Solved Examples & Key Derivations

Essential Derivations

To build a strong foundation for both Boards and Competitive exams, we must master these core derivations. They explain the 'why' behind the formulas we use.

A. Relation between Electric Field and Potential

Consider two closely spaced equipotential surfaces A and B with potentials $V$ and $V + dV$. Let $dl$ be the perpendicular distance between them.

1. Work done in moving a unit positive charge from B to A is: $$dW = V - (V + dV) = -dV$$
2. Also, $$dW = F \cdot dl = E \cdot dl$$ since $F = qE$ and $q = 1$.
3. Equating both, $$E \cdot dl = -dV$$ $$E = -\frac{dV}{dl}$$

B. Capacitance of a Parallel Plate Capacitor

Consider two plates of area $A$ with charges $+Q$ and $-Q$ separated by distance $d$.

1. Electric field between plates: $$E = \frac{\sigma}{\epsilon_0} = \frac{Q}{A\epsilon_0}$$
2. Potential difference: $$V = E \cdot d = \frac{Qd}{A\epsilon_0}$$
3. Capacitance: $$C = \frac{Q}{V} = \frac{Q}{Qd/(A\epsilon_0)} = \frac{\epsilon_0 A}{d}$$

C. Energy Stored in a Capacitor

1. Work done in adding a small charge $dq$ to a capacitor at potential $v$: $$dW = v \cdot dq = \frac{q}{C} \, dq$$
2. Total work done in charging it from 0 to $Q$: $$U = \int_0^Q \frac{q}{C} \, dq = \frac{1}{C} \left[ \frac{q^2}{2} \right]_0^Q$$
3. Result: $$U = \frac{Q^2}{2C} = \frac{1}{2}CV^2 = \frac{1}{2}QV$$

Example 1: Two charges $5 \times 10^{-8}$ C and $-3 \times 10^{-8}$ C are located 16 cm apart. At what point(s) on the line joining the two charges is the electric potential zero?

Solution:

1. Case 1 (Between the charges): Let the point P be at distance $x$ cm from the positive charge.
2. Equate potentials: $$k \frac{5 \times 10^{-8}}{x} + k \frac{-3 \times 10^{-8}}{16 - x} = 0$$
3. Simplify: $$\frac{5}{x} = \frac{3}{16-x}$$ $$80 - 5x = 3x$$ $$8x = 80$$ $$x = 10 \text{ cm}$$
4. Case 2 (Outside, on the side of the negative charge): Let P be at distance $x$ cm from the positive charge, where $x > 16$.
5. Equate potentials: $$\frac{5}{x} = \frac{3}{x-16}$$ $$5x - 80 = 3x$$ $$2x = 80$$ $$x = 40 \text{ cm}$$
6. Final Answer: The potential is zero at points 10 cm and 40 cm from the positive charge.

Example 2: A regular hexagon of side 10 cm has a charge $5 \mu C$ at each of its vertices. Calculate the potential at the center of the hexagon.

Solution:

1. In a regular hexagon, the distance from each vertex to the center is equal to the side length: $$r = 10 \text{ cm} = 0.1 \text{ m}$$
2. Potential is a scalar, so total potential is: $$V = 6 \times \left( 9 \times 10^9 \times \frac{5 \times 10^{-6}}{0.1} \right)$$
3. Simplify single-charge contribution: $$V_1 = 4.5 \times 10^5 \text{ V}$$
4. Therefore, $$V = 6 \times 4.5 \times 10^5 = 2.7 \times 10^6 \text{ V}$$
5. Final Answer: $$V = 2.7 \times 10^6 \text{ V}$$

Example 3: Two charges $+2 \mu C$ and $-2 \mu C$ are placed at points A and B 6 cm apart. (a) Identify an equipotential surface of the system. (b) What is the direction of the electric field at every point on this surface?

