A parallel plate capacitor is filled equally (half) with two dielectrics of dielectric constant varepsilon_1$\varepsilon_1$ and varepsilon_2$\varepsilon_2$, as shown in figures. The distance between the plates is d$d$ and area of each plate is A$A$. If capacitance in first configuration and second configuration are C_1$C_1$ and C_2$C_2$ respectively, then fracC_1C_2$\frac{C_1}{C_2}$ is:
Illustrates two parallel plate configurations: stacked horizontally (series) and stacked vertically (parallel).Illustrates two parallel plate configurations: stacked horizontally (series) and stacked vertically (parallel).
Keywords:#dielectric parallel plate capacitor#capacitance series and parallel#JEE Main 2025 Morning Q8#harmonic arithmetic mean capacitance#dielectric capacitor#series parallel capacitor#dielectric constant ratio
More Electrostatics Previous-Year Questions — Page 4
Q182025Electric Field Intensity
Two infinite identical charged sheets and a charged spherical body of charge density '
ho$
ho$' are arranged as shown in figure. Then the correct relation between the electrical fields at A, B, C and D points is :
The figure illustrates two parallel infinite charged plates with a uniform charge density along with an embedded solid spherical charge mass distributed near reference measurement tags labeled A, B, C, and D.
### Related Formula
Superposition of electric field tracks:
vecE*textnet = vecE*textsheets + vecE*textsphere$\vec{E}*{\text{net}} = \vec{E}*{\text{sheets}} + \vec{E}*{\text{sphere}}$
### Core Logic
Evaluate local positional tracking parameters:
* The fields due to the infinite plates add or subtract symmetrically across regions.
* The central sphere introduces a radially varying vector component (E propto frac1r^2$E \propto \frac{1}{r^2}$ outside or E propto r$E \propto r$ inside) whose structural mapping changes direction between symmetric tracking tags.
* At points C$C$ and D$D$, the directional orientation components of the spherical vector directly contrast each other, meaning vecE_C
e vecE_D$\vec{E}_C
e \vec{E}_D$.
* Sifting structural balances reveals field amplification profiles near point A$A$ exceeding local parameters at B$B$ due to positive alignment additions, so vecE_A > vecE_B$\vec{E}_A > \vec{E}_B$.
### Pattern Recognition
Vector fields demand both magnitude and coordinate direction vector compliance. Symmetries can ensure equal scalar values while completely breaking vector equivalence.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Electrostatics
Q192025Torque on an Electric Dipole
Two small spherical balls of mass 10g each with charges -2mumathrmC$-2mumathrm{C}$ and 2mumathrmC$2mumathrm{C}$, are attached to two ends of very light rigid rod of length 20 cm. The arrangement is now placed near an infinite non-conducting charge sheet with uniform charge density of 100mumathrmC/m^2$100mumathrm{C/m}^{2}$ such that length of rod makes an angle of 30^circ$30^{\circ}$ with electric field generated by charge sheet. Net torque acting on the rod is:
(Take epsilon*o = 8.85times10^-12mathrmC^2/mathrmNm^2$\epsilon*{o} = 8.85\times10^{-12}\mathrm{C}^{2}/\mathrm{Nm}^{2}$)
A. 112 Nm
B. 1.12 Nm
C. 2.24 Nm
D. 11.2 Nm
Solution
### Related Formula
Electric field due to an infinite non-conducting sheet:
E = fracsigma2epsilon_0$E = \frac{\sigma}{2\epsilon_0}$
Torque on an electric dipole configuration:
tau = p E sintheta$\tau = p E \sin\theta$
where p = q cdot d$p = q \cdot d$ (dipole moment).
