Electrostatic Potential and Capacitance Quiz Set-2 📘

Did you know? Your entire modern digital world—mobile phones, camera flashes, laptop motherboards—works smoothly only because tiny capacitors are quietly storing and releasing electric charge in microseconds. Mastering electrostatic potential and capacitance in Class 12 Physics is not just for exams; it’s literally learning the language of electronics.

1️⃣ Big Picture: Why This Chapter Is a Scoring Machine

Electrostatic Potential and Capacitance is one of the most algorithmic chapters in Class 12 Physics:

Well‑defined formulas
Standard types of derivations
Frequently asked numericals

Exam relevance:

CBSE Class 12 Board: Regularly 3–5 marks weightage, usually numerical + reasoning.
JEE Main/Advanced: Conceptual MCQs and tricky integer‑type questions from conductors, energy stored, and combination of capacitors.
NEET: Straight but conceptual questions (e.g., effect on capacitance when dielectric changes, or plate separation changes).

To do well on the Electrostatic Potential and Capacitance Quiz Set-2, you must be comfortable with:

Potential due to different charge configurations
Relations between E, V, Q, C
Series and parallel combinations of capacitors
Energy stored in capacitors and effect of inserting dielectrics

2️⃣ Concept Snapshot: Electrostatic Potential & Potential Difference ⚡

What is electrostatic potential?

Electrostatic potential at a point is the work done per unit positive test charge in bringing it from infinity to that point (against electrostatic forces).

On the formula level:

V = \frac{W}{q}

where

$V$ = electrostatic potential
$W$ = work done in bringing charge $q$ from infinity to that point

Potential due to a point charge

For a point charge $Q$ placed in free space (vacuum or air), potential at a distance $r$ from it is

V = \frac{1}{4\pi\epsilon_0}\frac{Q}{r}

Here $\epsilon_0$ is the permittivity of free space.

Quick Concept Table 🧠

Concept	Key Expression (in words)	Typical Question Pattern (CBSE/JEE)
Potential difference between A & B	Work done per unit charge from A to B	“Find work done in moving charge from A to B”
Potential due to point charge	Constant × charge ÷ distance	“Potential at a point due to multiple charges”
Relation between E and V	Electric field is negative potential gradient	“Given potential as function of x, find electric field”

Relation between electric field and potential

If potential varies along x‑axis, then:

E_x = -\frac{dV}{dx}

This is especially important for JEE questions where potential function is given and you have to find field.

3️⃣ Capacitance: The Art of Storing Charge 📦

Basic definition

Capacitance of a conductor is defined as the ratio of charge on it to potential of the conductor.

C = \frac{Q}{V}

Unit of capacitance is farad (F).

Parallel plate capacitor

For a parallel plate capacitor with

Plate area = $A$
Separation = $d$
Dielectric medium with permittivity $\epsilon$ between the plates

Capacitance is

C = \frac{\epsilon A}{d}

If medium is air/vacuum, $\epsilon = \epsilon_0$ .
If a dielectric of relative permittivity $K$ is inserted, $\epsilon = K\epsilon_0$ , so

C = \frac{K\epsilon_0 A}{d}

4️⃣ Worked Example: Potential and Energy in a Parallel Plate Capacitor 🧮

Example 1: Finding capacitance, charge and energy

A parallel plate capacitor has plate area 0.02 m², separation 1 mm, and is filled with air. It is connected across a 200 V battery. Calculate:

Capacitance
Charge on each plate
Energy stored in the electric field

Take $\epsilon_0 = 8.85 \times 10^{-12}\ \text{F m}^{-1}$ .