Solution:

1. This is an electric dipole.
2. The plane passing through the midpoint of AB and perpendicular to AB is the equatorial plane.
3. On this plane, potentials due to the two charges cancel, so it is an equipotential surface with: $$V = 0$$
4. The electric field is always perpendicular to an equipotential surface.
5. Here, the field is directed from the positive charge A toward the negative charge B, i.e. parallel to AB.
6. Final Answer:

• Equipotential surface: Equatorial plane of the dipole
• Direction of electric field: Perpendicular to the plane, from A to B

Example 4: A spherical conductor of radius 12 cm has a charge of $1.6 \times 10^{-7}$ C distributed uniformly on its surface. What is the electric field (a) inside the sphere (b) just outside the sphere (c) at a point 18 cm from the center?

Solution:

1. Inside a conductor: $$E = 0$$
2. Just outside the sphere $(r = 0.12 \text{ m})$: $$E = k \frac{Q}{R^2} = 9 \times 10^9 \times \frac{1.6 \times 10^{-7}}{(0.12)^2} = 10^5 \text{ N/C}$$
3. At $r = 0.18 \text{ m}$: $$E = k \frac{Q}{r^2} = 9 \times 10^9 \times \frac{1.6 \times 10^{-7}}{(0.18)^2} = 4.44 \times 10^4 \text{ N/C}$$
4. Final Answer:

• (a) $0$

• (b) $1.0 \times 10^5 \text{ N/C}$

• (c) $4.44 \times 10^4 \text{ N/C}$

Example 5: A parallel plate capacitor with air between the plates has a capacitance of 8 pF ($1 \text{ pF} = 10^{-12}$ F). What will be the capacitance if the distance between the plates is reduced by half, and the space between them is filled with a substance of dielectric constant 6?

Solution:

1. Initial capacitance: $$C_0 = \frac{\epsilon_0 A}{d} = 8 \text{ pF}$$
2. New separation: $$d' = \frac{d}{2}$$
3. With dielectric constant $K = 6$: $$C = \frac{K\epsilon_0 A}{d/2} = 2K \frac{\epsilon_0 A}{d} = 2KC_0$$
4. Therefore, $$C = 2 \times 6 \times 8 = 96 \text{ pF}$$
5. Final Answer: $$C = 96 \text{ pF}$$

Example 6: Three capacitors each of capacitance 9 pF are connected in series. (a) What is the total capacitance of the combination? (b) What is the potential difference across each capacitor if the combination is connected to a 120 V supply?

Solution:

1. For series combination: $$\frac{1}{C_s} = \frac{1}{9} + \frac{1}{9} + \frac{1}{9} = \frac{3}{9} = \frac{1}{3}$$ Hence, $$C_s = 3 \text{ pF}$$
2. Since the capacitors are identical, the total voltage divides equally: $$V_1 = V_2 = V_3 = \frac{120}{3} = 40 \text{ V}$$
3. Final Answer:

• (a) $C_s = 3 \text{ pF}$

• (b) $40 \text{ V across each capacitor}$

Example 7: Three capacitors of capacitances 2 pF, 3 pF and 4 pF are connected in parallel. (a) What is the total capacitance? (b) Determine the charge on each capacitor if the combination is connected to a 100 V supply.

Solution:

1. For parallel combination: $$C_p = 2 + 3 + 4 = 9 \text{ pF}$$
2. In parallel, the voltage across each capacitor is the same: $$V = 100 \text{ V}$$
3. Charges on each capacitor: $$Q_1 = C_1V = 2 \times 100 = 200 \text{ pC}$$ $$Q_2 = C_2V = 3 \times 100 = 300 \text{ pC}$$ $$Q_3 = C_3V = 4 \times 100 = 400 \text{ pC}$$
4. Final Answer:

• (a) $C_p = 9 \text{ pF}$

• (b) $Q_1 = 200 \text{ pC}, \; Q_2 = 300 \text{ pC}, \; Q_3 = 400 \text{ pC}$

Example 8: A 12 pF capacitor is connected to a 50 V battery. How much electrostatic energy is stored in the capacitor?

