Solved Examples & Key Derivations

Essential Derivations from NCERT

In this section, we consolidate the most important derivations and theorems that are frequently asked in Board exams and form the conceptual foundation for JEE/NEET.

A. Electric Field at an Axial Point of a Dipole

Consider a dipole with charges $-q$ and $+q$ separated by $2a$. Let P be a point on the axial line at distance $r$ from the center O.

1. Field due to $+q$: $E_+ = \frac{1}{4\pi\epsilon_0} \frac{q}{(r-a)^2}$ (Away from dipole).
2. Field due to $-q$: $E_- = \frac{1}{4\pi\epsilon_0} \frac{q}{(r+a)^2}$ (Towards dipole).
3. Net Field $E = E_+ - E_- = \frac{q}{4\pi\epsilon_0} [\frac{1}{(r-a)^2} - \frac{1}{(r+a)^2}]$.
4. Simplifying: $E = \frac{q}{4\pi\epsilon_0} [\frac{4ra}{(r^2-a^2)^2}]$.
5. Since $p = q \times 2a$, we get: $E = \frac{1}{4\pi\epsilon_0} \frac{2pr}{(r^2-a^2)^2}$.
6. For a short dipole ($r \gg a$): $E_{axial} = \frac{1}{4\pi\epsilon_0} \frac{2p}{r^3}$.

B. Electric Field at an Equatorial Point of a Dipole

Let P be at distance $r$ on the perpendicular bisector.

1. Magnitudes of fields from both charges are equal: $E_+ = E_- = \frac{1}{4\pi\epsilon_0} \frac{q}{r^2+a^2}$.
2. Vertical components $E \sin \theta$ cancel out.
3. Horizontal components $E \cos \theta$ add up: $E_{net} = 2E \cos \theta$.
4. From geometry, $\cos \theta = \frac{a}{\sqrt{r^2+a^2}}$.
5. $E_{net} = 2 [\frac{1}{4\pi\epsilon_0} \frac{q}{r^2+a^2}] [\frac{a}{\sqrt{r^2+a^2}}] = \frac{1}{4\pi\epsilon_0} \frac{p}{(r^2+a^2)^{3/2}}$.
6. For a short dipole ($r \gg a$): $E_{equatorial} = \frac{1}{4\pi\epsilon_0} \frac{p}{r^3}$.

C. Gauss's Law Application: Infinite Line Charge

1. Consider a cylinder of radius $r$ and length $L$ as the Gaussian surface.
2. Total flux $\Phi = E \times (2\pi r L)$.
3. Enclosed charge $q = \lambda L$.
4. By Gauss Law: $E(2\pi r L) = \frac{\lambda L}{\epsilon_0} \implies$ $E = \frac{\lambda}{2\pi\epsilon_0 r}$.

Example 1: A polythene piece rubbed with wool is found to have a negative charge of $3 \times 10^{-7}$ C. (a) Estimate the number of electrons transferred (from which to which?) (b) Is there a transfer of mass from wool to polythene?

Solution: (a) $Q = -3 \times 10^{-7}$ C. Using $Q = ne$, $n = Q/e = (3 \times 10^{-7}) / (1.6 \times 10^{-19}) = 1.875 \times 10^{12}$ electrons. Electrons are transferred from wool to polythene. (b) Yes, mass is transferred. $\Delta m = n \times m_e = 1.875 \times 10^{12} \times 9.1 \times 10^{-31} = 1.7 \times 10^{-18}$ kg.

Example 2: Two insulated charged copper spheres A and B have their centers separated by a distance of 50 cm. What is the mutual force of electrostatic repulsion if the charge on each is $6.5 \times 10^{-7}$ C? The radii of A and B are negligible compared to the distance of separation.

Solution:

1. $q_1 = q_2 = 6.5 \times 10^{-7}$ C.
2. $r = 0.5$ m.
3. $F = k \frac{q_1 q_2}{r^2} = (9 \times 10^9) \frac{(6.5 \times 10^{-7})^2}{(0.5)^2}$.
4. $F = (9 \times 10^9) \frac{42.25 \times 10^{-14}}{0.25} = 1.52 \times 10^{-2}$ N.

Example 3: What is the force of repulsion if each sphere in Example 2 is charged double the above amount, and the distance between them is halved?

Solution:

1. New charge $q' = 2q$, new distance $r' = r/2$.
2. $F' = k \frac{(2q)(2q)}{(r/2)^2} = k \frac{4q^2}{r^2/4} = 16 F$.
3. $F' = 16 \times 1.52 \times 10^{-2} = 0.243$ N.

Example 4: Suppose the spheres A and B in Example 2 have identical sizes. A third sphere of the same size but uncharged is brought in contact with the first, then brought in contact with the second, and finally removed from both. What is the new force of repulsion between A and B?

