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 6
Q212025Gauss's Law and Electric Flux
A square loop of sides a = 1$a = 1$ m is held normally in front of a point charge q = 1C$q = 1C$ The flux of the electric field through the shaded region is frac5p times frac1varepsilon_0 fracNm^2C$\frac{5}{p} \times \frac{1}{\varepsilon_0} \frac{Nm^2}{C}$ , where the value of p is . The diagram illustrates a square loop divided into 8 symmetric parts with a point charge positioned in front of its center.
Numerical Answer.Answer: 48 to 48
Solution
### Related Formula
By Gauss's Law, the total flux emitted by a point charge q$q$ through a completely enclosing symmetric cube container surface is:
Phitexttotal = fracqepsilon0 $\Phi{\text{total}} = \frac{q}{\epsilon{0}} $
### Core Logic
Assuming the charge resides at a symmetric center distance fraca2$\frac{a}{2}$ relative to the loop face [cite: 784, 786], this square loop represents one of the six identical faces of an enclosing cube system. Thus, the flux passing through the entire square loop face is[cite: 775, 776]:
Phitextsquare = frac16 Phitexttotal = fracq6epsilon0$\Phi{\text{square}} = \frac{1}{6} \Phi{\text{total}} = \frac{q}{6\epsilon{0}}$
### Step 1: Symmetric Partitioning
As shown in the solution schematic The diagram illustrates a square loop divided into 8 symmetric parts with a point charge positioned in front of its center., the square face can be divided into 8 identical symmetric right-angled triangle sections by drawing its diagonals and medians[cite: 771, 777]. Each individual part intercepts an equal portion of the flux field :
Phitextpart = frac18 Phitextsquare = frac18 left(fracq6epsilon_0
ight) = fracq48epsilon_0$\Phi{\text{part}} = \frac{1}{8} \Phi{\text{square}} = \frac{1}{8} \left(\frac{q}{6\epsilon_{0}}
ight) = \frac{q}{48\epsilon_{0}}$
The shaded region covers exactly 5 of these individual triangle parts [cite: 779, 780]:
Phi_textshaded = 5 times Phi_textpart = frac548 times fracqepsilon_0$\Phi_{\text{shaded}} = 5 \times \Phi_{\text{part}} = \frac{5}{48} \times \frac{q}{\epsilon_{0}}$
Comparing this result with the given expression frac5p times frac1epsilon_0$\frac{5}{p} \times \frac{1}{\epsilon_{0}}$ , we find:
p = 48$p = 48$
### Pattern Recognition
Exploit geometric symmetry to break solid angles down into equal fractions, avoiding complex surface integration.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Electrostatics
Q22025Energy Density in Capacitors
A parallel plate capacitor of capacitance 1mu mathrmF$1\mu \mathrm{F}$ is charged to a potential difference of 20mathrmV$20\mathrm{V}$. The distance between plates is 1mu mathrmm$1\mu \mathrm{m}$. The energy density between plates of capacitor is:
A.1.8 times 10^3 \, mathrmJ/m^3$1.8 \times 10^{3} \, \mathrm{J/m}^{3}$
B.2 times 10^-4 \, mathrmJ/m^3$2 \times 10^{-4} \, \mathrm{J/m^3}$
### Related Formula
The electric field E$E$ between the plates of a parallel plate capacitor is given by :
E = fracVd$E = \frac{V}{d}$
The electrostatic energy density U$U$ in a medium is given by:
U = frac12 epsilon_0 E^2$U = \frac{1}{2} \epsilon_0 E^2$
### Core Logic
Given values [cite: 662, 664, 665]:
* Capacitance, C = 1 \ mutextF$C = 1 \ \mu\text{F}$
* Potential difference, V = 20 text V$V = 20 \text{ V}$
* Plate separation, d = 1 \ mutextm = 10^-6 text m$d = 1 \ \mu\text{m} = 10^{-6} \text{ m}$
First, calculate the electric field E$E$ :
E = frac2010^-6 = 20 times 10^6 text V/m$E = \frac{20}{10^{-6}} = 20 \times 10^6 \text{ V/m}$
Now, calculate the energy density U$U$ using epsilon_0 approx 8.85 times 10^-12 text F/m$\epsilon_0 \approx 8.85 \times 10^{-12} \text{ F/m}$ :
U = frac12 times (8.85 times 10^-12) times (20 times 10^6)^2$U = \frac{1}{2} \times (8.85 \times 10^{-12}) \times (20 \times 10^6)^2$U = frac12 times 8.85 times 10^-12 times 400 times 10^12$U = \frac{1}{2} \times 8.85 \times 10^{-12} \times 400 \times 10^{12}$U = 8.85 times 200 = 1770 text J/m^3 = 1.77 times 10^3 text J/m^3$U = 8.85 \times 200 = 1770 \text{ J/m}^3 = 1.77 \times 10^3 \text{ J/m}^3$
Rounding to the nearest matching option gives 1.8 times 10^3 text J/m^3$1.8 \times 10^3 \text{ J/m}^3$.