Solution:

1. Given: $$C = 12 \times 10^{-12} \text{ F}, \quad V = 50 \text{ V}$$
2. Formula: $$U = \frac{1}{2}CV^2$$
3. Substitute: $$U = 0.5 \times (12 \times 10^{-12}) \times (50)^2$$
4. Calculate: $$U = 6 \times 10^{-12} \times 2500 = 1.5 \times 10^{-8} \text{ J}$$
5. Final Answer: $$U = 1.5 \times 10^{-8} \text{ J}$$

Example 9: A 600 pF capacitor is charged by a 200 V supply. It is then disconnected from the supply and connected to another uncharged 600 pF capacitor. How much electrostatic energy is lost in the process?

Solution:

1. Initial energy: $$U_i = \frac{1}{2} C V^2 = 0.5 \times 600 \times 10^{-12} \times (200)^2 = 1.2 \times 10^{-5} \text{ J}$$
2. Since the two capacitors are identical, final common potential is: $$V_f = \frac{200}{2} = 100 \text{ V}$$
3. Final energy of the system: $$U_f = \frac{1}{2}(C_1 + C_2)V_f^2 = 0.5 \times 1200 \times 10^{-12} \times (100)^2 = 0.6 \times 10^{-5} \text{ J}$$
4. Energy lost: $$\Delta U = U_i - U_f = 1.2 \times 10^{-5} - 0.6 \times 10^{-5} = 0.6 \times 10^{-5} \text{ J}$$
5. Final Answer: $$\Delta U = 6 \times 10^{-6} \text{ J}$$

Example 10: A parallel plate capacitor has plate area $A$ and separation $d$. A dielectric slab of thickness $t$ ($t < d$) and dielectric constant $K$ is inserted between the plates. Find the new capacitance.

Solution:

1. The arrangement behaves like two capacitors in series:

• Air part of thickness $(d-t)$
• Dielectric part of thickness $t$

1. Capacitances of the two parts: $$C_{air} = \frac{\epsilon_0 A}{d-t}, \quad C_{dielectric} = \frac{K\epsilon_0 A}{t}$$
2. Since they are in series: $$\frac{1}{C} = \frac{1}{C_{air}} + \frac{1}{C_{dielectric}} = \frac{d-t}{\epsilon_0 A} + \frac{t}{K\epsilon_0 A}$$
3. Simplifying: $$\frac{1}{C} = \frac{d - t + t/K}{\epsilon_0 A}$$ $$C = \frac{\epsilon_0 A}{d - t\left(1 - \frac{1}{K}\right)}$$
4. Final Answer: $$C = \frac{\epsilon_0 A}{d - t\left(1 - \frac{1}{K}\right)}$$

Example 11: The electric potential in a region is given by $V(x,y,z) = 6x - 8xy - 8y + 6yz$. Find the magnitude of the electric force acting on a 2 C charge at the origin $(0,0,0)$.

Solution:

1. Electric field components are obtained from: $$E_x = -\frac{\partial V}{\partial x}, \quad E_y = -\frac{\partial V}{\partial y}, \quad E_z = -\frac{\partial V}{\partial z}$$
2. Compute partial derivatives: $$\frac{\partial V}{\partial x} = 6 - 8y \Rightarrow E_x = -(6 - 8y)$$ $$\frac{\partial V}{\partial y} = -8x - 8 + 6z \Rightarrow E_y = -(-8x - 8 + 6z)$$ $$\frac{\partial V}{\partial z} = 6y \Rightarrow E_z = -6y$$
3. At the origin $(0,0,0)$: $$E_x = -6, \quad E_y = 8, \quad E_z = 0$$
4. Magnitude of electric field: $$E = \sqrt{(-6)^2 + 8^2 + 0^2} = 10 \text{ V/m}$$
5. Force on 2 C charge: $$F = qE = 2 \times 10 = 20 \text{ N}$$
6. Final Answer: $$F = 20 \text{ N}$$

Example 12: Two identical capacitors 1 and 2 are connected in series to a battery. Capacitor 2 is filled with a dielectric of constant $K$. Compare the charges and energy stored.

Solution:

1. In series, $$Q_1 = Q_2$$
2. Let original capacitance of each be $C$. Then after inserting dielectric: $$C_1 = C, \quad C_2 = KC$$
3. Energies stored: $$U_1 = \frac{Q^2}{2C}, \quad U_2 = \frac{Q^2}{2KC}$$
4. Therefore, $$U_1 = K U_2$$
5. Final Answer:

• Charges are equal: $$Q_1 = Q_2$$
• Energy relation: $$U_1 : U_2 = K : 1$$

Example 13: A capacitor $C$ is charged to potential $V$ and then disconnected. A second uncharged capacitor $2C$ is connected in parallel. Find the common potential.

Solution:

1. Initial charge on the first capacitor: $$Q = CV$$
2. Total final capacitance: $$C_{net} = C + 2C = 3C$$
3. Common final potential: $$V' = \frac{Q}{C_{net}} = \frac{CV}{3C} = \frac{V}{3}$$
4. Final Answer: $$V' = \frac{V}{3}$$

Example 14: A charge $q$ is placed at the center of the line joining two equal charges $Q$. Find the potential energy of the three-charge system if the outer charges are at distance $2a$ apart.

Solution:

1. Distances between pairs:

Between each $Q$ and $q$: $a$, and

Between the two outer charges $Q$ and $Q$: $2a$

2. Total potential energy is the sum over all pairs: $$U = k\frac{Qq}{a} + k\frac{Qq}{a} + k\frac{Q^2}{2a}$$
3. Simplify: $$U = \frac{kQ}{a}\left(2q + \frac{Q}{2}\right)$$
4. Final Answer: $$U = \frac{kQ}{a}\left(2q + \frac{Q}{2}\right)$$

Example 15: Find the equivalent capacitance between A and B if an infinite ladder contains 1 $\mu F$ and 2 $\mu F$ capacitors alternating.

Solution:

1. Let the total equivalent capacitance be: $X$

2. Because the network is infinite, adding one more repeating section does not change $X$.

3. The recurrence relation is: $$X = 1 + \frac{2X}{2+X}$$

4. Multiply through: $$X(2+X) = 2 + X + 2X$$ $$X^2 - X - 2 = 0$$

5. Factorize: $$(X-2)(X+1) = 0$$

6. Rejecting the negative root, $$X = 2 \mu F$$

7. Final Answer: $$C_{eq} = 2 \mu F$$

Example 16: A dipole of moment $p$ is rotated from stable to unstable equilibrium in a field $E$. Find the work done.

Solution:

1. Stable equilibrium corresponds to: $$\theta_1 = 0^{\circ}$$
2. Unstable equilibrium corresponds to: $$\theta_2 = 180^{\circ}$$
3. Work done by external agent: $$W = pE(\cos\theta_1 - \cos\theta_2)$$
4. Substitute values: $$W = pE(\cos 0^{\circ} - \cos 180^{\circ}) = pE(1 - (-1)) = 2pE$$
5. Final Answer: $$W = 2pE$$

Example 17: State the properties of equipotential surfaces.

Solution:

1. Work done in moving a charge along an equipotential surface is zero.
2. Electric field is always perpendicular to the equipotential surface.
3. Two equipotential surfaces never intersect.
4. Closer equipotential surfaces indicate a stronger electric field because: $$E = -\frac{dV}{dr}$$

Example 18: If a capacitor is charged to charge $Q$ and the plates are pulled apart so that the distance doubles, how does the energy change?

Solution:

1. Charge remains constant.
2. For a parallel plate capacitor, $$C \propto \frac{1}{d}$$ so when distance doubles, capacitance becomes: $$\frac{C}{2}$$
3. Since, $$U = \frac{Q^2}{2C}$$ halving $C$ doubles the energy.
4. Final Answer: The new energy becomes $$2U$$.

Example 19: Find the force of attraction between the plates of a parallel plate capacitor.

Solution:

1. Field due to one plate only: $$E_{\text{single plate}} = \frac{\sigma}{2\epsilon_0}$$
2. Since $\sigma = \frac{Q}{A}$, $$E_{\text{single plate}} = \frac{Q}{2A\epsilon_0}$$
3. Force on the other plate: $$F = Q E_{\text{single plate}} = Q \cdot \frac{Q}{2A\epsilon_0}$$
4. Therefore, $$F = \frac{Q^2}{2A\epsilon_0}$$
5. Final Answer: $$F = \frac{Q^2}{2A\epsilon_0}$$

Example 20: A system has two charges $7 \mu C$ and $-2 \mu C$ at $(-9,0,0)$ cm and $(9,0,0)$ cm. Find the potential energy.

Solution:

1. Given: $$q_1 = 7 \times 10^{-6} \text{ C}, \quad q_2 = -2 \times 10^{-6} \text{ C}$$
2. Separation between the charges: $$r = 18 \text{ cm} = 0.18 \text{ m}$$
3. Potential energy of the system: $$U = k\frac{q_1 q_2}{r} = 9 \times 10^9 \times \frac{(7 \times 10^{-6})(-2 \times 10^{-6})}{0.18}$$
4. Simplify: $$U = 9 \times 10^9 \times \frac{-14 \times 10^{-12}}{0.18} = -0.7 \text{ J}$$
5. Final Answer: $$U = -0.7 \text{ J}$$

Example 21: A sphere of radius $R$ has a volume charge density $\rho = \alpha r$. Find the potential at the center.

Solution:

1. Consider a thin spherical shell of radius $r$ and thickness $dr$.
2. Volume of shell: $$d\tau = 4\pi r^2 dr$$
3. Charge on shell: $$dq = \rho \, d\tau = \alpha r \cdot 4\pi r^2 dr = 4\pi \alpha r^3 dr$$
4. Potential contribution at the center due to this shell: $$dV = \frac{1}{4\pi\epsilon_0} \frac{dq}{r} = \frac{1}{4\pi\epsilon_0} \frac{4\pi \alpha r^3 dr}{r} = \frac{\alpha r^2 dr}{\epsilon_0}$$
5. Integrate from $0$ to $R$: $$V = \int_0^R \frac{\alpha r^2}{\epsilon_0} dr = \frac{\alpha}{\epsilon_0} \left[ \frac{r^3}{3} \right]_0^R = \frac{\alpha R^3}{3\epsilon_0}$$
6. Final Answer: $$V = \frac{\alpha R^3}{3\epsilon_0}$$

Example 22: A capacitor $C$ is charged by a battery to potential $V$. It is disconnected and connected to an identical uncharged capacitor. Find the ratio of initial to final energy.

Solution:

1. Initial energy: $$U_i = \frac{1}{2}CV^2$$
2. After connection to an identical uncharged capacitor, the common potential becomes: $$\frac{V}{2}$$
3. Final energy of the two-capacitor system: $$U_f = 2 \times \frac{1}{2}C\left(\frac{V}{2}\right)^2 = \frac{1}{4}CV^2$$
4. Ratio: $$\frac{U_i}{U_f} = \frac{\frac{1}{2}}{\frac{1}{4}} = 2$$
5. Final Answer: $$U_i : U_f = 2 : 1$$

Example 23: Why does the capacitance of a capacitor increase when a dielectric slab is inserted between its plates?

Solution:

1. Polarization of the dielectric creates an induced electric field opposing the original field.
2. Hence the net electric field between the plates decreases.
3. Since, $$V = Ed$$ the potential difference decreases for the same charge.
4. Therefore, using $$C = \frac{Q}{V}$$ the capacitance increases.
5. Final Answer: Capacitance increases because the dielectric reduces the potential difference for the same stored charge.

Example 24: Two metal spheres of radii $r_1$ and $r_2$ are charged to the same potential. Find the ratio of their surface charge densities.

Solution:

1. Same potential implies: $$\frac{kQ_1}{r_1} = \frac{kQ_2}{r_2} \Rightarrow \frac{Q_1}{Q_2} = \frac{r_1}{r_2}$$
2. Surface charge density is: $$\sigma = \frac{Q}{4\pi r^2}$$
3. Therefore, $$\frac{\sigma_1}{\sigma_2} = \frac{Q_1}{Q_2} \cdot \frac{r_2^2}{r_1^2}$$
4. Substitute $$\frac{Q_1}{Q_2} = \frac{r_1}{r_2}$$ $$\frac{\sigma_1}{\sigma_2} = \frac{r_1}{r_2} \cdot \frac{r_2^2}{r_1^2} = \frac{r_2}{r_1}$$
5. Final Answer: $$\frac{\sigma_1}{\sigma_2} = \frac{r_2}{r_1}$$

Example 25: The energy density in a parallel plate capacitor is $2.12 \times 10^{-8} \text{ J/m}^3$. Find the electric field between the plates.

Solution:

1. Formula for energy density: $$u = \frac{1}{2}\epsilon_0 E^2$$
2. Substitute values: $$2.12 \times 10^{-8} = \frac{1}{2} \times 8.85 \times 10^{-12} \times E^2$$
3. Solve for $E^2$: $$E^2 \approx 4.79 \times 10^3$$
4. Therefore, $$E \approx 69 \text{ V/m}$$
5. Final Answer: $$E \approx 69 \text{ V/m}$$

Example 26: A capacitor is made of two concentric spheres of radii $a$ and $b$ ($b > a$). Find the capacitance.

Solution:

1. Potential difference between the spheres: $$V = \frac{1}{4\pi\epsilon_0}\frac{Q}{a} - \frac{1}{4\pi\epsilon_0}\frac{Q}{b} = \frac{Q}{4\pi\epsilon_0}\left(\frac{1}{a} - \frac{1}{b}\right)$$
2. Simplify: $$V = \frac{Q}{4\pi\epsilon_0} \cdot \frac{b-a}{ab}$$
3. Capacitance: $$C = \frac{Q}{V} = \frac{4\pi\epsilon_0 ab}{b-a}$$
4. Final Answer: $$C = \frac{4\pi\epsilon_0 ab}{b-a}$$

Example 27: What is the work done in moving a test charge $q$ over an equipotential surface?

Solution:

1. On an equipotential surface, $$\Delta V = 0$$
2. Work done is: $$W = q\Delta V = 0$$
3. Final Answer: $$W = 0$$

Example 28: Three capacitors $C_1, C_2, C_3$ are connected so that $C_1$ is in series with the parallel combination of $C_2$ and $C_3$. If $C_1 = 2 \mu F$, $C_2 = 3 \mu F$, $C_3 = 3 \mu F$, find the net capacitance.

Solution:

1. Parallel part: $$C_p = 3 + 3 = 6 \mu F$$
2. Now $C_1 = 2 \mu F$ is in series with $C_p = 6 \mu F$: $$\frac{1}{C_{net}} = \frac{1}{2} + \frac{1}{6} = \frac{4}{6} = \frac{2}{3}$$
3. Therefore, $$C_{net} = 1.5 \mu F$$
4. Final Answer: $$C_{net} = 1.5 \mu F$$

Example 29: What is the net charge on a capacitor?

Solution:

1. A capacitor has equal and opposite charges on its two plates: $$+Q \text{ and } -Q$$
2. Net charge of the capacitor: $$+Q + (-Q) = 0$$
3. Final Answer: $$\text{Net charge} = 0$$

Example 30: Find the potential at the center of a charged ring of radius $R$ and total charge $Q$.

Solution:

1. Every small charge element $dq$ on the ring is at the same distance $R$ from the center.
2. Potential due to element $dq$: $$dV = \frac{k \, dq}{R}$$
3. Integrate over the ring: $$V = \int \frac{k \, dq}{R} = \frac{k}{R} \int dq = \frac{kQ}{R}$$
4. Final Answer: $$V = \frac{kQ}{R}$$