Revision Notes Chapter 2 Electrostatic Potential And Capacitance

Electrostatic Potential and Potential Energy Dynamics

The electrostatic potential ($V$) at any point in an electric field is physically defined as the work done per unit positive charge in bringing it from infinity to that point against the electrostatic forces without acceleration ($V = W/q_0$). Because the electrostatic force is a conservative force, this work done depends solely on the initial and final positions and is stored as electrostatic potential energy ($U$). For a discrete system of two point charges separated by distance $r$, the potential energy is $U = \frac{1}{4\pi\varepsilon_0}\frac{q_1 q_2}{r}$. For a single point charge $q$, the potential at a distance $r$ is expressed as $V = \frac{1}{4\pi\varepsilon_0}\frac{q}{r}$. An electric dipole of dipole moment $\vec{p}$ produces a potential that drops off more rapidly ($V \propto 1/r^2$): at a point $r$ making angle $\theta$ with the dipole axis, $V = \frac{1}{4\pi\varepsilon_0}\frac{p\cos\theta}{r^2}$. Crucially, the work done in moving a charge along any closed loop in an electrostatic field is zero.

Essential Chapter Glossary & Terminology

Electrostatic Potential (V)

The work done in bringing a unit positive charge from infinity to a point in an electric field against electrostatic forces.

Equipotential Surface

Any surface that has the same electrostatic potential at every point on it. The electric field is always perpendicular to this surface.

Capacitance (C)

The ratio of the magnitude of charge on either conductor plate to the potential difference between them ($C = Q/V$).

Dielectric Constant (K)

The ratio of the permittivity of the medium to the permittivity of free space, also equal to the factor by which capacitance increases when the medium is introduced.

Polarization (P)

The induced dipole moment per unit volume of a dielectric material when placed in an external electric field.

Potential Gradient

The rate of change of electric potential with respect to distance ($E = -dV/dr$). The negative sign indicates that the electric field points in the direction of decreasing potential.

Dielectric Slab vs. Conducting Slab inside a Capacitor

Key Points and Quick-Revision Formulas

• The relationship between electric field and potential is given by the differential form: $E = -\frac{dV}{dr}$.
• No work is done in moving a charge over an equipotential surface because the potential difference $\Delta V = 0$.
• The capacitance of an isolated spherical conductor of radius $R$ is $C = 4\pi\varepsilon_0 R$.
• Capacitance of a parallel plate capacitor in vacuum is $C_0 = \frac{\varepsilon_0 A}{d}$.
• Inserting a dielectric of constant $K$ that completely fills the gap increases the capacitance: $C = K C_0$.
• The energy stored in a charged capacitor is given by: $U = \frac{1}{2} C V^2 = \frac{1}{2} Q V = \frac{Q^2}{2 C}$.
• The electrostatic energy density (energy per unit volume) in free space is $u = \frac{1}{2} \varepsilon_0 E^2$.
• In series combination, the charge $Q$ is constant across all capacitors: $\frac{1}{C_s} = \frac{1}{C_1} + \frac{1}{C_2} + \dots$
• In parallel combination, the potential difference $V$ is constant across all capacitors: $C_p = C_1 + C_2 + \dots$

Step-by-Step Derivation of Parallel Plate Capacitor Capacitance

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Worked Mini-Examples for CBSE Revision

• {"title":"Example 1: Work Done on a Charge","bodyMarkdown":"Question: Calculate the work done in moving a charge of $3\\,\\mu\\text{C}$ from infinity to a point where the electric potential is $2 \\times 10^4\\,\\text{V}$.\n\nSolution:\nWe use the potential relation:\n$W = q \\cdot V$\nSubstituting the given values:\n$W = (3 \\times 10^{-6}\\,\\text{C}) \\times (2 \\times 10^4\\,\\text{V}) = 0.06\\,\\text{Joule}$\nTherefore, the work done is $0.06\\,\\text{J}$."}
• {"title":"Example 2: Energy Stored in a Capacitor","bodyMarkdown":"Question: A $10\\,\\mu\\text{F}$ capacitor is charged by a $100\\,\\text{V}$ battery. How much electrostatic energy is stored in it?\n\nSolution:\nUsing the formula for stored potential energy:\n$U = \\frac{1}{2} C V^2$\nSubstitute the values:\n$U = \\frac{1}{2} \\times (10 \\times 10^{-6}\\,\\text{F}) \\times (100)^2$\n$U = 5 \\times 10^{-6} \\times 10000 = 0.05\\,\\text{Joule}$\nThus, the energy stored is $0.05\\,\\text{J}$."}

Critical Board Exam Traps & Formula Shifts

Pay extreme attention to whether the battery is CONNECTED or DISCONNECTED when a dielectric slab is inserted! This is one of CBSE's most frequently tested conceptual traps:

1. Battery Remains Connected:

• Potential difference $V$ remains constant ($V = V_0$).
• Capacitance increases ($C = K C_0$).
• Charge increases ($Q = K Q_0$).
• Electric field remains constant ($E = E_0$).
• Energy stored increases ($U = K U_0$).

1. Battery is Disconnected (Isolated Capacitor):

• Charge $Q$ remains constant ($Q = Q_0$).
• Capacitance increases ($C = K C_0$).
• Potential difference decreases ($V = V_0 / K$).
• Electric field decreases ($E = E_0 / K$).
• Energy stored decreases ($U = U_0 / K$).

Marking Scheme Tip: When sketching equipotential surfaces for a point charge, make sure your surfaces are concentric circles and clearly show the spacing between surfaces increasing as you move further away, because $E = -dV/dr$ implies larger $dr$ for the same $dV$ as $E$ decreases.

Quick Revision Check

• Can two equipotential surfaces intersect each other? Explain. No. If they intersected, there would be two directions of electric field at the point of intersection, which is physically impossible.
• What is the electrostatic potential energy of a dipole placed perpendicular to an electric field? Zero. The potential energy is given by $U = -\vec{p} \cdot \vec{E} = -p E \cos\theta$. For a perpendicular orientation ($\theta = 90^\circ$), $\cos(90^\circ) = 0$, so $U = 0$.
• If the plate area of a parallel plate capacitor is doubled and the plate separation is halved, what happens to its capacitance? The capacitance increases by a factor of 4. Since $C = \frac{\varepsilon_0 A}{d}$, doubling $A$ and halving $d$ yields $C' = \frac{\varepsilon_0 (2A)}{d/2} = 4 C$.
• Why does the capacitance of a capacitor increase when a dielectric slab is introduced? The dielectric undergoes polarization, inducing opposite charges on its faces. This sets up an internal field opposing the external field, which reduces the potential difference $V$ between the plates for a given charge $Q$. Since $C = Q/V$, a lower $V$ results in a larger capacitance $C$.

Frequently Asked Questions