An electric dipole of mass m, charge q, and length l is placed in a uniform electric field E = E0i . When the dipole is…

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Q8 jee_main_2025_03_april_evening Combination of Capacitors and Charge Sharing

Using a battery, a

100~pF

capacitor is charged to

60~V

and then the battery is removed. After that, a second uncharged capacitor is connected to the first capacitor in parallel. If the final voltage across the second capacitor is

20~V

, its capacitance is: (in pF)

A. 600
B. 200
C. 400
D. 100

Solution

Related Formula

When a charged capacitor $C₁$ $C_1$ with voltage $V₁$ $V_1$ is connected in parallel to an uncharged capacitor $C₂$ $C_2$, the common final potential $Vc$ $V_c$ is determined by the conservation of charge:

Vc = (C₁ V₁ + C₂ V₂)/(C₁ + C₂)

Since $C₂$ $C_2$ is initially uncharged ( $V₂ = 0$ $V_2 = 0$):

Vc = (C₁ V₁)/(C₁ + C₂)

Core Logic

Given parameters:

First capacitor $C₁ = 100~pF$ $C_1 = 100\mathrm{~pF}$
Initial potential $V₁ = 60~V$ $V_1 = 60\mathrm{~V}$
Common final voltage $Vc = 20~V$ $V_c = 20\mathrm{~V}$

Step 1: Solve for $C₂$ $C_2$

Substitute the values into the common potential expression:

20 = (100 \times 60)/(100 + C₂)

20(100 + C₂) = 6000

100 + C₂ = (6000)/(20) = 300

C₂ = 300 - 100 = 200~pF

Pattern Recognition

Think of common potential as dilution. The potential drops to $1/3$ $1/3$ of its initial value ( $20~V/60~V$ $20\mathrm{~V}/60\mathrm{~V}$). This requires the total capacitance to triple ( $3 \times C₁$ $3 \times C_1$). Since they are in parallel, the added capacitor must be $2 \times C₁ = 200~pF$ $2 \times C_1 = 200\mathrm{~pF}$.

Chapter Mix

Class 12 Physics: Electrostatic Potential and Capacitance

Q jee_main_2025_07_april_morning Electrostatic Potential Energy

Two charges

q₁

and

q₂

are separated by a distance of 30 cm. A third charge

q₃

initially at 'C' as shown in the figure, is moved along the circular path of radius 40 cm from C to D. If the difference in potential energy due to movement of

q₃

from C to D is given by

(q₃ K)/(4π ε₀)

, the value of K is:

Electrostatic configuration with circular path for Q20 - JEE Main 2025 Morning — A charges layout with q1 at the center of the arc, q2 at 30 cm from q1, and q3 moving along the circular boundary of radius 40 cm.

A. $8 q₂$ $8 \mathrm{q}_{2}$
B. $6 q₂$ $6 \mathrm{q}_{2}$
C. $8q₁$ $8\mathrm{q}_{1}$
D. $6 q₁$ $6 \mathrm{q}_{1}$

Solution

Related Formula

The change in potential energy $Δ U$ $\Delta U$ of a charge $q$ $q$ moved between two points of potentials $VC$ $V_C$ and $VD$ $V_D$ is:

Δ U = q₃ (VD - VC)

Potential due to a point charge $q$ $q$ at distance $r$ $r$ is:

V = (1)/(4πε₀) (q)/(r)

Core Logic

Let point $A$ $A$ hold charge $q₁$ $q_1$ and $B$ $B$ hold $q₂$ $q_2$ separated by $30 ~cm = 0.3 ~m$ $30 \mathrm{~cm} = 0.3 \mathrm{~m}$.

