A parallel plate capacitor has capacitance C, when there is vacuum within the parallel plates. A sheet having thickness ((1/3))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 Alternatively, treat as two capacitors in series: Ceq = (C1 C2/C1 + C2) ### Core Logic Initial capacitance (vacuum): C = A epsilon0d. The system can be modelled as two capacitors in series: 1. A vacuum capacitor of thickness d - (d/3) = (2d/3) 2. A dielectric capacitor of thickness (d/3) and dielectric constant K. ### Step 1: Capacitors in Series The capacitances are: Now, equivalent capacitance in series: ### Pattern Recognition Partial dielectric filling of thickness t: Use formula Cnew = (epsilon0 A/d - t + t/K). Plugging t = d/3 directly gives (epsilon0 A/d - d/3 + d/3K) = (epsilon0 A/2d)3 + (d/3K) = (3K epsilon0 A/2Kd + d) = (3KC/2K+1). ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Electrostatics

A parallel plate capacitor has capacitance C, when there is vacuum within the parallel plates. A sheet having thickness ((1/3))rd of the separation between the plates and relative permittivity K is in

### Related Formula Alternatively, treat as two capacitors in series: Ceq = (C1 C2/C1 + C2) ### Core Logic Initial capacitance (vacuum): C = A epsilon0d. The system can be modelled as two capacitors in series: 1. A vacuum capacitor of thickness d - (d/3) = (2d/3) 2. A dielectric capacitor of thickness (d/3) and dielectric constant K. ### Step 1: Capacitors in Series The capacitances are: Now, equivalent capacitance in series: ### Pattern Recognition Partial dielectric filling of thickness t: Use

A parallel plate capacitor has capacitance C, when there is vacuum within the parallel plates. A sheet having thickness ((1/3))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 Alternatively, treat as two capacitors in series: Ceq = (C1 C2/C1 + C2) ### Core Logic Initial capacitance (vacuum): C = A epsilon0d. The system can be modelled as two capacitors in series: 1. A vacuum capacitor of thickness d - (d/3) = (2d/3) 2. A dielectric capacitor of thickness (d/3) and dielectric constant K. ### Step 1: Capacitors in Series The capacitances are: Now, equivalent capacitance in series: ### Pattern Recognition Partial dielectric filling of thickness t: Use formula Cnew = (epsilon0 A/d - t + t/K). Plugging t = d/3 directly gives (epsilon0 A/d - d/3 + d/3K) = (epsilon0 A/2d)3 + (d/3K) = (3K epsilon0 A/2Kd + d) = (3KC/2K+1). ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Electrostatics

A parallel plate capacitor has capacitance C, when there is vacuum within the parallel plates. A sheet having thickness ((1/3))rd of the separation between the plates and relative permittivity K is in

### Related Formula Alternatively, treat as two capacitors in series: Ceq = (C1 C2/C1 + C2) ### Core Logic Initial capacitance (vacuum): C = A epsilon0d. The system can be modelled as two capacitors in series: 1. A vacuum capacitor of thickness d - (d/3) = (2d/3) 2. A dielectric capacitor of thickness (d/3) and dielectric constant K. ### Step 1: Capacitors in Series The capacitances are: Now, equivalent capacitance in series: ### Pattern Recognition Partial dielectric filling of thickness t: Use

Solution & Explanation

### Related Formula

C = fracepsilon_0 Ad - t + fractK

Alternatively, treat as two capacitors in series:

C_eq = fracC_1 C_2C_1 + C_2

### Core Logic Initial capacitance (vacuum):

C = fracA epsilon_0d

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

2. A dielectric capacitor of thickness

fracd3

and dielectric constant

K

### Step 1: Capacitors in Series The capacitances are:

C_1 = fracepsilon_0 Aleft(frac2d3right) = frac32 left(fracepsilon_0 Adright) = frac32 C

C_2 = fracK epsilon_0 Aleft(fracd3right) = 3K left(fracepsilon_0 Adright) = 3KC

Now, equivalent capacitance in series:

C_texteq = fracC_1 C_2C_1 + C_2 = fracleft(frac32 Cright) times (3KC)frac32 C + 3KC

C_texteq = fracfrac92KC^2frac32C(1 + 2K)

C_texteq = frac3KC2K + 1

### Pattern Recognition Partial dielectric filling of thickness

t

: Use formula

C_textnew = fracepsilon_0 Ad - t + t/K

. Plugging

t = d/3

directly gives

fracepsilon_0 Ad - d/3 + d/3K = fracepsilon_0 Afrac2d3 + fracd3K = frac3K epsilon_0 A2Kd + d = frac3KC2K+1

. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Electrostatics

More Electrostatics Previous-Year Questions — Page 6

Q1 jee_main_2025_24_jan_morning Energy Stored in a Capacitor

Consider a parallel plate capacitor of area A (of each plate) and separation 'd' between the plates. If E is the electric field and

epsilon_0

is the permittivity of free space between the plates, then potential energy stored in the capacitor is :-

A. $frac12epsilon_0E^2Ad$ $\frac{1}{2}\epsilon_{0}E^{2}Ad$
B. $frac34epsilon_0E^2Ad$ $\frac{3}{4}\epsilon_{0}E^{2}Ad$
C. $frac14epsilon_0E^2Ad$ $\frac{1}{4}\epsilon_{0}E^{2}Ad$
D. $epsilon_0E^2Ad$ $\epsilon_{0}E^{2}Ad$

Solution

### Related Formula The electrostatic energy density

u

stored in an electric field

E

is given by:

u = frac12epsilon_0E^2

The total potential energy

U

stored in a volume

V

is:

