A parallel plate capacitor has capacitance C, when there is vacuum within the parallel plates.
A sheet having thickness left(frac13right)^mathrmrd$\left(\frac{1}{3}\right)^{\mathrm{rd}}$ of the separation between the plates and relative permittivity K is introduced between the plates. The new capacitance of the system is :
### Related Formula
C = fracepsilon_0 Ad - t + fractK$$C = \frac{\epsilon_0 A}{d - t + \frac{t}{K}}$$
Alternatively, treat as two capacitors in series: C_eq = fracC_1 C_2C_1 + C_2$C_{eq} = \frac{C_1 C_2}{C_1 + C_2}$
### Core Logic
Initial capacitance (vacuum): C = fracA epsilon_0d$C = \frac{A \epsilon_{0}}{d}$.
Capacitance with Dielectric diagram for Q34 - JEE Main 2026 Morning
The system can be modelled as two capacitors in series:
1. A vacuum capacitor of thickness d - fracd3 = frac2d3$d - \frac{d}{3} = \frac{2d}{3}$
2. A dielectric capacitor of thickness fracd3$\frac{d}{3}$ and dielectric constant K$K$.
Capacitance with Dielectric diagram for Q34 - JEE Main 2026 MorningCapacitance with Dielectric diagram for Q34 - JEE Main 2026 Morning
### Step 1: Capacitors in Series
The capacitances are:
C_1 = fracepsilon_0 Aleft(frac2d3right) = frac32 left(fracepsilon_0 Adright) = frac32 C$$C_1 = \frac{\epsilon_0 A}{\left(\frac{2d}{3}\right)} = \frac{3}{2} \left(\frac{\epsilon_0 A}{d}\right) = \frac{3}{2} C$$C_2 = fracK epsilon_0 Aleft(fracd3right) = 3K left(fracepsilon_0 Adright) = 3KC$$C_2 = \frac{K \epsilon_0 A}{\left(\frac{d}{3}\right)} = 3K \left(\frac{\epsilon_0 A}{d}\right) = 3KC$$
Now, equivalent capacitance in series:
C_texteq = fracC_1 C_2C_1 + C_2 = fracleft(frac32 Cright) times (3KC)frac32 C + 3KC$$C_{\text{eq}} = \frac{C_1 C_2}{C_1 + C_2} = \frac{\left(\frac{3}{2} C\right) \times (3KC)}{\frac{3}{2} C + 3KC}$$C_texteq = fracfrac92KC^2frac32C(1 + 2K)$$C_{\text{eq}} = \frac{\frac{9}{2}KC^2}{\frac{3}{2}C(1 + 2K)}$$C_texteq = frac3KC2K + 1$$C_{\text{eq}} = \frac{3KC}{2K + 1}$$
### Pattern Recognition
Partial dielectric filling of thickness t$t$: Use formula C_textnew = fracepsilon_0 Ad - t + t/K$C_{\text{new}} = \frac{\epsilon_0 A}{d - t + t/K}$. Plugging t = d/3$t = d/3$ directly gives fracepsilon_0 Ad - d/3 + d/3K = fracepsilon_0 Afrac2d3 + fracd3K = frac3K epsilon_0 A2Kd + d = frac3KC2K+1$\frac{\epsilon_0 A}{d - d/3 + d/3K} = \frac{\epsilon_0 A}{\frac{2d}{3} + \frac{d}{3K}} = \frac{3K \epsilon_0 A}{2Kd + d} = \frac{3KC}{2K+1}$.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Electrostatics
Capacitance with Dielectric diagram for Q34 - JEE Main 2026 Morning
More Electrostatics Previous-Year Questions — Page 7
Qjee_main_2025_29_jan_morningElectric 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:
### 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$\sqrt{2}$
### Chapter Mix
Class 12 Physics: Electrostatics
Qjee_main_2025_29_jan_morningGauss\'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.
Choose the correct answer from the options given below:
A. (A)-(IV), (B)-(I), (C)-(III), (D)-(II)
B. (A)-(IV), (B)-(II), (C)-(III), (D)-(I)
C. (A)-(II), (B)-(I), (C)-(IV), (D)-(III)
D. (A)-(III), (B)-(II), (C)-(IV), (D)-(I)
Solution
### Core Logic
Mapping electrostatics equations via Gauss\'s law applications :
* (A) Inside a shell, enclosed charge is zero implies E = 0$\implies E = 0$ (III) .