### Pattern Recognition
Notice that energy density is completely independent of the total capacitance value C$C$ if the voltage and spacing are directly provided. Always check for redundant parameters included to distract candidates.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Electrostatics
Q222025Electric Dipole
An electric dipole of dipole moment 6 times 10^-6 mathrmCm$6 \times 10^{-6} \mathrm{Cm}$ is placed in uniform electric field of magnitude 10^6 mathrm~V/m$10^{6} \mathrm{~V/m}$. Initially, the dipole moment is \parallel to electric field. The work that needs to be done on the dipole to make its dipole moment opposite to the field, will be ______ J.
Numerical Answer.Answer: 12
Solution
### Related Formula
The potential energy of an electric dipole aligned at an \angle theta$\theta$ inside a uniform electric field is given by:
U = -p E cos theta$U = -p E \cos \theta$
The work done by an external agent to rotate the dipole equals its change in potential energy:
W = Delta U = -pE (cos theta_f - cos theta_i)$W = \Delta U = -pE (\cos \theta_f - \cos \theta_i)$
### Core Logic
Given parameters [cite: 805, 806]:
* Dipole moment, p = 6 times 10^-6 text Cm$p = 6 \times 10^{-6} \text{ Cm}$
* Electric field, E = 10^6 text V/m$E = 10^6 \text{ V/m}$
* Initial \angle (\parallel state), theta_i = 0^circ$\theta_i = 0^{\circ}$
* Final \angle (opposite state), theta_f = 180^circ$\theta_f = 180^{\circ}$
Substitute the values into the work equation [cite: 807, 808]:
W = -pE (cos 180^circ - cos 0^circ)$W = -pE (\cos 180^{\circ} - \cos 0^{\circ})$W = -pE (-1 - 1) = 2 p E quad text$W = -pE (-1 - 1) = 2 p E \quad \text{}$W = 2 times (6 times 10^-6) times 10^6 = 12 text J quad text$W = 2 \times (6 \times 10^{-6}) \times 10^6 = 12 \text{ J} \quad \text{}$
### Pattern Recognition
Rotating a dipole from its most stable configuration (\parallel, theta=0^circ$\theta=0^{\circ}$) to its most unstable position (anti-\parallel, theta=180^circ$\theta=180^{\circ}$) always requires a maximum work value of exactly 2pE$2pE$.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Electrostatics
Q32025Electric Dipole
An electric dipole of mass m, charge q, and length l$l$ is placed in a uniform electric field vecmathrmE = mathrmE_0hatmathrmi$\vec{\mathrm{E}} = \mathrm{E}_0\hat{\mathrm{i}}$ . When the dipole is rotated slightly from its equilibrium position and released, the time period of its oscillations will be: [cite: 1, 5]
### Related Formula
tau = -pE sin theta$\tau = -pE \sin \theta$I = 2 m left(fracl2right)^2 = fracml^22$I = 2 m \left(\frac{l}{2}\right)^2 = \frac{ml^2}{2}$
### Core Logic
Restoring torque for small angle theta$\theta$ is given by:
tau = -q l E_0 theta$\tau = -q l E_0 \theta$
Equating with rotational inertia dynamics:
I omega^2 theta = q l E_0 theta$I \omega^2 \theta = q l E_0 \theta$fracm l^22 omega^2 = q l E_0 implies omega^2 = frac2 q E_0m l$\frac{m l^2}{2} \omega^2 = q l E_0 \implies \omega^2 = \frac{2 q E_0}{m l}$ [cite: 618, 620]
### Step 1: Compute Time Period
T = frac2piomega = 2pi sqrtfracml2qE_0$T = \frac{2\pi}{\omega} = 2\pi \sqrt{\frac{ml}{2qE_0}}$
### Pattern Recognition
For a two-particle system pivoting about midpoint, total moment of inertia drops to ml^2/2$ml^2/2$, scaling the period by a factor of sqrt2$\sqrt{2}$[cite: 618, 622].
### Chapter Mix
Class 12 Physics: Electrostatics
Q182025Gauss\'s Law
Match List-I with List-II.
List-I
List-II
(A) Electric field inside (distance r > 0 from center) of a uniformly charged spherical shell with surface charge density σ, and radius R.
(I) sigma / epsilon_0$\sigma / \epsilon_0$
(B) Electric field at distance r > 0 from a uniformly charged infinite plane sheet with surface charge density σ.
(II) sigma / 2epsilon_0$\sigma / 2\epsilon_0$
(C) Electric field outside (distance r > 0 from center) of a uniformly charged spherical shell with surface charge density σ, and radius R
(III) 0
(D) Electric field between 2 oppositely charged infinite plane parallel sheets with uniform surface charge density σ.
We Map Every Repeating Question in Competitive Exams.
Say goodbye to generic mock test fatigue. RankBit uses smart analysis to group past exam questions into their foundational Repeating Question Types. Find chapter weightage, track repeating questions, and score higher with targeted practice.
Select Your Target Exam
Choose an exam track below to find formulas per chapter and patterns.
Syncing Exam Intelligence
Mapping formulas and patterns across all tracks…
PATH A — FULL LENGTH PRACTICE
Full Mock Test Hub
Simulate real NTA exam conditions with fully tracked mocks. Time yourself against past papers.
Under Development
PATH B — TARGETED PRACTICE
Topic-wise Practice Hub
Practice past-year questions one chapter at a time. Pick an exam → subject → chapter and get every PYQ for that topic — pulled together from all past papers — with the chapter's key formulas alongside.
