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 5

Q1 jee_main_2025_07_april_evening Electrostatic 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

[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

Q10 jee_main_2025_07_april_evening Torque 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

from the axis, it tends to realign in the direction due to a torque. The value of torque is : [cite: 58, 59, 60]

A. $5times10^-9~mathrmNm$ $5\times10^{-9}~\mathrm{Nm}$ [cite: 60]
B. $2.5times10^-12~mathrmNm$ $2.5\times10^{-12}~\mathrm{Nm}$ [cite: 61]
C. $5times10^-3~mathrmNm$ $5\times10^{-3}~\mathrm{Nm}$ [cite: 62]
D. $2.5times10^-9~mathrmNm$ $2.5\times10^{-9}~\mathrm{Nm}$ [cite: 62]

Solution

### Related Formula

E = fracVd

[cite: 736]

tau = pEsintheta

[cite: 737]

p = q cdot a

[cite: 738] ### Core Logic First, calculate the electric field magnitude

E

between the capacitor plates: [cite: 58, 736]

E = frac50.5 times 10^-3 = 10^4\ textV/m

[cite: 58, 59, 736] Next, evaluate the dipole moment

p

: [cite: 58, 738]

p = (2 times 10^-6\ textC) times (0.5 times 10^-6\ textm) = 1 times 10^-12\ textCcdottextm

[cite: 58, 740] Now find the torque when rotated by

theta = 30^circ

: [cite: 59, 737]

tau = (1 times 10^-12) times 10^4 times sin 30^circ = 10^-8 times frac12 = 5 times 10^-9\ textNcdottextm

[cite: 743] ### Pattern Recognition Always convert parameters to pristine standard SI units before applying electrostatic expressions (

0.5\ mutextm = 5 times 10^-7\ textm

and

0.5\ textmm = 5 times 10^-4\ textm

) to secure zero conversion error[cite: 58, 59, 736, 738]. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Electrostatics

Q21 jee_main_2025_07_april_evening Dielectrics and Capacitance

A parallel plate capacitor has charge

5times10^-6mathrm~C

. A dielectric slab is inserted between the plates and almost fills the space between the plates. If the induced charge on one face of the slab is

4times10^-6mathrm~C

then the dielectric constant of the slab is _______. [cite: 183, 184]

Numerical Answer. Answer: 5 to 5

Solution

### Related Formula

Q_textind = Qleft(1 - frac1Kright)

[cite: 813] ### Core Logic Substitute the values given for free surface charge

Q = 5 times 10^-6\ textC

and bound induced charge

Q_textind = 4 times 10^-6\ textC

into the equation: [cite: 183, 184, 814]

4 times 10^-6 = 5 times 10^-6 left(1 - frac1Kright)

[cite: 814]

frac45 = 1 - frac1K implies frac1K = 1 - frac45 = frac15

[cite: 815]

K = 5

[cite: 815] ### Pattern Recognition The fraction of charge induced on the dielectric face scales structurally as

fracK-1K

[cite: 813, 815]. Observing a ratio of

4

parts out of

5

implies that the constant factor

K

must equal

5

directly[cite: 815]. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Electrostatics

Q24 jee_main_2025_07_april_evening Electric Flux

The electric field in a region is given by

vecmathrmE = (2hatmathrmi + 4hatmathrmj + 6hatmathrmk) times 10^3mathrmN / mathrmC

. The flux of the field through a rectangular surface parallel to x-z plane is

6.0mathrmNm^2mathrmC^-1

. The area of the surface is __________

mathrmcm^2

. [cite: 195, 196]

Numerical Answer. Answer: 15 to 15

Solution

### Related Formula

phi = vecE cdot vecA

[cite: 827] ### Core Logic A surface aligned parallel to the

xtext-z

plane possesses an area vector pointing completely orthogonal to it along the

y

-axis direction, meaning

vecA = Ahatj

[cite: 196, 827]. Performing the dot product: [cite: 827]

phi = left[(2hati + 4hatj + 6hatk) times 10^3right] cdot (Ahatj) = 4 times 10^3 A

[cite: 195, 827] Given that the net flux magnitude is

6.0\ textNm^2textC^-1

[cite: 196]:

6 = 4 times 10^3 A implies A = frac64 times 10^3 = 1.5 times 10^-3\ textm^2

[cite: 828, 829] Converting square meters to square centimeters (

1\ textm^2 = 10^4\ textcm^2

): [cite: 196, 830]

A = 1.5 times 10^-3 times 10^4 = 15\ textcm^2

[cite: 830] ### Pattern Recognition Always focus exclusively on the specific field component matched to the surface orientation normal[cite: 827]. For an

xtext-z

plane match, only the

hatj

coefficient creates flux[cite: 196, 827]. Do not miss the metric scale unit transition at the end (

m^2 rightarrow cm^2

)[cite: 196, 830]. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Electrostatics

Q18 jee_main_2025_24_jan_evening Coulomb's Law

A small uncharged conducting sphere is placed in contact with an identical sphere but having

4 times 10^-8

C charge and then removed to a distance such that the force of repulsion between them is

9 times 10^-3

N. The distance between them is (Take

frac14pivarepsilon_0

9 times 10^9

in SI units)

A. 2 cm
B. 3 cm
C. 4 cm
D. 1 cm

Solution

### Related Formula

F = frack q_1 q_2r^2

### Core Logic When two identical conducting spheres are brought into contact, the total initial charge splits equally between them:

q_1 = q_2 = frac4 times 10^-8\ mathrmC + 02 = 2 times 10^-8\ mathrmC

Given repulsion force,

F = 9 times 10^-3\ mathrmN

9 times 10^-3 = frac9 times 10^9 times (2 times 10^-8) times (2 times 10^-8)r^2

9 times 10^-3 = frac36 times 10^-7r^2 implies r^2 = frac36 times 10^-79 times 10^-3 = 4 times 10^-4

r = 2 times 10^-2\ mathrmm = 2\ mathrmcm

Charge redistribution schematic for two spheres Q18

### Pattern Recognition Identical spheres in contact distribute net charge equally due to symmetric capacitance sharing:

q' = Q_texttotal / 2

. ### 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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