🔙 Prerequisite Revision:
👉 Chapter 1: Electric Charges & Fields (Coulomb's Law)Chapter 2: Electrostatic Potential & Capacitance
Vol 1: Electric Potential ($V$)
Electric Potential: The work done in bringing a unit positive charge from infinity to that point against the electrostatic force.
$$V = \frac{1}{4\pi\epsilon_0} \frac{q}{r}$$
Unit: Volt (V) or J/C. Scalar Quantity.
Relation between E and V
Electric field is always in the direction of decreasing potential.
Equipotential Surfaces
A surface where the potential is constant at all points.
- Work Done = 0: No work is done in moving a charge on an equipotential surface.
- Direction: Electric field lines are always perpendicular to the equipotential surface.
- They never intersect.
Vol 2: Electrostatic Potential Energy ($U$)
The work done in assembling a system of charges by bringing them from infinity to their present locations.
$$U = \frac{1}{4\pi\epsilon_0} \frac{q_1 q_2}{r}$$
Dipole in Uniform Electric Field
- Potential Energy: $U = - \vec{p} \cdot \vec{E} = -pE \cos \theta$
• Stable Equilibrium ($\theta = 0^\circ$): $U_{min} = -pE$.
• Unstable Equilibrium ($\theta = 180^\circ$): $U_{max} = +pE$. - Work Done in Rotating: $W = pE(\cos \theta_1 - \cos \theta_2)$
Vol 3: Capacitors & Dielectrics
Capacitor: A device used to store electric charge and electrical energy.
Capacitance ($C$): Ability to store charge per unit voltage ($C = Q/V$). Unit: Farad (F).
Parallel Plate Capacitor
[attachment_0](attachment)$$C_0 = \frac{\epsilon_0 A}{d}$$
2. With Dielectric Slab (Constant K):
$$C = \frac{K \epsilon_0 A}{d} = K C_0$$
Inserting a dielectric increases capacitance by factor K.
Grouping of Capacitors
[attachment_1](attachment)| Series Combination | Parallel Combination |
|---|---|
| Charge ($Q$) is same. | Voltage ($V$) is same. |
| $\frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_2} + \dots$ | $C_{eq} = C_1 + C_2 + \dots$ |
| $C_{eq}$ is smaller than smallest $C$. | $C_{eq}$ is larger than largest $C$. |
Energy Stored in Capacitor ($U$)
Energy Density ($u$): $u = \frac{1}{2}\epsilon_0 E^2$ (Energy per unit volume)
Vol 4: Mission 100 Question Bank
Section A: MCQs
Q1. The work done in moving a charge of 5 C on an equipotential surface of potential 10 V is:
(a) 50 J (b) 2 J (c) Zero (d) 10 J
Ans: (c) Zero (Since $\Delta V = 0$, $W = q\Delta V = 0$).
Q2. If a dielectric slab is inserted between the plates of an isolated capacitor, its potential difference:
(a) Increases (b) Decreases (c) Remains same (d) Becomes zero
Ans: (b) Decreases ($V = V_0/K$ as $C$ increases and $Q$ is constant).
Section B: Numericals (Important)
Q3. Three capacitors of capacitance 9 pF each are connected in series. (a) What is the total capacitance? (b) If connected to 120 V supply, what is potential difference across each?
Solution:
(a) Series: $1/C_{eq} = 1/9 + 1/9 + 1/9 = 3/9 = 1/3$.
$C_{eq} = 3$ pF.
(b) Since all $C$ are equal, $V$ divides equally.
$V_1 = V_2 = V_3 = 120/3 = 40$ V.
Ans: 3 pF, 40 V.
Section C: Conceptual
Q4. Why must the electric field lines be normal to the surface at every point of a charged conductor?
Ans: If the field were not normal, it would have a tangential component along the surface. This component would cause the free charges on the surface to move (surface current), which contradicts the static nature of electrostatics.
Mission 100 Physics Series
Next Chapter: Current Electricity (Ohm's Law & Kirchhoff's Rules)
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