Questions are transcribed from NTA's official question paper. Diagrams are described in words; open the official PDF to see them. Every answer was solved twice independently and is shown only when both agree. Where they did not, we say so.
A parallel plate air capacitor has a capacitance C. When it is half filled as show in figure with a dielectric constant $K=5$, the percentage increase in the capacitance is _______. [Figure: A parallel plate capacitor with total separation $d$ between plates. The top half (thickness $d/2$) is filled with dielectric material $K=5$, and the bottom half (thickness $d/2$) is air.]
(1) 33.34
(2) 66.67
(3) 200
(4) 400
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Answer: Option (2)
Key idea: Capacitors in series with dielectric; Capacitance of parallel plate capacitor
Original capacitance: $C = \frac{\varepsilon_0 A}{d}$
With half-filled dielectric, treat as two capacitors in series: Air gap ($d/2$) and dielectric gap ($d/2$)
Two charged conducting spheres $S_1$ and $S_2$ of radii 8 cm and 18 cm are connected to each other by a wire. After equilibrium is established, the ratio of electric fields on $S_1$ and $S_2$ spheres are $E_{S_1}$ and $E_{S_2}$ respectively. The value of $\frac{E_{S_1}}{E_{S_2}}$ is _______.
(1) $\frac{3}{2}$
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Answer under review. Our two independent solutions did not agree, so we are not showing an answer yet. Check it against NTA's final answer key.
Key idea: Electric potential and field for conducting spheres in equilibrium
A parallel plate air capacitor is connected to a battery. The plates are pulled apart at uniform speed $v$. If $x$ is the separation between the plates at any instant, then the time rate of change of electrostatic energy of the capacitor is proportional to $x^\alpha$, where $\alpha$ is ______.
(1) $-2$
(2) $1$
(3) $-1$
(4) $2$
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Answer: Option (1)
Key idea: Energy stored in a capacitor connected to a battery; Rate of change of energy with varying plate separation
Write capacitance of parallel plate capacitor: $C = \frac{\varepsilon_0 A}{x}$
Since capacitor is connected to battery (constant voltage $V$), energy is: $U = \frac{1}{2}CV^2 = \frac{\varepsilon_0 A V^2}{2x}$
Find time rate of change: $\frac{dU}{dt} = \frac{\varepsilon_0 A V^2}{2} \cdot \frac{d}{dt}\left(\frac{1}{x}\right)$
From the circuit given below, the capacitance between terminals $A$ and $B$ shown in the circuit is __________ $\mu$F.
(take $C_1 = C_2 = C_3 = 1$ $\mu$F and $C_4 = 2$ $\mu$F.)
[Figure: Circuit diagram showing capacitors $C_1$, $C_2$, $C_3$ connected in series between terminals $A$ and $B$, with $C_4$ connected in parallel below them]
(1) 2
(2) 7/2
(3) 7/3
(4) 5/2
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Answer: Option (3)
Key idea: Series and Parallel Combination of Capacitors
Identify that $C_1$, $C_2$, and $C_3$ are connected in series between terminals $A$ and $B$
For series capacitors: $\frac{1}{C_{series}} = \frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3} = \frac{1}{1} + \frac{1}{1} + \frac{1}{1} = 3$
Therefore, $C_{series} = \frac{1}{3}$ μF
Capacitor $C_4$ is connected in parallel with the series combination
Given below are two statements: one is labelled as **Assertion A** and the other is labelled as **Reason R**
**Assertion A:** In electrostatics, a conductor does not store any net charge inside.
**Reason R:** Inside the capacitor (with no dielectric medium), the free charge carriers, if placed between the plates of capacitor, experience force and drift.
Choose the *correct* answer from the options given below
(1) Both A and R are true and R is the correct explanation of A
(2) Both A and R are true but R is NOT the correct explanation of A
(3) A is true but R is false
(4) A is false but R is true
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Answer: Option (2)
Key idea: Electrostatic properties of conductors and electric field in capacitors
**Analyze Assertion A**: In electrostatics, conductors have zero electric field inside, so all excess charge resides on the surface. Statement A is **TRUE**.
**Analyze Reason R**: Between capacitor plates, electric field $E = \sigma/\epsilon_0$ exists. Free charges placed here experience force $F = qE$ and drift. Statement R is **TRUE**.
**Check if R explains A**: Reason R describes charges between capacitor plates (empty space), while A describes charge distribution in conductors.
R does not explain why conductors store charge only on surface. The correct explanation involves internal field being zero and charge redistribution.
**Conclusion**: Both statements are true, but R is NOT the correct explanation of A.
A displacement current of $4.0$ A can be set up in the space between two parallel plates of $6$ µF capacitor. The rate of change of potential difference across the plates of the capacitor is nearly $a \times 10^5$ V/s. The value of $a$ is ______.
(1) 0.58
(2) 0.67
(3) 0.82
(4) 0.75
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Answer: Option (2)
Key idea: Displacement Current and Capacitor Charging
Write the relation between displacement current and capacitor: $I_d = C \frac{dV}{dt}$
Substitute given values: $4.0 = 6 \times 10^{-6} \times \frac{dV}{dt}$
Solve for rate of change: $\frac{dV}{dt} = \frac{4.0}{6 \times 10^{-6}} = \frac{2}{3} \times 10^{6}$ V/s
Express in scientific notation: $\frac{dV}{dt} = 0.67 \times 10^{6}$ V/s
Comparing with $a \times 10^{5}$ V/s (assuming typo, should be $10^6$): $a = 0.67$
A sphere of capacitance 100 pF is charged to a potential of 100 V. Another identical uncharged metal sphere is brought in contact with the charged sphere, then the change in the total energy stored on these spheres, when they touch is $\alpha \times 10^{-7}$ J. The value of $\alpha$ is __________. (combined capacitance of spheres is 200 pF)
(1) 5
(2) $\frac{5}{2}$
(3) $\frac{7}{2}$
(4) $\frac{9}{2}$
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Answer: Option (2)
Key idea: Energy loss in charge redistribution between capacitors
A parallel plate capacitor is having separation between plates 0.885 mm. It has a capacitance of 1 $\mu$F when the space between the plates is filled with an insulating material of resistivity $1\times 10^{13}$ $\Omega$m and resistance $17.7\times 10^{14}$ $\Omega$. Relative permittivity of the insulating material is $a\times 10^7$. The value of $a$ is __________.
(Take permittivity of free space $= 8.85\times 10^{-12}$ F/m)
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Answer: 2
Key idea: Parallel Plate Capacitor with Dielectric - Relating Capacitance, Resistance, and Material Properties
Write resistance formula: $R = \frac{\rho d}{A}$ where $A$ is the plate area
