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

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

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

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

i. Derive an expression for the capacitance of a parallel plate capacitor with air present between the two plates.
ii. Obtain the equivalent capacitance of the network shown in figure. For a 300 V supply, determine the charge on each capacitor.

Figure in the paper: Network across a 300 V supply: C1 = 100 pF in the top branch; C2 = 200 pF and C3 = 200 pF in series in the middle branch; C4 = 100 pF in the bottom branch; the three branches are in parallel across the 300 V source. 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

Between the plates, the fields of the two charged sheets add; find V from E and d.

Hint 2

Reduce the network step by step: series pair first, then parallel, then series with $C_{4}$ .

Hint 3

The total charge flows through $C_{4}$ ; use its voltage to find the voltage left for the rest.

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) Two parallel plates, each of area A, separated by distance d, carry charges +Q and −Q, with surface charge density $σ = Q / A$ . Each sheet produces field $\frac{σ}{2 ε _{0}}$ . Between the plates the two fields are in the same direction and add; outside they cancel. So the field between the plates is $E = \frac{σ}{ε _{0}} = \frac{Q}{A ε _{0}}$ .
1 mark
Potential difference: $V = E d = \frac{Q d}{A ε _{0}}$ . Capacitance: $C = \frac{Q}{V} = \frac{ε _{0} A}{d}$ .
1 mark
(ii) $C_{2}$ and $C_{3}$ in series: $\frac{200 \times 200}{400} = 100$ pF. This is in parallel with $C_{1}$ : $100 + 100 = 200$ pF. This combination is in series with $C_{4}$ :
$C_{e q} = \frac{200 \times 100}{300} = \frac{200}{3}$ pF ≈ 66.7 pF
1 mark
Charge drawn from supply = charge on $C_{4}$ : $Q_{4} = \frac{200}{3} \times 1 0^{- 12} \times 300 = 2 \times 1 0^{- 8}$ C. P.d. across $C_{4} = \frac{2 \times 1 0 ^{- 8}}{100 \times 1 0 ^{- 12}} = 200$ V.
½ mark
P.d. across $C_{1}$ = $300 - 200 = 100$ V, so $Q_{1} = 100 \times 1 0^{- 12} \times 100 = 1 \times 1 0^{- 8}$ C.
½ mark
The series pair $C_{2}$ , $C_{3}$ also has 100 V across it, i.e. 50 V across each (equal capacitances).
$Q_{2} = Q_{3} = 200 \times 1 0^{- 12} \times 50 = 1 \times 1 0^{- 8}$ C.
1 mark

Answer: $C = ε_{0} A / d$ ; $C_{e q} = 200/3$ pF; $Q_{1} = 1 0^{- 8}$ C, $Q_{2} = Q_{3} = 1 0^{- 8}$ C, $Q_{4} = 2 \times 1 0^{- 8}$ C

Where the marks are

Field between plates $E = σ / ε_{0} = Q / A ε_{0}$ (1 mark)
$V = E d$ and $C = ε_{0} A / d$ (1 mark)
Equivalent capacitance 200/3 pF (1 mark)
Charge on $C_{4} = 2 \times 1 0^{- 8}$ C (½ mark)
Charge on $C_{1} = 1 \times 1 0^{- 8}$ C (½ mark)
Charges on $C_{2}$ and $C_{3} = 1 \times 1 0^{- 8}$ C each (½ + ½ mark)

Revise the chapter

Chapter: Electrostatic Potential and Capacitance (notes, formulas, where marks are lost and every past question)
Topic: Parallel Plate Capacitor and Dielectrics

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.

The question is transcribed from the official paper; figures are described in words. The solution, hints and chapter tag are ours. Spotted an error? Write tosupport@questionbuddy.in.

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

Hints

Step-by-step solution

Revise the chapter

More questions from Electrostatic Potential and Capacitance