Step 1: Capacitance

Use

C = \frac{\epsilon_0 A}{d}

Here:

$A = 0.02\ \text{m}^2$
$d = 1\ \text{mm} = 1 \times 10^{-3}\ \text{m}$

C = \frac{8.85 \times 10^{-12} \times 0.02}{1 \times 10^{-3}}

C = 1.77 \times 10^{-10}\ \text{F}

Step 2: Charge on each plate

Use relation

Q = CV

Q = 1.77 \times 10^{-10} \times 200

Q = 3.54 \times 10^{-8}\ \text{C}

Step 3: Energy stored

Energy stored in capacitor is

U = \frac{1}{2}CV^2

Substitute:

U = \frac{1}{2} \times 1.77 \times 10^{-10} \times (200)^2

U = 3.54 \times 10^{-6}\ \text{J}

Exam Insight: This type of question is standard in CBSE and also forms the base for trickier JEE questions where plate separation or dielectric is changed after charging.

5️⃣ Series and Parallel Combinations: The “Effective C” Game 🔁

Most questions in quizzes and competitive exams revolve around finding equivalent capacitance of different arrangements.

5.1 Capacitors in series

For capacitors $C_1$ , $C_2$ , $C_3$ in series:

\frac{1}{C_{\text{eq}}} = \frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3}

Features:

Same charge on each capacitor
Potential differences add up

5.2 Capacitors in parallel

For capacitors $C_1$ , $C_2$ , $C_3$ in parallel:

C_{\text{eq}} = C_1 + C_2 + C_3

Features:

Same potential difference across each capacitor
Charge distributes according to capacitance

Quick Revision Box 📦

Series → Reciprocal adds
Parallel → Direct adds
Series → Same Q, different V
Parallel → Same V, different Q

6️⃣ Mixed Combination Example (Very JEE/NEET Friendly) 🧩

Three capacitors of 2 μF, 3 μF, and 6 μF are connected such that 2 μF and 3 μF are in series and this combination is in parallel with 6 μF capacitor. The combination is connected to a 12 V supply. Find:

Equivalent capacitance
Charge on each capacitor

Step 1: Series combination

For 2 μF and 3 μF in series:

\frac{1}{C_s} = \frac{1}{2} + \frac{1}{3}

\frac{1}{C_s} = \frac{3 + 2}{6} = \frac{5}{6}

C_s = \frac{6}{5} = 1.2\ \mu\text{F}

Step 2: Parallel with 6 μF

Now, $C_s = 1.2\ \mu\text{F}$ is in parallel with 6 μF:

C_{\text{eq}} = 1.2 + 6 = 7.2\ \mu\text{F}

Step 3: Total charge drawn from battery

Using $Q = C_{\text{eq}} V$ :

Q_{\text{total}} = 7.2\ \mu\text{F} \times 12\ \text{V} = 86.4\ \mu\text{C}

Step 4: Charge on 6 μF capacitor

Since it is directly across 12 V:

Q_6 = 6\ \mu\text{F} \times 12\ \text{V} = 72\ \mu\text{C}

Step 5: Charge on series combination

Charge on the parallel branch containing $C_s$ is:

Q_s = C_s V = 1.2\ \mu\text{F} \times 12\ \text{V} = 14.4\ \mu\text{C}

In series, same charge flows through both capacitors.
So charge on 2 μF and 3 μF is 14.4 μC each.

Pattern alert: This structure—two capacitors in series, then in parallel with a third—is very popular in JEE Main and high-level CBSE MCQs.

7️⃣ Energy Stored, Dielectrics, and “What If” Changes 🔋

7.1 Energy stored in a capacitor

Three common useful forms:

U = \frac{1}{2}CV^2

U = \frac{1}{2}QV

U = \frac{Q^2}{2C}

Choose the form depending on which quantities are constant in the problem (Q or V).

7.2 Battery connected vs battery removed

This is a favourite twist in quizzes:

Battery remains connected
- Potential difference stays constant
- Capacitance changes → charge and stored energy change
Battery disconnected
- Charge remains constant
- Capacitance changes → potential difference and energy change

Concept Comparison Table 🧾

Situation	Constant quantity	Changes when dielectric inserted	Typical question style
Battery connected	V	Q, U, E-field between plates	“Find new charge / energy”
Battery removed (isolated plates)	Q	V, U, E-field between plates	“Find new potential / energy”

8️⃣ “Did You Notice?” Conceptual Traps in This Chapter 🚨

Common Mistake 1: Confusing potential and potential energy

Potential (V): Work done per unit charge.
Potential energy (U): Total work done in bringing the actual charge configuration.