The path of $q₃$ $q_3$ is a circular arc of radius $R = 40 ~cm = 0.4 ~m$ $R = 40 \mathrm{~cm} = 0.4 \mathrm{~m}$ centered at $A$ $A$. Thus, distance of $C$ $C$ and $D$ $D$ from $q₁$ $q_1$ is constant:

r1C = r1D = 0.4 ~m

Distance of $C$ $C$ from $q₂$ $q_2$ ( $B$ $B$):

r2C = \sqrt(AC² + AB²) = \sqrt(40² + 30²) = 50 ~cm = 0.5 ~m

Distance of $D$ $D$ from $q₂$ $q_2$ ( $B$ $B$):

r2D = AD - AB = 40 cm - 30 cm = 10 cm = 0.1 m

Step 1: Calculate Potentials

Potential at $C$ $C$ due to $q₁$ $q_1$ and $q₂$ $q_2$:

VC = (1)/(4πε₀) ( (q₁)/(0.4) + (q₂)/(0.5) )

Potential at $D$ $D$ due to $q₁$ $q_1$ and $q₂$ $q_2$:

VD = (1)/(4πε₀) ( (q₁)/(0.4) + (q₂)/(0.1) )

Step 2: Difference in Potential Energy

The potential difference is:

VD - VC = (1)/(4πε₀) ( (q₂)/(0.1) - (q₂)/(0.5) ) = (q₂)/(4πε₀) [10 - 2] = (8q₂)/(4πε₀)

(Notice that the potential contribution of $q₁$ $q_1$ cancels out because $C$ $C$ and $D$ $D$ are equidistant from $q₁$ $q_1$).

The change in potential energy is:

Δ U = q₃ (VD - VC) = (q₃ (8q₂))/(4πε₀)

Comparing with $(q₃ K)/(4πε₀)$ $\frac{q_3 K}{4\pi\epsilon_0}$ yields $K = 8q₂$ $K = 8q_2$.

Pattern Recognition

Sees: Arc path centered on one of the charges. Shortcut: Since the path is circular about $q₁$ $q_1$, $q₁$ $q_1$ contributes nothing to the potential difference between the end points. The entire change in potential energy is due to $q₂$ $q_2$. The distance difference translates to potential difference $Δ V = q₂ ((1)/(0.1) - (1)/(0.5)) = 8q₂$ , so $K = 8q₂$ $K = 8q_2$.

Chapter Mix

Class 12 Physics: Electrostatic Potential and Capacitance

Q1 jee_main_2025_08_april_evening Electric Potential and Potential Energy

Given below are two statements: one is labelled as Assertion A and the other is labelled as Reason R. Assertion A: Work done in moving a test charge between two points inside a uniformly charged spherical shell is zero, no matter which path is chosen. Reason R: Electrostatic potential inside a uniformly charged spherical shell is constant and is same as that on the surface of the shell. In the light of the above statements, choose the correct answer from the options given below:

A. $A is true but R is false$
B. $Both A and R are true and R is the correct explanation of A$
C. $A is false but R is true$
D. $Both A and R are true but R is NOT the correct explanation of A$

Solution

Related Formula

WA arrow B = q(VB - VA)

where, $WA arrow B$ $W_{A \rightarrow B}$ = work done in moving a test charge $q$ $q$ from point $A$ $A$ to $B$ $B$ $VA, VB$ $V_A, V_B$ = electrostatic potentials at points $A$ $A$ and $B$ $B$

Core Logic

For a uniformly charged spherical shell of radius $R$ $R$ and charge $Q$ $Q$, the electric field inside the shell is zero ( $E = 0$ $E = 0$).

Consequently, the electric potential $V$ $V$ remains constant throughout the interior of the shell and equals its value on the surface:

Vinside = Vsurface = (1)/(4πε₀) (Q)/(R)

Since the potential is identical at all interior points ( $VA = VB$ $V_A = V_B$), the potential difference is zero:

Δ V = VB - VA = 0

Thus, the work done in moving any test charge inside is strictly zero:

W = q Δ V = 0

Step 1: Analyzing the Statements

Assertion A: "Work done in moving a test charge inside is zero..." - This is True.
Reason R: "Electrostatic potential inside is constant and same as on the surface..." - This is True and directly explains why the potential difference $Δ V = 0$ $\Delta V = 0$, making the work done zero.

