Two charges 7mu C and -2mu C are placed at (-9,0,0) cm and (9,0,0) cm respectively in an external field E=(A/r2)r , where A=9x105N/C.m2 . Considering the potential at infinity is 0, the electrostatic energy of the configuration is ____ J.

First, derive the potential function V(r) from the given external electric field. Both charges are at distance r = 9 , cm = 0.09 , m from the origin (where the field points radiate from). Energy of q1 in the external field: Energy of q2 in the external field: Sum of self energies: Distance between charges r12 = 9 - (-9) = 18 , cm = 0.18 , m. When an external field is present, total energy is the sum of the potential energies of each charge independently in that external field, plus the mutual interaction energy between the charges themselves. Class 12 Physics: Electrostatics

Two charges 7mu C and -2mu C are placed at (-9,0,0) cm and (9,0,0) cm respectively in an external field E=(A/r2)r , where A=9x105N/C.m2 . Considering the potential at infinity is 0, the electrostatic

First, derive the potential function V(r) from the given external electric field. Both charges are at distance r = 9 , cm = 0.09 , m from the origin (where the field points radiate from). Energy of q1 in the external field: Energy of q2 in the external field: Sum of self energies: Distance between charges r12 = 9 - (-9) = 18 , cm = 0.18 , m. When an external field is present, total energy is the sum of the potential energies of each charge independently in that external field, plus the mutual in

Two charges 7mu C and -2mu C are placed at (-9,0,0) cm and (9,0,0) cm respectively in an external field E=(A/r2)r , where A=9x105N/C.m2 . Considering the potential at infinity is 0, the electrostatic energy of the configuration is ____ J.

First, derive the potential function V(r) from the given external electric field. Both charges are at distance r = 9 , cm = 0.09 , m from the origin (where the field points radiate from). Energy of q1 in the external field: Energy of q2 in the external field: Sum of self energies: Distance between charges r12 = 9 - (-9) = 18 , cm = 0.18 , m. When an external field is present, total energy is the sum of the potential energies of each charge independently in that external field, plus the mutual interaction energy between the charges themselves. Class 12 Physics: Electrostatics

Two charges 7mu C and -2mu C are placed at (-9,0,0) cm and (9,0,0) cm respectively in an external field E=(A/r2)r , where A=9x105N/C.m2 . Considering the potential at infinity is 0, the electrostatic

First, derive the potential function V(r) from the given external electric field. Both charges are at distance r = 9 , cm = 0.09 , m from the origin (where the field points radiate from). Energy of q1 in the external field: Energy of q2 in the external field: Sum of self energies: Distance between charges r12 = 9 - (-9) = 18 , cm = 0.18 , m. When an external field is present, total energy is the sum of the potential energies of each charge independently in that external field, plus the mutual in

Two charges 7mu C and -2mu C are placed at (-9,0,0) cm and (9,0,0) cm respectively in an external field E=(A/r2)r…

Year	2026	2025	2024	Total
Questions	24	39	16	79

Q9 jee_main_2025_08_april_evening Electric Charge and Properties

Two metal spheres of radius

R

and

3R

have same surface charge density

σ

. If they are brought in contact and then separated, the surface charge density on smaller and bigger sphere becomes

σ₁

and

σ₂

, respectively. The ratio

σ₁σ₂

is:

A. $(1)/(9)$ $\frac{1}{9}$
B. $9$ $9$
C. $(1)/(3)$ $\frac{1}{3}$
D. $3$ $3$

Solution

Related Formula

V = (σ r)/(ε₀)

where, $V$ $V$ = electrostatic potential of a conducting sphere $σ$ $\sigma$ = surface charge density $r$ $r$ = radius of the sphere

Core Logic

For any conducting sphere, the potential on its surface is related to its surface charge density by:

V = (k Q)/(r) = (1)/(4πε₀) (σ (4π r²))/(r) = (σ r)/(ε₀)

When the two spheres of radii $r₁ = R$ $r_1 = R$ and $r₂ = 3R$ $r_2 = 3R$ are brought into contact, charge flows between them until they reach an identical electric potential:

$V₁ = V₂$ $V_1 = V_2$

Step 1: Ratio Calculation

Equate the potentials of the two spheres after separation:

(σ₁ r₁)/(ε₀) = (σ₂ r₂)/(ε₀)

σ₁ R = σ₂ (3R) (σ₁)/(σ₂) = (3R)/(R) = 3

Pattern Recognition

Sees: "Conducting spheres brought in contact" → Electric potentials become equal: $V₁ = V₂$ $V_1 = V_2$. Shortcut: Since $V \propto σ r$ $V \propto \sigma r$, equal potential directly implies $σ₁ r₁ = σ₂ r₂$ $\sigma_1 r_1 = \sigma_2 r_2$. Thus, the ratio of final densities is simply the inverse ratio of their radii: $(σ₁)/(σ₂) = (r₂)/(r₁) = (3)/(1) = 3$ . This bypasses computing the individual final charges entirely! ✓

Chapter Mix

Class 12 Physics: Electrostatics

Q23 jee_main_2025_08_april_evening Capacitance

Space between the plates of a parallel plate capacitor of plate area

4~cm²

and separation of (d)

1.77~mm

, is filled with uniform dielectric materials with dielectric constants (

3

and

5

) as shown in figure. Another capacitor of capacitance

7.5~pF

is connected in parallel with it. The effective capacitance of this combination is ________

~pF

.

