A point charge +q is placed at the origin. A second point charge +9q is placed at (d, 0, 0) in Cartesian coordinate system. The point in between them where the electric field vanishes is:

Let the null point where the electric field is zero be at (x, 0, 0) where 0 < x < d. At this point, the fields due to both charges are equal in magnitude and opposite in direction: Taking the square root on both sides: Thus, the coordinates of the null point are ((d/4), 0, 0). The point where the electric field vanishes is ((d/4), 0, 0). For two like charges, the zero-field null point always lies along the line joining them and is closer to the smaller charge. Use the standard shortcut: x = (d/1 + sqrt(q2/q1)) measured from charge q1. Class 12 Physics: Electrostatics

A point charge +q is placed at the origin. A second point charge +9q is placed at (d, 0, 0) in Cartesian coordinate system. The point in between them where the electric field vanishes is:

Let the null point where the electric field is zero be at (x, 0, 0) where 0 < x < d. At this point, the fields due to both charges are equal in magnitude and opposite in direction: Taking the square root on both sides: Thus, the coordinates of the null point are ((d/4), 0, 0). The point where the electric field vanishes is ((d/4), 0, 0). For two like charges, the zero-field null point always lies along the line joining them and is closer to the smaller charge. Use the standard shortcut: x = (d/1

A point charge +q is placed at the origin. A second point charge +9q is placed at (d, 0, 0) in Cartesian coordinate system. The point in between them where the electric field vanishes is:

Let the null point where the electric field is zero be at (x, 0, 0) where 0 < x < d. At this point, the fields due to both charges are equal in magnitude and opposite in direction: Taking the square root on both sides: Thus, the coordinates of the null point are ((d/4), 0, 0). The point where the electric field vanishes is ((d/4), 0, 0). For two like charges, the zero-field null point always lies along the line joining them and is closer to the smaller charge. Use the standard shortcut: x = (d/1 + sqrt(q2/q1)) measured from charge q1. Class 12 Physics: Electrostatics

A point charge +q is placed at the origin. A second point charge +9q is placed at (d, 0, 0) in Cartesian coordinate system. The point in between them where the electric field vanishes is:

Let the null point where the electric field is zero be at (x, 0, 0) where 0 < x < d. At this point, the fields due to both charges are equal in magnitude and opposite in direction: Taking the square root on both sides: Thus, the coordinates of the null point are ((d/4), 0, 0). The point where the electric field vanishes is ((d/4), 0, 0). For two like charges, the zero-field null point always lies along the line joining them and is closer to the smaller charge. Use the standard shortcut: x = (d/1

A point charge +q is placed at the origin. A second point charge +9q is placed at (d, 0, 0) in Cartesian coordinate…

Year	2026	2025	2024	Total
Questions	24	39	16	79

Q27 jee_main_2026_24_january_morning Electric Dipole

Three charges

+2q

,

+3q

and

-4q

are situated at

(0,-3a)

,

(2a, 0)

and

(-2a,0)

respectively in the xy plane. The resultant dipole moment about origin is

A. $2qa(3 j- i)$
B. $2qa(3 i-7 j)$ $2qa\left(3\hat{i}-7\hat{j}\right)$
C. $2qa(7 i-3 j)$
D. $2qa(3 j-7 i)$

Solution

Related Formula

p = Σ qᵢ rᵢ

Core Logic

The resultant dipole moment for a system of charges is given by:

p = q₁ r₁ + q₂ r₂ + q₃ r₃

Substituting the given values:

p = (2q)(-3a) j + (3q)(2a) i + (-4q)(-2a) i

p = -6qa j + 6qa i + 8qa i

p = 14qa i - 6qa j

p = 2qa(7 i - 3 j)

Pattern Recognition

For a system of point charges where the net charge is non-zero, the dipole moment depends on the origin. However, taking the standard formula $Σ qᵢ rᵢ$ $\sum q_i r_i$ yields the required mathematical expression directly.

