.7\ \mu$J.

Working (b): the field just outside is $E_s=\dfrac{\sigma}{\varepsilon_0}=1\times10^{6}$ N/C, so

$$P=\dfrac{(8.85\times10^{-6})^{2}}{2(8.85\times10^{-12})}=4.425\ \text{N/m}^2$$

Answer: (a) $u=17.7$ J/m$^3$ and $U=17.7\ \mu$J; (b) $P=4.425$ N/m$^2$. The one expression $\tfrac12\varepsilon_0 E^{2}$ serves both as an energy per unit volume and as a force per unit area.

Example 6 — Capacitor combination and stored energy

Problem: A $4\ \mu$F and a $12\ \mu$F capacitor are joined in series, and that pair is placed in parallel with a $5\ \mu$F capacitor across a $50$ V supply. Find the equivalent capacitance, the total charge drawn and the total energy stored.

Formula: series $C_s=\dfrac{C_1C_2}{C_1+C_2}$; parallel adds; then $Q=C_{eq}V$ and $U=\tfrac12 C_{eq}V^{2}$.

Working: the series pair is

$$C_s=\dfrac{4\times12}{4+12}=\dfrac{48}{16}=3\ \mu\text{F}$$

In parallel with $5\ \mu$F, $C_{eq}=3+5=8\ \mu$F. Then

$$Q=C_{eq}V=(8\times10^{-6})(50)=4\times10^{-4}\ \text{C}=400\ \mu\text{C}$$

$$U=\tfrac12 C_{eq}V^{2}=\tfrac12(8\times10^{-6})(50)^{2}=1.0\times10^{-2}\ \text{J}=10\ \text{mJ}$$

Answer: $C_{eq}=8\ \mu$F, $Q=400\ \mu$C and $U=10$ mJ. Reduce the network in stages — the series pair first, then the parallel step — and apply $Q=C_{eq}V$ only to the final single capacitance.

Example 7 — Charge sharing and the energy lost

Problem: A $2\ \mu$F capacitor charged to $100$ V is connected, positive plate to positive plate, across a $3\ \mu$F capacitor charged to $50$ V. Find the common potential and the heat dissipated.

Formula: common potential $V=\dfrac{C_1V_1+C_2V_2}{C_1+C_2}$; heat lost $\Delta U=\dfrac{1}{2}\dfrac{C_1C_2}{C_1+C_2}(V_1-V_2)^{2}$.

Working:

$$V=\dfrac{(2)(100)+(3)(50)}{2+3}=\dfrac{350}{5}=70\ \text{V}$$

$$\Delta U=\tfrac12\cdot\dfrac{(2\times10^{-6})(3\times10^{-6})}{5\times10^{-6}}(100-50)^{2}=\tfrac12(1.2\times10^{-6})(2500)=1.5\times10^{-3}\ \text{J}$$

Answer: common potential $70$ V and heat lost $1.5$ mJ. As a check, $U_i=\tfrac12(2\times10^{-6})(100)^{2}+\tfrac12(3\times10^{-6})(50)^{2}=13.75$ mJ falls to $U_f=\tfrac12(5\times10^{-6})(70)^{2}=12.25$ mJ, a drop of $1.5$ mJ — the stored energy always falls on reconnection, never rises.