You can have zero potential at a point and still have non-zero electric field, and vice versa.

Common Mistake 2: Treating series combination like resistors for charge

Students often think: “In resistors in series, current is same; in capacitors in series, maybe potential is same?”
Wrong. In capacitors in series, charge is same on each plate; potential adds up.

Common Mistake 3: Forgetting the role of dielectric constant K

For a parallel plate capacitor:

Inserting dielectric increases capacitance by factor K.
But what happens to energy depends on whether battery is connected or not.

For example, if battery is connected (V constant):

C increases → Q increases → energy
- $U = \frac{1}{2}CV^2$ → increases.

If battery is removed (Q constant):

C increases → V decreases → energy
- $U = \frac{Q^2}{2C}$ → decreases.

Exactly this style appears in NEET conceptual MCQs.

9️⃣ Visualizing with “Imaginary Diagrams” 🧭 (Mental Picture Aid)

Since you may not have the diagram in front of you while practicing, here is how to visualize key configurations:

Parallel Plate Capacitor
Imagine two large metallic plates facing each other, separated by a small air gap. Electric field lines are straight from positive plate to negative plate, almost uniform in the middle.
Series Capacitors
Think of several plates back-to-back:
- Outer plates connected to battery terminals
- Inner plates forming pairs with induced opposite charges
Combination Networks
Visualize complex circuits as “islands” of plates grouped either side-by-side (parallel) or end-to-end (series). Redrawing the circuit cleanly is often half the solution in JEE/NEET numericals.

🔟 Rapid-Fire Revision Points for Last-Minute Preparation ⚡📚

Electrostatic potential is scalar; potential due to multiple charges is algebraic sum.
At infinity, potential is usually taken as zero for isolated charges.
For a point charge, potential decreases as 1/r.
Equipotential surfaces are always perpendicular to electric field lines.
Work done in moving a charge on an equipotential surface is zero.
Capacitance depends only on geometry and medium, not on charge or potential.
In series combination:
- Same charge
- Equivalent capacitance is less than smallest individual capacitance.
In parallel combination:
- Same potential difference
- Equivalent capacitance is greater than largest individual capacitance.
Energy stored in capacitor can be asked in all three forms: $\frac{1}{2}CV^2$ , $\frac{1}{2}QV$ , $\frac{Q^2}{2C}$ .
Dielectric insertion problems always ask: Is battery connected or disconnected? Answer that mentally first.

1️⃣1️⃣ Strategy Tips to Ace Electrostatic Potential and Capacitance Quiz Set-2 🎯

Start with formula warm‑up: Before taking the quiz, write down on a rough sheet:
- Expressions for potential, capacitance, series and parallel combinations, and energy stored.
Re-draw circuits: For combination problems, always redraw the network to clearly see series/parallel groups.
Unit check: For numericals, quickly check if your final answer has correct units:
- V for potential
- C or μC for charge
- F or μF for capacitance
- J for energy
Think in “battery connected vs disconnected” mode: Especially for dielectric and plate‑separation questions.
Don’t rush conceptual options: Many MCQs in CBSE/JEE/NEET are designed with “almost correct” statements. Read each word carefully (isolated, grounded, connected to battery, etc.).

1️⃣2️⃣ Ready to Practise? Take the Chapter Quiz Now 🧪

Put your understanding of Electrostatic Potential and Capacitance to the test with a focused, exam-style quiz. Solve under timed conditions and then review your mistakes to strengthen weak areas.

Start Quiz: Electrostatic Potential and Capacitance

Electrostatic Potential and Capacitance Quiz Set-2