Chapter Mix

Class 12 Physics: Electric Charges and Fields

Q40 jee_main_2026_24_january_morning Gauss's Law

The electrostatic potential in a charged spherical region of radius r varies as

V = ar³ + b

, where a and b are constants. The total charge in the sphere of unit radius is

α \times π a in₀

. The value of

α

is ____. (permittivity of vacuum is

in₀

)

A. $-12$ $-12$
B. $-6$ $-6$
C. $-9$ $-9$
D. $-8$ $-8$

Solution

Related Formula

E = -(dV)/(dr)

\oint E \cdot d A = qencε₀

Core Logic

Given potential $V = ar³ + b$ $V = ar^3 + b$. The electric field $E$ $E$ is given by the negative gradient of potential:

E = -(dV)/(dr) = -(d)/(dr)(ar³ + b) = -3ar²

Using Gauss's Law to find the enclosed charge for a spherical region:

Φclosed = E \cdot A = qencε₀

Step 1: Enclosed Charge Calculation

The surface area of a sphere of radius $r=1$ $r=1$ is $A = 4π(1)² = 4π$ $A = 4\pi(1)^2 = 4\pi$. The electric field at $r=1$ $r=1$ is:

E = -3a(1)² = -3a

So,

qenc = ε₀ \cdot E \cdot A = ε₀ (-3a) (4π) = -12π a ε₀

Comparing with the given expression $α \times π a ε₀$ $\alpha \times \pi a \epsilon_0$, we get: $α = -12$ $\alpha = -12$

Pattern Recognition

For spherical symmetry, extracting total charge enclosed is fastest using Gauss's Law at the boundary surface rather than integrating local charge density $ρ(r)$ $\rho(r)$ using Poisson's equation.

Chapter Mix

Class 12 Physics: Electrostatics

Q44 jee_main_2026_24_january_morning Electric Potential of Spherical Shells

There are three co-centric conducting spherical shells A, B and C of radii a, b and c respectively. The potential of the spheres A, B and C respectively, are :

A. $14π in₀( q₁ + q₂ + q₃a), 14π in₀( q₁ + q₂ + q₃b), 14π in₀( q₁ + q₂ + q₃c)$
B. $14π in₀( q₁ + q₂ + q₃a), 14π in₀( q₁ + q₂b + q₃c), 14π in₀( q₁a + q₂b + q₃c)$
C. $14π in₀( q₁a + q₂b + q₃c), 14πin₀( q₁ + q₂b + q₃c), 14πin₀( q₁ + q₂ + q₃c)$
D. $(1)/(4πε₀)( q₁a + q₂b + q₃c),(1)/(4πε₀)( q₁ + q₂ + q₃b),(1)/(4πε₀)( q₁ + q₂ + q₃c)$

Solution

Related Formula

V = Σ K qᵢreffective

where $reffective = (rshell, robservation)$ .

Core Logic

Three cocentric conducting spherical shells

Potential at surface of inner sphere A (radius $a$ $a$):

VA = (K q₁)/(a) + (K q₂)/(b) + (K q₃)/(c) = (1)/(4 π ε₀) ((q₁)/(a) + (q₂)/(b) + (q₃)/(c))

Potential at surface of middle sphere B (radius $b$ $b$): For charge $q₁$ $q_1$, this is an outside point. For $q₂$ $q_2$, it's on the surface. For $q₃$ $q_3$, it's inside.

VB = (K q₁)/(b) + (K q₂)/(b) + (K q₃)/(c) = (1)/(4 π ε₀) ((q₁ + q₂)/(b) + (q₃)/(c))

Potential at surface of outer sphere C (radius $c$ $c$): For charges $q₁$ $q_1$ and $q₂$ $q_2$, this is an outside point.

VC = (K q₁)/(c) + (K q₂)/(c) + (K q₃)/(c) = (1)/(4 π ε₀) ((q₁ + q₂ + q₃)/(c))

Step 1: Selection of Correct Option

Matching these results shows option (3) represents the potentials perfectly.

Pattern Recognition

Potential on a sphere from inner charges uses its own radius, whereas potential from outer shells uses the outer shell's radius. Inner charges "collapse" computationally to the sphere's center.

Chapter Mix

Class 12 Physics: Electrostatics

Q35 jee_main_2026_24_january_evening Capacitors with Dielectrics

Three parallel plate capacitors each with area A and separation d are filled with two dielectric (

k₁

and

k₂

) in the following fashion. Which of the following is true? (

k₁ > k₂

)

Capacitors with Dielectrics diagram for Q35 - JEE Main 2026 Evening — Three different capacitor configurations showing dielectrics k1 and k2 split vertically or horizontally.

A. $CB > CC > CA$ $\mathrm{C_B > C_C > C_A}$
B. $CC > CB > CA$ $\mathrm{C_C > C_B > C_A}$
C. $CC > CA > CB$ $\mathrm{C_C > C_A > C_B}$
D. $CA > CC > CB$ $\mathrm{C_A > C_C > C_B}$

Solution

Related Formula

C = (ε₀ A)/(d)

Core Logic

Let $C = (ε₀ A)/(d)$ $C = \frac{\epsilon_0 A}{d}$. We decompose the configurations into equivalent circuits.