U = u cdot V

### Core Logic For a parallel plate capacitor, the volume between the plates where the electric field exists is the product of the plate area

A

and the plate separation

d

V = Ad

### Step 1: Calculating Stored Energy Substitute the volume expression into the total energy equation:

U = left(frac12epsilon_0E^2 ight)(Ad)

U = frac12epsilon_0E^2Ad

### Pattern Recognition Energy density times volume is a universal relation for field fields. Remember that volume is simply cross-sectional area multiplied by distance (

Ad

). ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Electrostatics

Q18 jee_main_2025_24_jan_morning Capacitance of a Parallel Plate Capacitor

A parallel plate capacitor was made with two rectangular plates, each with a length of

l = 3

cm and breath of

b = 1

cm. The distance between the plates is

3mu m

Out of the following, which are the ways to increase the capacitance by a factor of 10? A.

l = 30

cm, b = 1 cm,

d=1mu m

l = 3

cm,

b=1

cm,

d=30~mu m

l = 6

cm,

b=5

cm,

d=3~mu m

l = 1

cm,

b=1textcm

d=10~mu m

l = 5text cm

b=2

cm,

d=1mu m

Choose the correct answer from the options given below :

A. C and E only
B. B and D only
C. A only
D. C only

Solution

### Related Formula The capacitance of a parallel plate system is given by :

C = fracepsilon_0Ad = fracepsilon_0lbd

where

l

is length,

b

is breadth, and

d

is separation distance. ### Core Logic Evaluate the initial capacitance base scaling parameter :

C_0 = fracepsilon_0 times 3text cm times 1text cm3mutextm = 1 times epsilon_0text units

We want to increase this initial baseline capacitance value by a factor of 10, meaning our target capacitance is

10epsilon_0

. ### Step 1: Audit Options Let's check the capacitance for options C and E : * Option C:

l=6text cm, b=5text cm, d=3mutextm

C_C = fracepsilon_0 times 6 times 53 = 10epsilon_0text units (Correct)

* Option E:

l=5text cm, b=2text cm, d=1mutextm

C_E = fracepsilon_0 times 5 times 21 = 10epsilon_0text units (Correct)

### Pattern Recognition Capacitance scales matching the geometric factor

fracl cdot bd

. Look for options where this ratio scales up to exactly 10 times the initial baseline value. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Electrostatics

Q21 jee_main_2025_24_jan_morning Gauss's Law and Electric Flux

A square loop of sides

a = 1

m is held normally in front of a point charge

q = 1C

The flux of the electric field through the shaded region is

frac5p times frac1varepsilon_0 fracNm^2C

, where the value of p is .

Gauss's Law and Electric Flux diagram for Q21 - JEE Main 2025 Morning — 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

through a completely enclosing symmetric cube container surface is:

Phitexttotal = fracqepsilon0

### Core Logic Assuming the charge resides at a symmetric center distance

fraca2

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

### Step 1: Symmetric Partitioning As shown in the solution schematic

, 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

The shaded region covers exactly 5 of these individual triangle parts [cite: 779, 780]:

Phi_textshaded = 5 times Phi_textpart = frac548 times fracqepsilon_0

Comparing this result with the given expression

frac5p times frac1epsilon_0

, we find:

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

Q2 jee_main_2025_28_jan_evening Energy Density in Capacitors

A parallel plate capacitor of capacitance

1mu mathrmF

is charged to a potential difference of

20mathrmV

. The distance between plates is

1mu mathrmm

. 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. $2 times 10^-4 \, mathrmJ/m^3$ $2 \times 10^{-4} \, \mathrm{J/m^3}$
C. $2 times 10^2 mathrm~J / mathrmm^3$
D. $1.8 times 10^5 mathrm~J / mathrmm^3$

Solution

### Related Formula The electric field

E

between the plates of a parallel plate capacitor is given by :

E = fracVd

The electrostatic energy density

U

in a medium is given by:

U = frac12 epsilon_0 E^2

### Core Logic Given values [cite: 662, 664, 665]: * Capacitance,

C = 1 \ mutextF

* Potential difference,

V = 20 text V

* Plate separation,

d = 1 \ mutextm = 10^-6 text m

First, calculate the electric field

E

E = frac2010^-6 = 20 times 10^6 text V/m

Now, calculate the energy density

U

using

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 = frac12 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

Rounding to the nearest matching option gives

1.8 times 10^3 text J/m^3

. ### Pattern Recognition Notice that energy density is completely independent of the total capacitance value

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

Q22 jee_main_2025_28_jan_evening Electric Dipole

An electric dipole of dipole moment

6 times 10^-6 mathrmCm

is placed in uniform electric field of magnitude

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

inside a uniform electric field is given by:

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)

### Core Logic Given parameters [cite: 805, 806]: * Dipole moment,

p = 6 times 10^-6 text Cm

* Electric field,

E = 10^6 text V/m

* Initial \angle (\parallel state),

theta_i = 0^circ

* Final \angle (opposite state),

theta_f = 180^circ

Substitute the values into the work equation [cite: 807, 808]:

W = -pE (cos 180^circ - cos 0^circ)

W = -pE (-1 - 1) = 2 p E 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

) to its most unstable position (anti-\parallel,

theta=180^circ

) always requires a maximum work value of exactly

2pE

. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Electrostatics

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A parallel plate capacitor has capacitance C, when there is vacuum within the parallel plates. A sheet having thickness $left(frac13right)^mathrmrd$ of the separation between the plates and relative permittivity K is introduced between the plates. The new capacitance of the system is :

Solution & Explanation

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