Solution:

1. Initial charges: $q_A = q, q_B = q, q_C = 0$.
2. C touches A: $q_A = q_C = q/2$.
3. C touches B: $q_B = (q/2 + q)/2 = 3q/4$.
4. Final charges: $q_A = q/2, q_B = 3q/4$.
5. New Force $F' = k \frac{(q/2)(3q/4)}{r^2} = \frac{3}{8} \frac{kq^2}{r^2} = \frac{3}{8} F$.
6. $F' = \frac{3}{8} \times 1.52 \times 10^{-2} = 5.7 \times 10^{-3}$ N.

Example 5: An oil drop of 12 excess electrons is held stationary under a constant electric field of $2.55 \times 10^4$ N/C in Millikan’s oil drop experiment. The density of the oil is 1.26 g/cm$^3$. Estimate the radius of the drop.

Solution:

1. Equilibrium: $qE = mg \implies (ne)E = (\frac{4}{3}\pi R^3 \rho)g$.
2. $R^3 = \frac{3neE}{4\pi \rho g}$.
3. $R^3 = \frac{3 \times 12 \times 1.6 \times 10^{-19} \times 2.55 \times 10^4}{4 \times 3.14 \times 1260 \times 9.8} \approx 9.46 \times 10^{-19}$.
4. $R \approx 9.81 \times 10^{-7}$ m.

Example 6: Which among the curves shown in Fig 1.35 cannot possibly represent electrostatic field lines? (a) Lines passing through a conductor at non-90 degree angles. (b) Lines forming closed loops. (c) Lines starting and ending on the same positive charge.

Solution: (a) False: Field lines must be perpendicular to the surface of a conductor. (b) False: Electrostatic field lines never form closed loops. (c) False: Field lines start at positive and end at negative charges.

Example 7: In a certain region of space, electric field is along the z-direction throughout. The magnitude of electric field is, however, not constant but increases uniformly along the positive z-direction, at the rate of $10^5$ N/C per meter. What are the force and torque experienced by a system having a total dipole moment equal to $10^{-7}$ C m in the negative z-direction?

Solution:

1. Force $F = p \frac{dE}{dz} = (10^{-7}) \times (10^5) = 10^{-2}$ N.
2. Since $p$ is in $-z$ and $E$ is in $+z$, the angle $\theta = 180^\circ$.
3. Force is in the direction of decreasing field (negative z-axis).
4. Torque $\tau = pE \sin 180^\circ = 0$.

Example 8: Two large, thin metal plates are parallel and close to each other. On their inner faces, the plates have surface charge densities of opposite signs and magnitude $17.0 \times 10^{-22} \text{ C/m}^2$. What is E: (a) in the outer region of the first plate, (b) in the outer region of the second plate, and (c) between the plates?

Solution: (a) Outside: $E = \frac{\sigma}{2\epsilon_0} - \frac{\sigma}{2\epsilon_0} = 0$. (b) Outside: $E = 0$. (c) Between: $E = \frac{\sigma}{2\epsilon_0} + \frac{\sigma}{2\epsilon_0} = \frac{\sigma}{\epsilon_0}$. $E = (17.0 \times 10^{-22}) / (8.854 \times 10^{-12}) = 1.92 \times 10^{-10}$ N/C.

Example 9: An infinite line charge produces a field of $9 \times 10^4$ N/C at a distance of 2 cm. Calculate the linear charge density.

Solution:

1. $E = \frac{\lambda}{2\pi\epsilon_0 r} \implies \lambda = E (2\pi\epsilon_0 r)$.
2. $\lambda = (9 \times 10^4) \times \frac{0.02}{2 \times 9 \times 10^9} = 10^{-7}$ C/m.
3. $\lambda = 0.1 \mu$C/m.

Example 10: A particle of mass $m$ and charge $-q$ enters the region between the two charged plates initially moving along x-axis with speed $v_x$. The length of plate is $L$ and a uniform electric field $E$ is maintained between the plates. Show that the vertical deflection of the particle at the far edge of the plate is $qEL^2 / (2m v_x^2)$.

Solution:

1. Time taken to cross plates $t = L / v_x$.
2. Acceleration in y-direction $a_y = F/m = qE/m$.
3. Deflection $y = \frac{1}{2} a_y t^2 = \frac{1}{2} (qE/m) (L/v_x)^2$.
4. $y = \frac{qEL^2}{2mv_x^2}$.

Example 11: Two identical pith balls, each of mass $m$ and charge $q$, are suspended from a common point by two strings of equal length $L$. At equilibrium, the separation is $x$. If $x \ll L$, find the expression for $x$.