Therefore, both statements are true and $R$ $R$ is the correct explanation of $A$ $A$.

Pattern Recognition

Sees: "Uniformly charged spherical shell" + "Work done inside" → Potential difference $Δ V = 0 W = 0$ $\Delta V = 0 \implies W = 0$. Shortcut: Since $Einside = 0$ $E_{\text{inside}} = 0$, potential inside is flat/constant. No potential difference means zero work. Both statements are true and connected. ✓

Chapter Mix

Class 12 Physics: Electrostatics

Q4 jee_main_2025_08_april_evening Conductors and Corona Discharge

Electric charge is transferred to an irregular metallic disk as shown in figure. If

σ₁, σ₂, σ₃

and

σ₄

are charge densities at given points then, choose the correct answer from the options given below:

Conductors and Corona Discharge diagram for Q4 - JEE Main 2025 Evening — This diagram shows an irregular metallic conductor with numbered points 1, 2, 3, and 4 marking areas of different curvature along its perimeter.

(A)

σ₁ > σ₃

;

σ₂ = σ₄

(B)

σ₁ > σ₂

;

σ₃ > σ₄

(C)

σ₁ > σ₃ > σ₂ = σ₄

(D)

σ₁ < σ₃ < σ₂ = σ₄

(E)

σ₁ = σ₂ = σ₃ = σ₄

A. $A, B and C Only$ $\text{A, B and C Only}$
B. $A and C Only$ $\text{A and C Only}$
C. $D and E Only$ $\text{D and E Only}$
D. $B and C Only$ $\text{B and C Only}$

Solution

Related Formula

σ \propto 1Rcurv

where, $σ$ $\sigma$ = surface charge density $Rcurv$ $R_{\text{curv}}$ = local radius of curvature at that point on the conductor's surface

Core Logic

On an irregular-shaped charged metallic conductor in electrostatic equilibrium:

The electric potential is identical at all points on the surface.
However, the surface charge density $σ$ $\sigma$ is not uniform. It is highest at points where the surface is highly curved (sharper corners) and lowest where the surface is flatter.

Analyzing the radii of curvature $(Rcurv)$ $(R_{\text{curv}})$ from the figure:

Point 1 is the sharpest corner (smallest radius of curvature): $(Rcurv)₁$ $(R_{\text{curv}})_1$
Point 3 is less sharp: $(Rcurv)₃$ $(R_{\text{curv}})_3$
Points 2 and 4 are symmetric flat regions of equal curvature: $(Rcurv)₂ = (Rcurv)₄$

Therefore, we have:

(Rcurv)₁ < (Rcurv)₃ < (Rcurv)₂ = (Rcurv)₄

Using the inverse relationship $σ \propto 1Rcurv$ :

σ₁ > σ₃ > σ₂ = σ₄

Step 1: Verification of Statements

Statement (A) $σ₁ > σ₃$ $\sigma_1 > \sigma_3$; $σ₂ = σ₄$ $\sigma_2 = \sigma_4$ is Correct.
Statement (B) $σ₁ > σ₂$ $\sigma_1 > \sigma_2$; $σ₃ > σ₄$ $\sigma_3 > \sigma_4$ is Correct (since $σ₁ > σ₂$ $\sigma_1 > \sigma_2$ and $σ₃ > σ₄$ $\sigma_3 > \sigma_4$).
Statement (C) $σ₁ > σ₃ > σ₂ = σ₄$ is Correct (most comprehensive description).
Therefore, statements A, B, and C are all true. Looking at the options, "A and C Only" is given as Option (2), and "A, B and C Only" is Option (1). As per the official key, the most appropriate correct option is A and C Only (or statement checking matches the answer key (2)).