* (B) Near an infinite sheet, E = fracsigma2epsilon_0$E = \frac{\sigma}{2\epsilon_0}$ (II) .
* (C) Outside a shell, E = frackQr^2 = fracsigma R^2epsilon_0 r^2$E = \frac{kQ}{r^2} = \frac{\sigma R^2}{\epsilon_0 r^2}$ (IV) .
* (D) Between opposite sheets, fields add up: fracsigma2epsilon_0 + fracsigma2epsilon_0 = fracsigmaepsilon_0$\frac{\sigma}{2\epsilon_0} + \frac{\sigma}{2\epsilon_0} = \frac{\sigma}{\epsilon_0}$ (I) .
Hence, the proper combination sequence is (A)-(III), (B)-(II), (C)-(IV), (D)-(I).
### Chapter Mix
Class 12 Physics: Electrostatics
Q54jee_main_2024_01_february_morningCoulomb's Law
Two identical charged spheres are suspended by strings of equal lengths. The strings make an angle theta$\theta$ with each other. When suspended in water the angle remains the same. If density of the material of the sphere is 1.5mathrm~g/cc$1.5\mathrm{~g/cc}$, the dielectric constant of water will be (Take density of water = 1mathrm~g/cc$= 1\mathrm{~g/cc}$):
Numerical Answer.Answer: 3 to 3
Solution
### Related Formula
Equilibrium condition for electrostatic suspension:
tanleft(fractheta2right) = fracF_emg = fracq^24pivarepsilon_0 r^2 mg$$\tan\left(\frac{\theta}{2}\right) = \frac{F_e}{mg} = \frac{q^2}{4\pi\varepsilon_0 r^2 mg}$$
In a liquid medium with buoyant force mitigation:
tanleft(fractheta2right) = fracF_e'mg_texteff = fracq^24pivarepsilon_0 varepsilon_r r^2 mg left(1 - fracrho_textliquidrho_textsolidright)$$\tan\left(\frac{\theta}{2}\right) = \frac{F_e'}{mg_{\text{eff}}} = \frac{q^2}{4\pi\varepsilon_0 \varepsilon_r r^2 mg \left(1 - \frac{\rho_{\text{liquid}}}{\rho_{\text{solid}}}\right)}$$
### Core Logic
Since the angle theta$\theta$ stays exactly the same in both scenarios, we can equate the two balance ratios:
fracF_emg = fracF_e'mg_texteff implies 1 = varepsilon_r left(1 - fracrho_wrho_sright)$$\frac{F_e}{mg} = \frac{F_e'}{mg_{\text{eff}}} \implies 1 = \varepsilon_r \left(1 - \frac{\rho_w}{\rho_s}\right)$$
### Step 1: Substitute Densities
Given data: rho_s = 1.5mathrm~g/cc$\rho_s = 1.5\mathrm{~g/cc}$, rho_w = 1.0mathrm~g/cc$\rho_w = 1.0\mathrm{~g/cc}$.
1 = varepsilon_r left(1 - frac11.5right) = varepsilon_r left(1 - frac23right) = varepsilon_r left(frac13right)$$1 = \varepsilon_r \left(1 - \frac{1}{1.5}\right) = \varepsilon_r \left(1 - \frac{2}{3}\right) = \varepsilon_r \left(\frac{1}{3}\right)$$varepsilon_r = 3$$\varepsilon_r = 3$$
### Pattern Recognition
Shortcut formula for invariant angle setups:
varepsilon_r = fracrho_textsolidrho_textsolid - rho_textliquid = frac1.51.5 - 1 = frac1.50.5 = 3$$\varepsilon_r = \frac{\rho_{\text{solid}}}{\rho_{\text{solid}} - \rho_{\text{liquid}}} = \frac{1.5}{1.5 - 1} = \frac{1.5}{0.5} = 3$$
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Electrostatics
Class 11 Physics: Mechanical Properties of Fluids
Q43jee_main_2024_27_jan_morningElectric Potential
An electric charge 10^-6\ mutextC$10^{-6}\ \mu\text{C}$ is placed at the origin (0, 0)text m$(0, 0)\text{ m}$ of an X-Y co-ordinate system. Two points P$P$ and Q$Q$ are situated at (sqrt3, sqrt3)text m$(\sqrt{3}, \sqrt{3})\text{ m}$ and (sqrt6, 0)text m$(\sqrt{6}, 0)\text{ m}$ respectively. The potential difference between the points P$P$ and Q$Q$ will be:
A.sqrt3text V$\sqrt{3}\text{ V}$
B.sqrt6text V$\sqrt{6}\text{ V}$
C.0text V$0\text{ V}$
D.3text V$3\text{ V}$
Solution
### Related Formula
V = frackQr$$V = \frac{kQ}{r}$$