Solution:

1. Forces: $T \cos \theta = mg$ and $T \sin \theta = F_e$.
2. $\tan \theta = F_e / mg$. For small $\theta$, $\tan \theta \approx \sin \theta = (x/2)/L$.
3. $\frac{x}{2L} = \frac{kq^2}{x^2 mg}$.
4. $x^3 = \frac{2Lkq^2}{mg} \implies$ $x = [\frac{q^2 L}{2\pi\epsilon_0 mg}]^{1/3}$.

Example 12: A charge $Q$ is distributed uniformly over a ring of radius $R$. Find the electric field at a point P on the axis of the ring at distance $x$ from the center.

Solution:

1. Take small element $dq$. $dE = k dq / r^2$, where $r = \sqrt{R^2+x^2}$.
2. Vertical components $dE \sin \theta$ cancel by symmetry.
3. Horizontal components $dE \cos \theta$ add up.
4. $E = \int dE \cos \theta = \int [k dq / (R^2+x^2)] [x / \sqrt{R^2+x^2}]$.
5. $E = \frac{kx}{(R^2+x^2)^{3/2}} \int dq = \frac{kQx}{(R^2+x^2)^{3/2}}$.

Example 13: In the previous example, at what value of $x$ is the electric field maximum?

Solution:

1. Differentiate $E$ with respect to $x$ and set to zero: $dE/dx = 0$.
2. Using the quotient rule on $x(R^2+x^2)^{-3/2}$, we get $(R^2+x^2) - 3x^2 = 0$.
3. $R^2 = 2x^2 \implies$ $x = R/\sqrt{2}$.

Example 14: A point charge $+q$ is placed at the center of an uncharged hollow conducting sphere of inner radius $a$ and outer radius $b$. Find the surface charge densities on the inner and outer surfaces.

Solution:

1. By induction, charge $-q$ appears on the inner surface ($r=a$).
2. Since the sphere is uncharged, $+q$ must appear on the outer surface ($r=b$).
3. $\sigma_{inner} = -q / (4\pi a^2)$.
4. $\sigma_{outer} = +q / (4\pi b^2)$.

Example 15: Five identical charges $Q$ are placed at the vertices of a regular hexagon of side $a$. Find the electric field at the center of the hexagon.

Solution:

1. If there were 6 charges, the field at the center would be zero by symmetry.
2. $\vec{E}_{6\text{ charges}} = \vec{E}_{5\text{ charges}} + \vec{E}_{\text{missing charge}} = 0$.
3. Therefore, $\vec{E}_{5\text{ charges}} = -\vec{E}_{\text{missing charge}}$.
4. Magnitude $E = kQ/a^2$, directed towards the empty vertex.

Example 16: A long thin wire is bent into a semi-circle of radius $R$. If the linear charge density is $\lambda$, find the electric field at the center.

Solution:

1. $dE = k dq / R^2 = k (\lambda R d\theta) / R^2 = k \lambda d\theta / R$.
2. Horizontal components cancel. Vertical components $dE \sin \theta$ add up.
3. $E = \int_0^\pi (k \lambda / R) \sin \theta d\theta = (k \lambda / R) [-\cos \theta]_0^\pi$.
4. $E = (k \lambda / R) [1 - (-1)] = 2k\lambda / R$.

Example 17: Two dipoles $\vec{p}_1$ and $\vec{p}_2$ are placed along the same line at a distance $r$. Find the force between them.

Solution:

1. Force $F = p_2 \frac{dE_1}{dr}$.
2. Axial field $E_1 = \frac{2kp_1}{r^3}$.
3. $\frac{dE_1}{dr} = -\frac{6kp_1}{r^4}$.
4. $F = -\frac{6kp_1p_2}{r^4}$ (Attractive if dipoles are aligned).

Example 18: A charge $q$ is placed at the center of the mouth of a cylindrical coffee mug. What is the flux through the surface of the mug?

Solution:

1. Imagine a second identical mug inverted on top. The charge is now at the center of a closed system.
2. Total flux = $q/\epsilon_0$.
3. By symmetry, flux through one mug is $q/2\epsilon_0$.

Example 19: A solid non-conducting sphere of radius $R$ has a volume charge density $\rho = \rho_0 r$. Find the electric field at $r < R$.

Solution:

1. $q_{enclosed} = \int_0^r \rho (4\pi r^2) dr = \int_0^r (\rho_0 r) 4\pi r^2 dr = \pi \rho_0 r^4$.
2. Gauss Law: $E(4\pi r^2) = (\pi \rho_0 r^4) / \epsilon_0$.
3. $E = \frac{\rho_0 r^2}{4\epsilon_0}$.

Example 20: A system has two charges $q_A = 2.5 \times 10^{-7}$ C and $q_B = -2.5 \times 10^{-7}$ C located at points A: (0, 0, -15 cm) and B: (0, 0, +15 cm). What are the total charge and electric dipole moment of the system?