### Core Logic
Given parameters:
* Charge, q = 2 times 10^-6mathrm~C$q = 2 \times 10^{-6}\mathrm{~C}$
* Separation length, d = 20mathrm~cm = 0.2mathrm~m$d = 20\mathrm{~cm} = 0.2\mathrm{~m}$
* Charge density, sigma = 100 times 10^-6mathrm~C/m^2$\sigma = 100 \times 10^{-6}\mathrm{~C/m}^2$
* Orientation angle, theta = 30^circ$\theta = 30^{\circ}$
### Step 1: Compute Field and Torque Value
First, evaluate the field matrix strength value:
E = frac100 times 10^-62 times 8.85 times 10^-12$E = \frac{100 \times 10^{-6}}{2 \times 8.85 \times 10^{-12}}$
Now insert everything into the torque expression:
tau = (q cdot d) cdot E cdot sin(30^circ)$\tau = (q \cdot d) \cdot E \cdot \sin(30^{\circ})$tau = left[ (2 times 10^-6) cdot (0.2)
ight] cdot left[ frac100 times 10^-62 times 8.85 times 10^-12
ight] cdot left( frac12
ight)$\tau = \left[ (2 \times 10^{-6}) \cdot (0.2)
ight] \cdot \left[ \frac{100 \times 10^{-6}}{2 \times 8.85 \times 10^{-12}}
ight] \cdot \left( \frac{1}{2}
ight)$tau = frac108.85 approx 1.12mathrm~Nm$\tau = \frac{10}{8.85} \approx 1.12\mathrm{~Nm}$Dipole torque vector field distribution alignment for Q19 - JEE Main 2025 Morning
### Pattern Recognition
Equal and opposite charges separated by a fixed distance form an electric dipole. The torque in a uniform electric field depends exclusively on the dipole moment value, field strength, and the sine of the orientation angle.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Electrostatics
Q242025Combination of Capacitors
Four capacitor each of capacitance 16mumathrmF$16mumathrm{F}$ are connected as shown in the figure. The capacitance between points A and B is: (in mumathrmF$mumathrm{F}$).
The figure details an interconnected array layout composed of four identical storage components branching outwards between primary measurement terminals A and B.
Numerical Answer.Answer: 64 to 64
Solution
### Related Formula
Equivalent capacitance for a parallel configuration network:
C_texteq = C_1 + C_2 + C_3 + dots$C_{\text{eq}} = C_1 + C_2 + C_3 + dots$
### Core Logic
By tracing node connectivity potentials carefully throughout the short-circuit wire paths, we can label the plates of all four capacitors.
The figure details an interconnected array layout composed of four identical storage components branching outwards between primary measurement terminals A and B.
Redrawing the circuit connection matrix map reveals that all 4 capacitors are connected directly in parallel between node terminal A and node terminal B.
### Step 1: Calculate Equivalent Value
Since they are in parallel:
C_texteq = 4 cdot C$C_{\text{eq}} = 4 \cdot C$
Given each individual element carries C = 16mumathrmF$C = 16mumathrm{F}$:
C_texteq = 4 times 16 = 64mumathrmF$C_{\text{eq}} = 4 \times 16 = 64mumathrm{F}$The figure details an interconnected array layout composed of four identical storage components branching outwards between primary measurement terminals A and B.
### Pattern Recognition
Shorting loops across alternate terminals routinely unravels interlocking rows back into straightforward parallel grids. Always trace and label nodes first.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Electrostatics
Q12025Electrostatic Shielding
Given below are two statements: one is labelled as Assertion (A) and the other is labelled as Reason (R).
Assertion (A): The outer body of an air craft is made of metal which protects persons sitting inside from lightning-strikes. [cite: 12]
Reason (R): The electric field inside the cavity enclosed by a conductor is zero. [cite: 13]
In the light of the above statements, chose the most appropriate answer from the options given below: [cite: 14]
A. Both (A) and (R) are correct and (R) is the correct explanation of (A) [cite: 15]
B. (A) is correct but (R) is not correct [cite: 16]
C. Both (A) and (R) are correct but (R) is not correct explanation of (A) [cite: 17]
D. (A) is not correct but (R) is correct [cite: 18]
Solution
### Core Logic
According to electrostatic shielding, the electric field inside a cavity of a conductor is always zero, regardless of the size and shape of the cavity and regardless of any charges located outside or on the conductor's surface[cite: 660]. Therefore, when lightning strikes a metal aircraft, the entire charge stays on the outer metallic surface and flows down without producing an electric field inside, keeping passengers safe[cite: 12].
### Step 1: Statement Evaluation
* **Assertion (A):** Correct, passengers are protected from lightning because of the metallic body shield [cite: 12].
* **Reason (R):** Correct, the field inside a cavity of a conductor is zero [cite: 13].
* **Explanation:** Since the zero field property is precisely why passengers are protected, (R) correctly explains (A)[cite: 15].
### Pattern Recognition
Metallic shell / shield configuration always establishes E_textinside = 0$E_{\text{inside}} = 0$[cite: 660]. This shielding mechanism directly underpins safety features in lightning scenarios for cars and airplanes.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Electrostatics
Q102025Torque on a Dipole
A dipole with two electric charges of 2 µC magnitude each, with separation distance 0.5 µm, is placed between the plates of a capacitor such that its axis is parallel to an electric field established between the plates when a potential difference of 5 V is applied.
Separation between the plates is 0.5 mm. If the dipole is rotated by 30^circ$30^{circ}$ from the axis, it tends to realign in the direction due to a torque.
The value of torque is : [cite: 58, 59, 60]
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