PhysicsCBSE Class 12 · Sample paper, session 2024-25 · 3 marks · Short answer

CBSE Class 12 Physics sample paper 2024-25: question 23

From the official CBSE sample question paper for 2024-25, section C, question 23, for 3 marks. Try it first; the hints and our step-by-step solution are below.

Q23 · 3 marks · Short answerElectrostatic Potential and Capacitance

Find the expression for the capacitance of a parallel plate capacitor of plate area A and plate separation d when (I) a dielectric slab of thickness t and (II) a metallic slab of thickness t, where (t < d) are introduced one by one between the plates of the capacitor. In which case would the capacitance be more and why?

Source: official sample paper from cbseacademic.nic.in (PDF). See it in the full Sample paper 2024-25.

Hints

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

Hint 1

Add up the potential drops across the air gaps and across the slab.

Hint 2

Inside a dielectric the field is reduced by a factor K; inside a conductor it is zero.

Hint 3

Compare the two denominators.

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 the plates carry charges $\pm q$ , with surface density $σ = q / A$ . In the air gaps the field is $E_{0} = \frac{σ}{ε _{0}} = \frac{q}{ε _{0} A}$ . Inside the slab, polarisation reduces the field to $E = \frac{E _{0}}{K}$ .
½ mark
Potential difference between the plates, with air thickness $(d - t)$ and slab thickness $t$ : $V = E_{0} (d - t) + \frac{E _{0}}{K} t = \frac{q}{ε _{0} A} (d - t + \frac{t}{K})$
½ mark
$C = \frac{q}{V} = \frac{ε _{0} A}{d - t + \frac{t}{K}}$
½ mark
(II) Metallic slab: the field inside a conductor is zero, so $V = E_{0} (d - t) = \frac{q}{ε _{0} A} (d - t)$ and $C = \frac{ε _{0} A}{d - t}$ . This is the same as putting $K \to \infty$ in (I).
½ mark
The capacitance is greater with the metallic slab. For the dielectric, the term $t / K$ is still present in the denominator. For the metal it is zero, so the denominator $(d - t)$ is smaller and C is larger.
1 mark

Answer: (I) $C = \frac{ε _{0} A}{d - t + t / K}$ ; (II) $C = \frac{ε _{0} A}{d - t}$ ; capacitance is more with the metallic slab (field inside metal is zero, $K \to \infty$ )

Where the marks are

Fields: $E_{0} = q / ε_{0} A$ in air, $E_{0} / K$ in the slab (½ mark)
Potential difference $V = E_{0} (d - t) + E_{0} t / K$ (½ mark)
Expression for C with the dielectric slab (½ mark)
Expression for C with the metallic slab (½ mark)
Concludes the metallic slab gives more capacitance, with reason (1 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 2024-25: every question, by section.

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CBSE Class 12 Physics sample paper 2024-25: question 23

Hints

Step-by-step solution

Revise the chapter

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