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✓ Board-verified
Physics · 2023 · 4 marks
CBSE 2023 · Region 5 · Set 2 · Q34
A capacitor is a system of two conductors separated by an insulator. The two conductors have equal and opposite charges with a potential difference…
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 $\displaystyle 2$ C . Find an expression for (a) the potential difference across the combination and (b) the charge lost by capacitor A.
OR
(iii)
Two slabs of dielectric constants $\displaystyle 2$ K 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.
Marking-scheme solution
(i)
$\displaystyle C_{net} = C + C$
$\displaystyle = 2C$
(ii)
Within the dielectric slab, induced electric field due to polarization, decreases the electric field.
Alternatively: $\displaystyle E = E_0 - E_p$
Alternatively: $\displaystyle E = \dfrac{E_0}{K}$
(iii)(a)
$\displaystyle V' = \dfrac{Q_{Total}}{C_{eqi}} = \dfrac{Q}{3C}$
$\displaystyle V' = \dfrac{Q}{3C} = \dfrac{V}{3}$
(b)
$\displaystyle Q_A^{'} = C \times \dfrac{V}{3} = \dfrac{Q}{3}$
$\displaystyle Q_A = CV = Q$
Charge lost by capacitor A is
$\displaystyle \Delta Q = Q - \dfrac{Q}{3} = \dfrac{2Q}{3}$
ORCapacitance of left portion, $\displaystyle C_1 = \dfrac{6K \varepsilon_0 A}{d}$
Capacitance of right portion, $\displaystyle C_2 = \dfrac{3K \varepsilon_0 A}{2d}$
As the capacitors are in series
$\displaystyle \dfrac{1}{C_{eqi}} = \dfrac{1}{C_1} + \dfrac{1}{C_2}$
$\displaystyle \dfrac{1}{C_{eqi}} = \dfrac{d}{6K \varepsilon_O A} + \dfrac{2d}{3K \varepsilon_O A} = \dfrac{5d}{6KA \varepsilon_O}$
$\displaystyle C_{eqi} = \dfrac{6KA \varepsilon_O}{5d}$
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CBSE Class 12 Physics past-paper question from the 2023 board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.