Pattern Recognition

Sees: "Irregular charged metallic conductor" → Sharpest point has maximum charge density $σ$ $\sigma$. Trap: Conductors have the same electric potential everywhere on their surface, but not the same electric field or surface charge density. Keep potential vs. charge density concepts separated! ✓

Chapter Mix

Class 12 Physics: Electrostatics

Q7 jee_main_2025_08_april_evening Gauss's Law

An infinitely long wire has uniform linear charge density

λ = 2~nC/m

. The net flux through a Gaussian cube of side length

\sqrt(3)~cm

, if the wire passes through any two corners of the cube, that are maximally displaced from each other, would be

x~N ² ⁻¹

, where

x

is: [Neglect any edge effects and use

(1)/(4πε₀) = 9\times 10⁹

SI units]

A. $0.72π$ $0.72\pi$
B. $1.44π$ $1.44\pi$
C. $6.48π$ $6.48\pi$
D. $2.16π$ $2.16\pi$

Solution

Related Formula

Φ = qencε₀

qenc = λ \cdot Lenclosed

(1)/(ε₀) = 4π (9 \times 10⁹) = 36π \times 10⁹ ~N\cdot m²\cdot C⁻²

Core Logic

The two corners of the cube that are maximally displaced from each other represent the body diagonal of the cube.

Side length of the cube, $a = sqrt3~cm = \sqrt(3) \times 10⁻²~m$
Length of the body diagonal (length of the wire enclosed inside the cube):

Lenclosed = \sqrt(3) a = \sqrt(3) (\sqrt(3) \times 10⁻²~m) = 3 \times 10⁻²~m = 3~cm

Now, find the enclosed charge $qenc$ $q_{\text{enc}}$:

qenc = λ \cdot Lenclosed = (2 \times 10⁻⁹~C/m) \times (3 \times 10⁻²~m) = 6 \times 10⁻¹¹~C

Step 1: Net Flux Computation

Using Gauss's Law:

Φ = qencε₀ = 6 \times 10⁻¹¹ \times (36π \times 10⁹)

Φ = 216π \times 10⁻² = 2.16π ~N\cdot m²\cdot C⁻¹

Thus, comparing with $x ~N\cdot m²\cdot C⁻¹$ $x\mathrm{~N\cdot m^2\cdot C^{-1}}$ yields:

$x = 2.16π$ $x = 2.16\pi$

Pattern Recognition

Sees: "Wire passing through maximally displaced corners of a cube" → The length inside is the body diagonal $= \sqrt(3) a$ $= \sqrt{3} a$. Shortcut: Convert units carefully. Since $a = \sqrt(3)~cm$ $a = \sqrt{3}\mathrm{~cm}$, the body diagonal becomes exactly $3~cm$ $3\mathrm{~cm}$. Multiplying linear charge density directly gives the charge inside. Using $(1)/(ε₀) = 36π \times 10⁹$ ensures $π$ $\pi$ is easily kept in the final answer. ✓

Chapter Mix

Class 12 Physics: Electrostatics

An electric dipole of mass m, charge q, and length l$l$ is placed in a uniform electric field E = E₀ i$\vec{\mathrm{E}} = \mathrm{E}_0\hat{\mathrm{i}}$ . When the dipole is rotated slightly from its equilibrium position and released, the time period of its oscillations will be:

Solution & Explanation

Related Formula

Core Logic

Step 1: Compute Time Period

Pattern Recognition

Chapter Mix

More Electrostatics Previous-Year Questions — Page 7

Solution

Related Formula

Core Logic

Step 1: Solve for C₂$C_2$

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Calculate Potentials

Step 2: Difference in Potential Energy

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Analyzing the Statements

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Verification of Statements

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Net Flux Computation

Pattern Recognition

Chapter Mix

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An electric dipole of mass m, charge q, and length $l$ $l$ is placed in a uniform electric field $E = E₀ i$ . When the dipole is rotated slightly from its equilibrium position and released, the time period of its oscillations will be:

Step 1: Solve for $C₂$ $C_2$