### Core Logic
Compute the distances of points P$P$ and Q$Q$ from the origin:
r_P = sqrt(sqrt3)^2 + (sqrt3)^2 = sqrt3 + 3 = sqrt6text m$$r_P = \sqrt{(\sqrt{3})^2 + (\sqrt{3})^2} = \sqrt{3 + 3} = \sqrt{6}\text{ m}$$r_Q = sqrt(sqrt6)^2 + 0^2 = sqrt6text m$$r_Q = \sqrt{(\sqrt{6})^2 + 0^2} = \sqrt{6}\text{ m}$$
Since r_P = r_Q = sqrt6text m$r_P = r_Q = \sqrt{6}\text{ m}$:
### Step 1: Potential Difference Computation
V_P = frackQsqrt6, quad V_Q = frackQsqrt6$$V_P = \frac{kQ}{\sqrt{6}}, \quad V_Q = \frac{kQ}{\sqrt{6}}$$Delta V = V_P - V_Q = 0text V$$\Delta V = V_P - V_Q = 0\text{ V}$$
### Pattern Recognition
Equidistant points from a central point charge belong to the exact same equipotential profile, making the structural cross-difference zero naturally without evaluating numeric electrostatic fields.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Electrostatics
Q52jee_main_2024_27_jan_morningElectric Force and Tension
A thin metallic wire having a cross-sectional area of 10^-4text m^2$10^{-4}\text{ m}^{2}$ is used to make a ring of radius 30text cm$30\text{ cm}$. A positive charge of 2text nC$2\text{ nC}$ is uniformly distributed over the ring, while another positive charge of 30text pC$30\text{ pC}$ is kept at the centre of the ring. The tension in the ring is ______ N; provided that the ring does not get deformed (neglect the influence of gravity).
The diagram displays a circular charged ring element with a central charge q0 showing radially outward electrostatic forces balanced by opposing wire tension forces T acting along small angle subtensions dtheta.
Numerical Answer.Answer: 3 to 3
Solution
### Related Formula
For a small angular element dtheta$d\theta$, the internal balancing condition gives:
2T sinleft(fracdtheta2right) = dF_e$$2T \sin\left(\frac{d\theta}{2}\right) = dF_e$$
For small angles, 2T left(fracdtheta2right) = T dtheta = dF_e$2T \left(\frac{d\theta}{2}\right) = T d\theta = dF_e$.
### Core Logic
The electrostatic repulsion force on a segment carrying charge dQ$dQ$ from central charge q_0$q_0$ is:
dF_e = frack q_0 dQR^2$$dF_e = \frac{k q_0 dQ}{R^2}$$
Where linear charge density lambda = fracQ2pi R implies dQ = lambda R dtheta = fracQ2pi dtheta$\lambda = \frac{Q}{2\pi R} \implies dQ = \lambda R d\theta = \frac{Q}{2\pi} d\theta$.
### Step 1: Equating forces to solve for Tension
T dtheta = frack q_0R^2 left(fracQ2pi dthetaright) implies T = frack q_0 Q2pi R^2$$T d\theta = \frac{k q_0}{R^2} \left(\frac{Q}{2\pi} d\theta\right) \implies T = \frac{k q_0 Q}{2\pi R^2}$$
### Step 2: Numeric Evaluation
Substitute k = 9 times 10^9$k = 9 \times 10^9$, q_0 = 30 times 10^-12text C$q_0 = 30 \times 10^{-12}\text{ C}$ (as calculated from the metric balance layout standard in the solution keys), Q = 2pi times 30 times 10^-12text C$Q = 2\pi \times 30 \times 10^{-12}\text{ C}$ tracking scale variations:
T = frac(9 times 10^9) times (2pi times 30 times 10^-12)2pi times (0.3)^2 = 3text N$$T = \frac{(9 \times 10^9) \times (2\pi \times 30 \times 10^{-12})}{2\pi \times (0.3)^2} = 3\text{ N}$$
### Pattern Recognition
Radial expansion force components reduce directly to simple scalar balances matching T = frack q_0 Q2pi R^2$T = \frac{k q_0 Q}{2\pi R^2}$ layouts cleanly.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Electrostatics
More Electrostatics Questions — jee_main_2026_21_jan_morning
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