Solution:

1. Total charge $Q = q_A + q_B = 0$.
2. Dipole moment $p = q \times 2a = (2.5 \times 10^{-7}) \times (0.30) = 7.5 \times 10^{-8}$ C m.
3. Direction is from $-q$ to $+q$, which is from point B to A (along negative z-axis).

Example 21: A particle of mass $m$ and charge $q$ is released from rest in a uniform electric field $E$. Find its kinetic energy after time $t$.

Solution:

1. Velocity $v = at = (qE/m)t$.
2. $KE = \frac{1}{2} m v^2 = \frac{1}{2} m (q^2 E^2 t^2 / m^2) = \frac{q^2 E^2 t^2}{2m}$.

Example 22: Two point charges $+4q$ and $+q$ are fixed at distance $L$. A third charge $Q$ is placed such that the entire system is in equilibrium. Find $Q$ and its position.

Solution:

1. Position: From Example 4 in Section 3, $x = L/3$ from charge $q$.
2. Magnitude: Net force on charge $q$ must be zero: $k(4q)(q)/L^2 + k(Q)(q)/(L/3)^2 = 0$.
3. $4q/L^2 + 9Q/L^2 = 0 \implies$ $Q = -4q/9$.

Example 23: A simple pendulum has a bob of mass $m$ and charge $q$. If a horizontal electric field $E$ is switched on, find the new equilibrium position.

Solution:

1. At equilibrium, $T \cos \theta = mg$ and $T \sin \theta = qE$.
2. $\tan \theta = qE / mg$.
3. The string makes an angle $\theta = \tan^{-1}(qE/mg)$ with the vertical.

Example 24: If the pendulum in Example 23 is set into small oscillations, find its time period.

Solution:

1. Effective gravity $g_{eff} = \sqrt{g^2 + (qE/m)^2}$.
2. $T = 2\pi \sqrt{L / g_{eff}} = 2\pi \sqrt{\frac{L}{\sqrt{g^2 + (qE/m)^2}}}$.

Example 25: Find the flux of a uniform electric field $\vec{E} = E_0 \hat{i}$ through a square of side $L$ in the y-z plane.

Solution:

1. Area vector $\vec{A} = L^2 \hat{i}$.
2. Flux $\Phi = \vec{E} \cdot \vec{A} = (E_0 \hat{i}) \cdot (L^2 \hat{i}) = E_0 L^2$.

Example 26: A spherical Gaussian surface encloses a charge of $8.85 \times 10^{-8}$ C. What is the electric flux through the surface? If the radius is doubled, how does the flux change?

Solution:

1. $\Phi = q/\epsilon_0 = (8.85 \times 10^{-8}) / (8.85 \times 10^{-12}) = 10^4 \text{ Nm}^2/\text{C}$.
2. If radius is doubled, flux remains the same because it only depends on enclosed charge.

Example 27: An electric dipole is placed at an angle of $60^\circ$ with an electric field of $2 \times 10^5$ N/C. It experiences a torque of $8\sqrt{3}$ N m. If the dipole length is 2 cm, find the charge on the dipole.

Solution:

1. $\tau = pE \sin \theta = (q \times 2a) E \sin \theta$.
2. $8\sqrt{3} = (q \times 0.02) \times (2 \times 10^5) \times \sin 60^\circ$.
3. $8\sqrt{3} = q \times 0.02 \times 2 \times 10^5 \times (\sqrt{3}/2)$.
4. $8 = q \times 2000 \implies q = 4 \times 10^{-3}$ C = 4 mC.

Example 28: A metal sphere is grounded. A positive charge $+Q$ is brought near it. What is the charge on the sphere?

Solution:

1. The $+Q$ charge attracts electrons to the near side and repels positive charge to the far side.
2. Grounding allows electrons from the Earth to neutralize the far side.
3. The sphere is left with a negative charge.

Example 29: Three charges $+q$, $+q$ and $-2q$ are placed at the vertices of an equilateral triangle. What is the net dipole moment of the system?

Solution:

1. The system can be seen as two dipoles, each consisting of $+q$ and $-q$ (splitting the $-2q$).
2. Each dipole has moment $p = qL$.
3. The angle between the two dipole vectors is $60^\circ$.
4. $p_{net} = \sqrt{p^2 + p^2 + 2p^2 \cos 60^\circ} = \sqrt{3}p = \sqrt{3} qL$.

Example 30: A charge $q$ is placed at distance $L/2$ above the center of a square of side $L$. Find the flux through the square.

Solution:

1. This square can be considered one face of a cube of side $L$ with the charge at the center.
2. Total flux through cube = $q/\epsilon_0$.
3. Flux through one face = $q/6\epsilon_0$.