For $CA$ $C_A$:

CA = (K₁ C)/(2) + (K₁ K₂ C)/(K₁ + K₂) = K₁ C [ (K₁ + 2K₂)/(2(K₁ + K₂)) ]

Step 1: Calculate CB

For $CB$ $C_B$:

CB = (K₂ C)/(2) + (K₁ K₂ C)/(K₁ + K₂) = K₂ C [ (K₁ + 2K₂)/(2(K₁ + K₂)) ]

Step 2: Calculate CC

For $CC$ $C_C$:

CC = (2K₁ K₂ C)/((K₁ + K₂))

Step 3: Comparison

Since $K₁ > K₂$ $K_1 > K_2$: Comparing $CA$ $C_A$ and $CC$ $C_C$, and $CC$ $C_C$ and $CB$ $C_B$, algebraic manipulation proves: $CA > CC > CB$ $C_A > C_C > C_B$

Pattern Recognition

Symmetry dictates the capacity. Adding more of the higher dielectric constant material ( $K₁$ $K_1$) in parallel paths effectively boosts total capacitance significantly, while stacking lower $K₂$ $K_2$ diminishes it.

Chapter Mix

Class 12 Physics: Electrostatics

Q47 jee_main_2026_24_january_evening Coulomb's Law and Continuous Charge Distribution

A point charge

q = 1 μ C

is located at a distance 2 cm from one end of a thin insulating wire of length 10 cm having a charge

Q = 24 μ C

, distributed uniformly along its length, as shown in figure. Force between q and wire is ____ N.

( Use 14 π ε_ 0 = 9 \times 1 0 ^ 9 N.m ^ 2 / C ^ 2)

Coulomb's Law and Continuous Charge Distribution diagram for Q47 - JEE Main 2026 Evening — A point charge placed 2 cm coaxially from a uniformly charged wire of length 10 cm.

Numerical Answer. Answer: 90 to 90

Solution

Related Formula

dF = (k \cdot q \cdot dQ)/(x²)

where $dQ = λ dx$ $dQ = \lambda dx$

Core Logic

Linear charge density of the wire:

λ = (Q)/(L) = 24 \times 10⁻⁶0.1 C/m

Taking a small element $dx$ $dx$ at distance $x$ $x$ from point charge $q$ $q$, the force is:

F = \int dF = \int2 cm12 cm (k q λ dx)/(x²)

Step 1: Integration

F = k q λ [ -(1)/(x) ]0.020.12

F = k q λ ( (1)/(0.02) - (1)/(0.12) )

F = k q λ ( 50 - (100)/(12) )

Step 2: Values Substitution

F = (9 \times 10⁹) (10⁻⁶) ( 24 \times 10⁻⁶10⁻¹) ( 12 \times 10⁻² - 112 \times 10⁻²)

F = 9 \times 10³ \times 2.4 \times 10⁻⁴ \times ( 12-224 \times 10⁻² )

Wait, computing directly from the exact equation provided:

F = (9 \times 10⁹) (10⁻⁶) ( 24 \times 10⁻⁶10⁻¹) ((5)/(12)) \times 10²

F = 9 \times 24 \times (5)/(12) = 90 N

Pattern Recognition

For a point charge $q$ $q$ acting on a line charge of length $L$ $L$ separated by distance $a$ $a$, $F = (k q Q)/(a(a+L))$ $F = \frac{k q Q}{a(a+L)}$. Directly applying this bypasses the integration completely.

Chapter Mix

Class 12 Physics: Electrostatics

A point charge +q$+q$ is placed at the origin. A second point charge +9q$+9q$ is placed at (d, 0, 0)$(d, 0, 0)$ in Cartesian coordinate system. The point in between them where the electric field vanishes is:

Solution & Explanation

Related Formula

Core Logic

Step 1: Final Conclusion

Pattern Recognition

Chapter Mix

Reference Study Guides

More Electrostatics Previous-Year Questions — Page 4

Solution

Related Formula

Core Logic

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Enclosed Charge Calculation

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Selection of Correct Option

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Calculate CB

Step 2: Calculate CC

Step 3: Comparison

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Integration

Step 2: Values Substitution

Pattern Recognition

Chapter Mix

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A point charge $+q$ $+q$ is placed at the origin. A second point charge $+9q$ $+9q$ is placed at $(d, 0, 0)$ $(d, 0, 0)$ in Cartesian coordinate system. The point in between them where the electric field vanishes is: