NCERT Class 12 Physics: Electrostatic Potential and Capacitance

The Physics of Electrostatic Potential and Energy

To understand electrostatic potential, consider a test charge $q$ placed in an electric field $\vec{E}$ generated by a source charge $Q$. When we move this test charge from point A to point B against the repulsive electrostatic force, we must do external work. Because the electrostatic force is conservative, this work done is stored entirely as electrostatic potential energy ($U$).

The electrostatic potential ($V$) at any point in an electric field is defined as the work done by an external agent in bringing a unit positive charge from infinity to that point without acceleration. Mathematically, potential is given by:

$V = \frac{W_{\infty \to P}}{q}$

For a point charge $q$ at a distance $r$, the potential is derived as:

$V = \frac{1}{4\pi\varepsilon_0} \frac{q}{r}$

Unlike electric fields, electrostatic potential is a scalar quantity, which greatly simplifies complex calculations. When dealing with a system of multiple charges, you can find the net potential at a point by taking the simple algebraic sum of individual potentials, without worrying about vector directions.

Core Terms and Core Concepts

Electrostatic Potential Difference

The work done by an external force in moving a unit positive charge from one point to another in an electric field, represented as V_B - V_A = W_AB / q.

Equipotential Surface

Any surface that has the same electrostatic potential at every point on it. No work is done in moving a charge along an equipotential surface.

Capacitance

The ability of a conductor to store electric charge and potential energy, quantified as C = Q / V, measured in Farads (F).

Dielectric Material

An insulating material that polarizes when placed in an external electric field, reducing the net electric field inside and increasing the capacitance of a capacitor.

Deriving the Capacitance of a Parallel-Plate Capacitor

1. Setup Plate Geometry and Charges — Consider two parallel conducting plates, each of area $A$, separated by a small distance $d$. Let the plates carry equal and opposite surface charge densities, $+\sigma$ and $-\sigma$ respectively, where $\sigma = Q/A$.
2. Determine the Electric Field Between Plates — Using Gauss's Law, the electric field in the outer regions is zero. In the inner region between the plates, the fields due to both plates add up: $E = \frac{\sigma}{2\varepsilon_0} + \frac{\sigma}{2\varepsilon_0} = \frac{\sigma}{\varepsilon_0} = \frac{Q}{\varepsilon_0 A}$.
3. Relate Potential Difference to Electric Field — Since the field is uniform, the potential difference $V$ between the plates is given by: $V = E \cdot d = \frac{Q \cdot d}{\varepsilon_0 A}$.
4. Calculate the Final Capacitance — Substitute $V$ into the definition of capacitance $C = \frac{Q}{V}$. This yields: $C = \frac{Q}{\left(\frac{Q \cdot d}{\varepsilon_0 A}\right)} = \frac{\varepsilon_0 A}{d}$.

Step-by-Step Worked Examples

• Example 1: Potential of a Dipole Calculate the electrostatic potential at an axial point distance $r = 10\text{ cm}$ from the center of an electric dipole of dipole moment $p = 9 \times 10^{-9}\text{ C}\cdot\text{m}$. Step 1: Identify the formula for the axial potential of a dipole: $V = \frac{1}{4\pi\varepsilon_0} \frac{p}{r^2}$ (for $r \gg a$). Step 2: Plug in the known values. $1/(4\pi\varepsilon_0) = 9 \times 10^9\text{ N}\cdot\text{m}^2/\text{C}^2$, $p = 9 \times 10^{-9}\text{ C}\cdot\text{m}$, and $r = 0.1\text{ m}$. Step 3: Calculate: $V = \frac{(9 \times 10^9) \times (9 \times 10^{-9})}{(0.1)^2} = \frac{81}{0.01} = 8100\text{ V}$. Answer: The electrostatic potential is $8100\text{ V}$ (or $8.1\text{ kV}$).
• Example 2: Equivalent Capacitance and Energy Two capacitors of capacities $3\,\mu\text{F}$ and $6\,\mu\text{F}$ are connected in series across a $12\text{ V}$ battery. Find the equivalent capacitance and total energy stored. Step 1: Use the series formula: $\frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_2} = \frac{1}{3} + \frac{1}{6} = \frac{2+1}{6} = \frac{3}{6} = \frac{1}{2}$. Thus, $C_{eq} = 2\,\mu\text{F}$. Step 2: Calculate the total stored energy: $U = \frac{1}{2} C_{eq} V^2$. Step 3: Substitute values: $U = \frac{1}{2} \times (2 \times 10^{-6}\text{ F}) \times (12)^2 = 10^{-6} \times 144 = 1.44 \times 10^{-4}\text{ J}$. Answer: The equivalent capacitance is $2\,\mu\text{F}$ and the total stored energy is $1.44 \times 10^{-4}\text{ J}$.

Board Exam Traps and Shortcuts

Here are critical traps and formulas to keep in mind for your CBSE Class 12 board exam:

• Dielectric Insertions: If a dielectric slab of dielectric constant $K$ is inserted, check if the battery remains connected or is disconnected:
• Battery Connected: Potential difference $V$ remains constant ($V = V_0$). Charge increases ($Q = KQ_0$), and capacitance increases ($C = KC_0$).
• Battery Disconnected: Charge $Q$ remains constant ($Q = Q_0$). Potential difference decreases ($V = V_0 / K$), and capacitance increases ($C = KC_0$).
• Conductors in Equilibrium: The electric field inside a charged hollow conductor is zero, but the electrostatic potential remains constant throughout its volume and is equal to its value on the surface ($V = \frac{kq}{R}$). Do not mark potential as zero inside!

Practice Questions with Solutions

• Q: A $4\,\mu\text{F}$ capacitor is charged by a $200\text{ V}$ supply. It is then disconnected from the supply and is connected to another uncharged $2\,\mu\text{F}$ capacitor. How much electrostatic energy is lost in the process? A: Step 1: Calculate initial energy stored: $U_i = \frac{1}{2} C_1 V_1^2 = \frac{1}{2} \times (4 \times 10^{-6}) \times (200)^2 = 2 \times 10^{-6} \times 40000 = 0.08\text{ J}$. Step 2: Find common potential $V$ when connected: $V = \frac{C_1 V_1 + C_2 V_2}{C_1 + C_2} = \frac{(4 \times 10^{-6} \times 200) + 0}{(4 + 2) \times 10^{-6}} = \frac{8 \times 10^{-4}}{6 \times 10^{-6}} = 133.3\text{ V}$. Step 3: Calculate final energy stored: $U_f = \frac{1}{2} (C_1 + C_2) V^2 = \frac{1}{2} \times (6 \times 10^{-6}) \times (133.3)^2 = 3 \times 10^{-6} \times 17768.9 = 0.0533\text{ J}$. Step 4: Energy lost: $\Delta U = U_i - U_f = 0.08 - 0.0533 = 0.0267\text{ J}$. Final answer: $2.67 \times 10^{-2}\text{ J}$ is lost as heat and electromagnetic radiation.
• Q: Derive the work done in bringing three charges $q_1$, $q_2$, and $q_3$ from infinity to the vertices of an equilateral triangle of side $r$. A: Step 1: Work done to bring $q_1$ to its vertex is zero ($W_1 = 0$) because there is no external electric field. Step 2: Work done to bring $q_2$ in the field of $q_1$: $W_2 = \frac{1}{4\pi\varepsilon_0} \frac{q_1 q_2}{r}$. Step 3: Work done to bring $q_3$ in the fields of both $q_1$ and $q_2$: $W_3 = \frac{1}{4\pi\varepsilon_0} \left( \frac{q_1 q_3}{r} + \frac{q_2 q_3}{r} \right)$. Step 4: Total work (electrostatic potential energy $U$): $W_{total} = W_1 + W_2 + W_3 = \frac{1}{4\pi\varepsilon_0 r} (q_1 q_2 + q_2 q_3 + q_3 q_1)$. Final answer: $U = \frac{1}{4\pi\varepsilon_0 r} (q_1 q_2 + q_2 q_3 + q_3 q_1)$.
• Q: What is the shape of equipotential surfaces for: (a) a single point charge, and (b) a uniform electric field along the z-axis? A: Step 1: For a single point charge, the potential is constant at a constant distance $r$ from the charge. Thus, the equipotential surfaces are concentric spheres centered on the point charge. Step 2: For a uniform electric field along the z-axis, the potential changes only in the z-direction. Therefore, the equipotential surfaces are planes perpendicular to the z-axis (i.e., planes parallel to the x-y plane). Final answer: (a) Concentric spheres, (b) Planes parallel to the x-y plane.
• Q: A parallel-plate capacitor with air between plates has a capacitance of $8\text{ pF}$. 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 $K = 6$? A: Step 1: Write down the initial capacitance: $C_0 = \frac{\varepsilon_0 A}{d} = 8\text{ pF}$. Step 2: Write the expression for the new capacitance $C'$ with distance $d' = d/2$ and dielectric constant $K = 6$: $C' = \frac{K \varepsilon_0 A}{d'} = \frac{6 \varepsilon_0 A}{(d/2)}$. Step 3: Simplify the expression: $C' = 12 \left( \frac{\varepsilon_0 A}{d} \right) = 12 \cdot C_0$. Step 4: Solve: $C' = 12 \times 8\text{ pF} = 96\text{ pF}$. Final answer: The new capacitance is $96\text{ pF}$.

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