PhysicsCBSE Class 12 · Sample paper, session 2023-24 · 5 marks · Long answer

CBSE Class 12 Physics sample paper 2023-24: question 32(b)

From the official CBSE sample question paper for 2023-24, section E, question 32(b), for 5 marks. Try it first; the hints and our step-by-step solution are below.

Q32(b) · 5 marks · Long answerElectrostatic Potential and Capacitance

i. A dielectric slab of thickness 't' is kept between the plates of a parallel plate capacitor with plate separation 'd' (t < d). Derive the expression for the capacitance of the capacitor.
ii. A capacitor of capacity $C_{1}$ is charged to the potential of $V_{o}$ . On disconnecting with the battery, it is connected with an uncharged capacitor of capacity $C_{2}$ as shown in the adjoining figure. Find the ratio of energies before and after the connection of switch S.

Figure in the paper: Circuit with capacitor C1 (charged to V0) and uncharged capacitor C2 connected in parallel through a switch S in the top wire. See it in the official paper.

Source: official sample paper from cbseacademic.nic.in (PDF). See it in the full Sample paper 2023-24.

Hints

Open one hint at a time and try again before the next.

Hint 1

In (i), the field is not the same everywhere between the plates: treat the air part and the slab part separately.

Hint 2

Add the potential drops across the air gap $(d - t)$ and the slab $t$ , where the field in the slab is reduced by $K$ .

Hint 3

In (ii), total charge $C_{1} V_{0}$ is conserved and shared at a common potential; then compare $\frac{1}{2} C V^{2}$ before and after.

Step-by-step solution

Each step shows the marks it earns, as an examiner would award them. Checked by an independent re-solve from the question paper.

(i) Let each plate have area $A$ and charge density $σ = Q / A$ . In the air gap (total thickness $d - t$ ) the field is $E_{0} = \frac{σ}{ε _{0}} = \frac{Q}{A ε _{0}}$ .
½ mark
Inside the dielectric slab of constant $K$ , polarisation reduces the field to $E = \frac{E _{0}}{K} = \frac{Q}{K A ε _{0}}$ .
1 mark
Potential difference between the plates: $V = E_{0} (d - t) + E t = \frac{Q}{A ε _{0}} (d - t + \frac{t}{K})$ .
1 mark
Capacitance: $C = \frac{Q}{V} = \frac{ε _{0} A}{d - t + \frac{t}{K}}$ . This is greater than $\frac{ε _{0} A}{d}$ because $K > 1$ .
½ mark
(ii) Before S is closed, only $C_{1}$ is charged: $U_{i} = \frac{1}{2} C_{1} V_{0}^{2}$ .
½ mark
Charge is conserved when S is closed. With the capacitors in parallel, the common potential is $V = \frac{C _{1} V _{0}}{C _{1} + C _{2}}$ .
½ mark
Final energy: $U_{f} = \frac{1}{2} (C_{1} + C_{2}) V^{2} = \frac{C _{1}^{2} V _{0}^{2}}{2 ( C _{1} + C _{2} )}$ .
½ mark
Ratio: $\frac{U _{f}}{U _{i}} = \frac{C _{1}}{C _{1} + C _{2}}$ , i.e. $U_{i} : U_{f} = (C_{1} + C_{2}) : C_{1}$ . The final energy is less; the difference is lost as heat and radiation during charge redistribution.
½ mark

Answer: (i) $C = \frac{ε _{0} A}{d - t + t / K}$ ; (ii) $U_{i} : U_{f} = (C_{1} + C_{2}) : C_{1}$ (i.e. $U_{f} / U_{i} = C_{1} / (C_{1} + C_{2})$ )

Where the marks are

Writes the field in the air gap $E_{0} = σ / ε_{0}$ (½ mark)
Writes the field in the dielectric $E_{0} / K$ (1 mark)
Gives $V = E_{0} (d - t) + E_{0} t / K$ (1 mark)
Obtains $C = ε_{0} A / (d - t + t / K)$ (½ mark)
Initial energy $\frac{1}{2} C_{1} V_{0}^{2}$ (½ mark)
Common potential $C_{1} V_{0} / (C_{1} + C_{2})$ (½ mark)
Final energy $C_{1}^{2} V_{0}^{2} /2 (C_{1} + C_{2})$ (½ mark)
Ratio $U_{f} : U_{i} = C_{1} : (C_{1} + C_{2})$ (½ mark)

Revise the chapter

Chapter: Electrostatic Potential and Capacitance (notes, formulas, where marks are lost and every past question)

Practise this chapterBoard-style questions from Electrostatic Potential and Capacitance, one at a time, marked against the answer key and marking points.Back to the full paperSample paper 2023-24: every question, by section.

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CBSE Class 12 Physics sample paper 2023-24: question 32(b)

Hints

Step-by-step solution

Revise the chapter

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