A capacitor is a system of two conductors separated by an insulator. The two conductors have equal and opposite charges with a potential difference between them. The capacitance of a capacitor depends on the geometrical configuration (shape, size and separation) of the system and also on the nature of the insulator separating the two conductors. They are used to store charges. Like resistors, capacitors can be arranged in series or parallel or a combination of both to obtain desired value of capacitance. (i) Find the equivalent capacitance between points A and B in the given diagram. (ii) A dielectric slab is inserted between the plates of a parallel plate capacitor. The electric field between the plates decreases. Explain. (iii) A capacitor A of capacitance C, having charge Q is connected across another uncharged capacitor B of capacitance 2C. Find an expression for (a) the potential difference across the combination and (b) the charge lost by capacitor A. (iii) Two slabs of dielectric constants 2K and K fill the space between the plates of a parallel plate capacitor of plate area A and plate separation d as shown in figure. Find an expression for capacitance of the system.

PhysicsCBSE Class 12 · Board exam 2023, set 55/5/1 · 4 marks · Case-based

CBSE Class 12 Physics 2023: question 35

Set in the CBSE Class 12 board exam of 2023, paper 55/5/1, section E, question 35, for 4 marks. Try it first; the hints and our step-by-step solution are below.

Q35 · 4 marks · Case-basedCompetency-basedElectrostatic Potential and Capacitance

A capacitor is a system of two conductors separated by an insulator. The two conductors have equal and opposite charges with a potential difference between them. The capacitance of a capacitor depends on the geometrical configuration (shape, size and separation) of the system and also on the nature of the insulator separating the two conductors. They are used to store charges. Like resistors, capacitors can be arranged in series or parallel or a combination of both to obtain desired value of capacitance.

(i)
Find the equivalent capacitance between points A and B in the given diagram.
0 marks
(ii)
A dielectric slab is inserted between the plates of a parallel plate capacitor. The electric field between the plates decreases. Explain.
0 marks
(iii)
A capacitor A of capacitance C, having charge Q is connected across another uncharged capacitor B of capacitance 2C. Find an expression for (a) the potential difference across the combination and (b) the charge lost by capacitor A.
0 marks
OR
(iii)
Two slabs of dielectric constants 2K and K fill the space between the plates of a parallel plate capacitor of plate area A and plate separation d as shown in figure. Find an expression for capacitance of the system.
0 marks

Figure in the paper: (i) Circuit of five capacitors each C between A and B: a top capacitor C bridging over the first two series capacitors C and C, then a third C in series to B; a capacitor C connected from the junction between the first two capacitors down to the B end; and a bottom capacitor C connecting A-side to B-side. (iii OR) Parallel plate capacitor of plate separation d with two dielectric slabs side by side along the thickness: slab 2K of thickness d/3 and slab K of thickness 2d/3. See it in the official paper.

Source: official question paper from cbse.gov.in (PDF). See it in the full 2023 board paper, set 55/5/1.

Hints

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

Hint 1

In (i), look for capacitors whose ends are joined to the same points; they can be combined or ignored.

Hint 2

When a charged capacitor is joined to an uncharged one, total charge is conserved and both end up at the same potential.

Hint 3

Dielectric slabs stacked along the plate separation behave like capacitors in series.

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) Trace the connections in the figure. The bridging and cross capacitors reduce the network to two effective branches between A and B, each of capacitance $C$ , connected in parallel. So $C_{A B} = C + C = 2 C$ .
1 mark
(ii) The external field $E_{0}$ polarises the dielectric slab, so bound charges appear on its two faces. These induced charges set up a field $E_{p}$ inside the slab that opposes $E_{0}$ . The net field is therefore $E = E_{0} - E_{p} = \frac{E _{0}}{K}$ , which is smaller than $E_{0}$ .
1 mark
(iii)(a) Potential difference. Total charge is conserved and the two capacitors are in parallel, so $C_{e q} = C + 2 C = 3 C$ . Then $V = \frac{Q _{t o t a l}}{C _{e q}} = \frac{Q}{3 C}$ .
1 mark
(iii)(b) Charge lost by A. The charge left on A is $Q_{A}^{'} = C V = C \cdot \frac{Q}{3 C} = \frac{Q}{3}$ . Charge lost by A $= Q - \frac{Q}{3} = \frac{2 Q}{3}$ , and this charge flows to B.
1 mark
(iii) OR alternative (2 marks if this choice is attempted). The two slabs lie one after the other between the plates, so they act as two capacitors in series. For the $2 K$ slab of thickness $d /3$ : $C_{1} = \frac{2 K ε _{0} A}{d /3} = \frac{6 K ε _{0} A}{d}$ . For the $K$ slab of thickness $2 d /3$ : $C_{2} = \frac{K ε _{0} A}{2 d /3} = \frac{3 K ε _{0} A}{2 d}$ .
Combining in series: $\frac{1}{C} = \frac{d}{6 K ε _{0} A} + \frac{2 d}{3 K ε _{0} A} = \frac{5 d}{6 K ε _{0} A}$ . Hence $C = \frac{6 K ε _{0} A}{5 d}$ .

Answer: (i) 2C; (ii) the induced polarisation field opposes the applied field, so $E = E_{0} / K$ ; (iii)(a) $V = Q /3 C$ , (b) charge lost = $2 Q /3$ [OR $C = \frac{6 K ε _{0} A}{5 d}$ ]

Where the marks are

(i) Equivalent capacitance 2C (1 mark)
(ii) Explains that the induced field due to polarisation opposes the applied field, giving $E = E_{0} - E_{p} = E_{0} / K$ (1 mark)
(iii)(a) Uses charge conservation to get $V = Q /3 C$ (1 mark)
(iii)(b) Finds the charge on A as $Q /3$ and the charge lost as $2 Q /3$ (1 mark)
(iii) OR: Writes $C_{1} = 6 K ε_{0} A / d$ and $C_{2} = 3 K ε_{0} A /2 d$ (1 mark), then combines them in series to get $6 K ε_{0} A /5 d$ (1 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 paper2023 board paper, set 55/5/1: 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 2023: question 35

Hints

Step-by-step solution

Revise the chapter

More questions from Electrostatic Potential and Capacitance