Capacitor Without Dielectric (Advanced Level) Concept Quiz

 

⚡ Capacitor Without Dielectric (Advanced Level)

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🧠 Section A: Conceptual (10 Questions)

Q1. Two identical parallel plate capacitors are connected in series. One plate of each is suddenly earthed. What happens to the equivalent capacitance?

Q2. A capacitor is charged and then disconnected. If the plate separation is increased, how do charge, voltage, and energy change?

Q3. A metal slab (not touching plates) is inserted between plates. How does capacitance change and why?

Q4. If one plate of a capacitor is earthed while the other is kept at potential V, what is the charge distribution?

Q5. A charged capacitor is connected to an identical uncharged capacitor. What is the final energy? Is energy conserved?

Q6. Why does energy decrease when two capacitors are connected? Where does the energy go?

Q7. A parallel plate capacitor is connected to a battery. If plate area is doubled, what happens to stored energy?

Q8. A charged capacitor is pulled apart slowly vs suddenly. Compare work done.

Q9. Why is capacitance independent of charge and voltage?

Q10. In a system of capacitors, why is potential same in parallel but charge same in series?

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🔢 Section B: Numerical (JEE/NEET Advanced Level)

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Q11. Variable Plate Separation

A parallel plate capacitor has plate area $A$, separation $d$, connected to battery $V$. Plate separation is increased to $2d$.

Find:

• New capacitance
• Work done by external agent

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Q12. Energy Loss in Sharing

A capacitor $C$ charged to $V$ is connected to identical uncharged capacitor.

Find:

• Final voltage
• Energy lost

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Q13. Capacitor + Earth

One plate of a capacitor $C$ is earthed. Other plate is given charge $Q$.

Find:

• Potential of system
• Effective capacitance

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Q14. Force Between Plates

A capacitor has plate area $A$, separation $d$, voltage $V$.

Find force between plates.

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Q15. Work to Increase Separation

A capacitor is disconnected after charging. Plate separation is increased from $d$ to $3d$.

Find:

• Final energy
• Work done

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Q16. Capacitor Network Reduction

Three capacitors each $C$ arranged in triangle. Find equivalent capacitance between two vertices.

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Q17. Sliding Plate Problem

A plate capacitor has overlapping area $A$. One plate is slid reducing overlap to $A/2$.

Find:

• Change in capacitance
• Work done

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Q18. Time-dependent Charging Energy

A capacitor is charged slowly from 0 to $V$.

Find total energy supplied by battery vs stored energy.

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Q19. Series + Battery Removal

Two capacitors in series connected to battery. Battery removed.

Now separation of one capacitor is doubled. Find new voltage distribution.

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Q20. Advanced JEE Problem

A capacitor $C$ charged to $V$ is disconnected. A conducting slab of thickness $t$ is inserted between plates (no contact).

Find:

• New capacitance
• New energy

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✅ Answer Key (Final Results Only)

Q	Answer
11	$C/2$, Work = $\frac{1}{2}CV^2$
12	$V/2$, Loss = $\frac{1}{4}CV^2$
13	$V = Q/C$, same C
14	$F = \frac{1}{2} \frac{\epsilon_0 A V^2}{d^2}$
15	Energy increases 3×
16	$\frac{2C}{3}$
17	Capacitance halves
18	Supplied = $CV^2$, Stored = $\frac{1}{2}C{V}^2$
19	Redistribution occurs
20	$C' = \frac{\epsilon_0 A}{d-t}$

🧾 Detailed Solutions 

✅ SECTION A: CONCEPTUAL SOLUTIONS

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Q1. Two capacitors in series, one plate earthed

• Earthing fixes potential = 0 for that plate.
• Charges redistribute due to grounding.
• Effective system behaves like a single capacitor of reduced separation.

✅ Answer: Equivalent capacitance increases

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Q2. Increasing plate separation (disconnected capacitor)

• Charge $Q$ = constant
• $C = \frac{\varepsilon_0 A}{d}$ ⇒ decreases
• $V = \frac{Q}{C}$ ⇒ increases
• $U = \frac{Q^2}{2C}$ ⇒ increases

✅ Answer:
Charge = constant, Voltage ↑, Energy ↑

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Q3. Metal slab inserted (no contact)

Effective separation reduces:

$$d_{eff} = d - t$$

Capacitance:

$$C = \frac{\varepsilon_0 A}{d - t}$$

✅ Answer: Capacitance increases

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Q4. One plate earthed, other at V

• Potential difference = $V$
  
• Charge:

$$Q = CV$$

• Induced charge appears on grounded plate

✅ Answer: Charges equal & opposite, magnitude $CV$

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Q5. Charged capacitor + identical uncharged

Final voltage:

$$V_f = \frac{V}{2}$$

Energy reduces.

✅ Answer: Final energy = $\frac{1}{4}CV^2$, not conserved

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Q6. Energy loss explanation

Energy lost due to:

• Heat in wires (Joule heating)
• EM radiation (minor)

✅ Answer: Lost as heat

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Q7. Area doubled (battery connected)

$$C \propto A \Rightarrow C' = 2C$$

Energy:

$$U = \frac{1}{2}CV^2 \Rightarrow U' = 2U$$

✅ Answer: Energy doubles

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Q8. Slow vs sudden separation

• Slow: reversible → max work extracted
• Sudden: non-reversible → energy loss

✅ Answer: Slow process requires more controlled work, sudden wastes energy

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Q9. Why capacitance independent?

$$C = \frac{\varepsilon_0 A}{d}$$

Depends only on geometry

✅ Answer: Property of system, not charge

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Q10. Series vs Parallel

• Series: same charge (single path)
• Parallel: same voltage (common nodes)

✅ Answer: Due to circuit constraints

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🔢 SECTION B: NUMERICAL SOLUTIONS

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Q11. Plate separation doubled (battery connected)

$$C = \frac{\varepsilon_0 A}{d}, \quad C' = \frac{C}{2}$$

Energy:

$$U_i = \frac{1}{2}CV^2,\quad U_f = \frac{1}{2}\cdot\frac{C}{2}V^2 = \frac{1}{4}CV^2$$

Work done:

$$W = U_f - U_i = -\frac{1}{4}CV^2$$

External work = $+\frac{1}{4}CV^2$

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Q12. Energy sharing

$$V_f = \frac{V}{2}$$ $$U_f = \frac{1}{4}CV^2$$

Loss:

$$\frac{1}{4}CV^2$$

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Q13. One plate earthed

$$V = \frac{Q}{C}$$

Capacitance unchanged

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Q14 Solution (Force Between Plates)

Using energy method:

$$F = \frac{dU}{dd}$$ $$U = \frac{1}{2} \frac{\epsilon_0 A V^2}{d}$$ $$F = \frac{1}{2} \frac{\epsilon_0 A V^2}{d^2}$$

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Q15. Separation increased to 3d (disconnected)

$$C' = \frac{C}{3}$$ $$U' = \frac{Q^2}{2C'} = 3U$$

Work done:

$$W = 2U$$

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Q16. Triangle capacitors

Using symmetry:

$$C_{eq} = \frac{2C}{3}$$

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Q17. Sliding plate

$$C \propto A \Rightarrow C' = \frac{C}{2}$$

Energy (battery connected):

$$U' = \frac{1}{2}C'V^2 = \frac{1}{2}U$$

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Q18 Solution (Energy Supplied vs Stored)

Battery supplies:

$$W = \int V dq = CV^2$$

Stored energy:

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

👉 Half energy lost in circuit

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Q19. Series capacitor modification

• Initially equal charge
• After separation change → capacitance decreases
• Voltage redistributes

Final result:

• Larger voltage across modified capacitor

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Q20. Conducting slab inserted

$$C' = \frac{\varepsilon_0 A}{d - t}$$

Energy:

$$U' = \frac{Q^2}{2C'} \Rightarrow decreases$$

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🚀 Summary Insight (Important for JEE Advanced)

• Battery connected → Voltage constant
• Disconnected → Charge constant
• Energy loss → always heat
• Insert conductor → reduces effective distance
• Capacitance depends only on geometry