Capacitance parallel plate dielectric diagram for Q23 - JEE Main 2025 Evening — This diagram shows a parallel plate capacitor filled with two layers of dielectric constant k=5 and k=3, each having thickness d/2.

(Given

ε₀ = 8.85\times 10⁻¹²~F/m

)

Numerical Answer. Answer: 15 to 15

Solution

Related Formula

C = k (ε₀ A)/(t)

1Cₛₑᵣᵢₑₛ = (1)/(C₁) + (1)/(C₂)

Cparallel = Ceq + Cₚ

Core Logic

The dielectrics $k₁ = 5$ $k_1 = 5$ and $k₂ = 3$ $k_2 = 3$ split the capacitor separation horizontally into two layers, each of thickness $(d)/(2)$ $\frac{d}{2}$. Thus, they act as two capacitors connected in series:

Capacitor 1: $C₁ = k₁ (ε₀ A)/(d/2) = 5 \times (2 ε₀ A)/(d) = 10 (ε₀ A)/(d)$
Capacitor 2: $C₂ = k₂ (ε₀ A)/(d/2) = 3 \times (2 ε₀ A)/(d) = 6 (ε₀ A)/(d)$

Step 1: Compute Base Capacitance Factor

First, find the term $(ε₀ A)/(d)$ $\frac{\varepsilon_0 A}{d}$ in SI units:

$A = 4~cm² = 4 \times 10⁻⁴~m²$
$d = 1.77~mm = 1.77 \times 10⁻³~m$
$ε₀ = 8.85 \times 10⁻¹²~F/m$

(ε₀ A)/(d) = 8.85 \times 10⁻¹² \times 4 \times 10⁻⁴1.77 \times 10⁻³ = 35.4 \times 10⁻¹⁶1.77 \times 10⁻³ = 20 \times 10⁻¹³~F = 2~pF

Now find $C₁$ $C_1$ and $C₂$ $C_2$:

$C₁ = 10 \times 2~pF = 20~pF$
$C₂ = 6 \times 2~pF = 12~pF$

Calculate their series equivalent $Ceq$ $C_{\text{eq}}$:

Ceq = (C₁ C₂)/(C₁ + C₂) = (20 \times 12)/(20 + 12) = (240)/(32) = 7.5~pF

Step 2: Add Parallel Capacitor

Another capacitor of $Cₚ = 7.5~pF$ $C_p = 7.5\mathrm{~pF}$ is connected in parallel with the combination:

Cfinal = Ceq + Cₚ = 7.5~pF + 7.5~pF = 15~pF

Pattern Recognition

Sees: Dielectric boundary parallel to the plates → Series capacitors. Trap: Don't treat horizontally split layers as parallel; split in distance $d$ $d$ means series, while split in area $A$ $A$ means parallel. Shortcut: Notice $(35.4)/(1.77)$ $\frac{35.4}{1.77}$ is exactly $20$ $20$. This makes the numerical calculations incredibly clean! ✓

Chapter Mix

Class 12 Physics: Electrostatics

Q3 jee_main_2025_29_jan_evening Electric Dipole in Uniform Electric Field

An electric dipole is placed at a distance of

2~cm

from an infinite plane sheet having positive charge density

σ₀

. Choose the correct option from the following.

Electric Dipole in Uniform Electric Field diagram for Q3 - JEE Main 2025 Evening — The diagram illustrates an electric dipole with charges -q and +q aligned parallel to an infinite plane sheet of positive charge density.

A. $Torque on dipole is zero and net force is directed away from the sheet.$
B. $Torque on dipole is zero and net force acts towards the sheet.$
C. $Potential energy of dipole is minimum and torque is zero.$
D. $Potential energy and torque both are maximum$

Solution

Related Formula

E = (σ₀)/(2ε₀)

τ = p \times E

U = - p \cdot E

Core Logic

An infinite plane sheet produces a uniform electric field $E$ $\vec{E}$ directed normally away from the sheet.

As shown in the image layout, the dipole moment vector $p$ $\vec{p}$ (pointing from $-q$ $-q$ to $+q$ $+q$) is oriented parallel to the electric field vectors $E$ $\vec{E}$:

θ = 0°

Torque evaluation:

τ = pE (0°) = 0

Potential Energy evaluation:

U = -pE (0°) = -pE (Minimum)

Net Force evaluation:

Since the electric field is uniform, the force on $+q$ $+q$ balances the force on $-q$ $-q$, meaning $Fₙₑₜ = 0$ $F_{\text{net}} = 0$.

Pattern Recognition

When a dipole aligns perfectly with a uniform electric field ( $p ∥ E$ $\vec{p} \parallel \vec{E}$), it reaches stable equilibrium. Stable equilibrium fundamentally means minimum potential energy ( $U = -pE$ $U = -pE$) and zero torque.

Chapter Mix

Class 12 Physics: Electrostatics

Q5 jee_main_2025_29_jan_evening Gauss's Law and Flux

A point charge causes an electric flux of

-2 \times 10⁴ ~Nm²C⁻¹

to pass through a spherical Gaussian surface of

8.0~cm

radius, centred on the charge. The value of the point charge is :

(Given ε₀ = 8.85 \times 10⁻¹² ~C²N⁻¹m⁻²)

A. $-17.7 \times 10⁻⁸ ~C$ $-17.7 \times 10^{-8} \mathrm{~C}$
B. $-15.7 \times 10⁻⁸ ~C$ $-15.7 \times 10^{-8} \mathrm{~C}$
C. $17.7 \times 10⁻⁸ ~C$ $17.7 \times 10^{-8} \mathrm{~C}$
D. $15.7 \times 10⁻⁸ ~C$ $15.7 \times 10^{-8} \mathrm{~C}$

Solution

Related Formula

φ = qenclosedε₀

where, $φ$ $\phi$ = net electric flux through the closed surface $qenclosed$ $q_{\text{enclosed}}$ = net charge enclosed by the surface $ε₀$ $\epsilon_0$ = permittivity of free space

Core Logic

According to Gauss's Law, the total electric flux through a closed surface depends only on the charge enclosed inside it, completely independent of the radius of the surface.

Rearranging the formula to solve for $q$ $q$:

q = φ \cdot ε₀

Substitute the given values:

q = (-2 \times 10⁴ ~Nm²C⁻¹) \times (8.85 \times 10⁻¹² ~C²N⁻¹m⁻²)

q = -17.7 \times 10⁻⁸ ~C

Pattern Recognition

Distractor alert: The radius ( $8.0~cm$ $8.0\mathrm{~cm}$) is extra data meant to mislead. Gauss's flux depends entirely on the magnitude of the enclosed charge, not the physical surface area configuration.

Chapter Mix

Class 12 Physics: Electrostatics

Q14 jee_main_2025_29_jan_evening Sharing of Charges and Loss of Energy

A capacitor,

C₁ = 6

is charged to a potential difference of

V₀ = 5V

using a

5V

battery. The battery is removed and another capacitor,

C₂ = 12

is inserted in place of the battery. When the switch 'S' is closed, the charge flows between the capacitors for some time until equilibrium condition is reached. What are the charges

(q₁

and

q₂)

on the capacitors

C₁

and

C₂

when equilibrium condition is reached.

Sharing of Charges and Loss of Energy diagram for Q14 - JEE Main 2025 Evening — The circuit diagram displays two capacitors C1 and C2 with a switch S used to establish parallel connectivity between them.

A. $q₁ = 15 , q₂ = 30$
B. $q₁ = 30 , q₂ = 15$
C. $q₁ = 10 , q₂ = 20$
D. $q₁ = 20 , q₂ = 10$

Solution

Related Formula

Qtotal = C₁ V₀

Vc = QtotalC₁ + C₂

$q = C \cdot Vc$ $q = C \cdot V_c$

Core Logic

Initial Charge Calculation:

Before closing the switch, capacitor $C₁$ $C_1$ accumulates total charge:

q₁' = C₁ \cdot V₀ = 6 \times 5V = 30

Uncharged capacitor $C₂$ $C_2$ holds $q₂' = 0$ $q_2' = 0$.

Sharing of Charges Initial State diagram for Q14 - JEE Main 2025 Evening — The circuit diagram displays two capacitors C1 and C2 with a switch S used to establish parallel connectivity between them.

Redistribution at Equilibrium:

Closing the switch causes parallel redistribution until they reach a common potential $Vc$ $V_c$:

Vc = 30 + 06 + 12 = (30)/(18) = (5)/(3)~V

Final Charges:

q₁ = C₁ \cdot Vc = 6 \times (5)/(3) = 10

q₂ = C₂ \cdot Vc = 12 \times (5)/(3) = 20

Hence, the charges are $10$ $10\mu\mathrm{C}$ and $20$ $20\mu\mathrm{C}$ respectively.

Pattern Recognition

For parallel combination setups, the final charge splits in the exact direct ratio of their capacitances: $q₁ : q₂ = C₁ : C₂ = 6 : 12 = 1 : 2$ . Out of $30$ $30\mu\mathrm{C}$, the breakdown must cleanly be $10$ $10\mu\mathrm{C}$ and $20$ $20\mu\mathrm{C}$.

Chapter Mix

Class 12 Physics: Electrostatic Potential and Capacitance

Solution & Explanation

Related Formula

Core Logic

Step 1: Calculate Energy of Individual Charges in External Field

Step 2: Calculate Interaction Energy Between Charges

Step 3: Total Energy

Pattern Recognition

Chapter Mix

Reference Study Guides

More Electrostatics Previous-Year Questions — Page 8

Solution

Related Formula

Core Logic

Step 1: Ratio Calculation

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Compute Base Capacitance Factor

Step 2: Add Parallel Capacitor

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Pattern Recognition

Chapter Mix

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