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

Year

2026

2025

2024

Total

Questions

Solution & Explanation

Related Formula

E = (k Q)/(r²)

x = d1 + \sqrt((q₂)/(q₁))

Core Logic

Let the null point where the electric field is zero be at $(x, 0, 0)$ $(x, 0, 0)$ where $0 < x < d$ $0 < x < d$.

At this point, the fields due to both charges are equal in magnitude and opposite in direction:

(k q)/(x²) = (k (9q))/((d - x)²)

Taking the square root on both sides:

(1)/(x) = (3)/(d - x) d - x = 3x

4x = d x = (d)/(4)

Thus, the coordinates of the null point are $((d)/(4), 0, 0)$ $\left(\frac{d}{4}, 0, 0\right)$.

Step 1: Final Conclusion

The point where the electric field vanishes is $((d)/(4), 0, 0)$ $\left(\frac{d}{4}, 0, 0\right)$.

Pattern Recognition

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: $x = d1 + \sqrt(q₂/q₁)$ $x = \frac{d}{1 + \sqrt{q_2/q_1}}$ measured from charge $q₁$ $q_1$.

Chapter Mix

Class 12 Physics: Electrostatics

Null point calculation coordinates diagram

More Electrostatics Previous-Year Questions — Page 2

Q42 jee_main_2026_22_january_morning Electric Potential and Field

Electric field in a region is given by

E = Ax i + By j

, where

A = 10~V / m²

and

B = 5~V / m²

. If the electric potential at a point (10, 20) is

500~V

, then the electric potential at origin is \_\_\_\_ V.

A. 1000
B. 500
C. 2000
D. 0

Solution

Related Formula

V₂ - V₁ = -\int E \cdot d r

Core Logic

Using potential difference relation:

500 - V₀ = -\int(0,0)(10,20) (10x i + 5y j) \cdot (dx i + dy j)

500 - V₀ = -[5x² + (5y²)/(2)](0,0)(10,20)

V₀ - 500 = 500 + 1000 V₀ = 2000 V

Pattern Recognition

Sees: Electric field vector function given, find potential at origin. Shortcut: Integrate line integral of electric field from origin to given point. Check: Matches option (3). ✓

Chapter Mix

Class 12 Physics: Electrostatics

Q43 jee_main_2026_22_january_morning Charged Pendulum in Electric Field

A simple pendulum has a bob with mass m and charge q. The pendulum string has negligible mass. When a uniform and horizontal electric field

E

is applied, the tension in the string changes. The final tension in the string, when pendulum attains an equilibrium position is \_\_\_\_. (g : acceleration due to gravity)

A. $mg - qE$ $mg - qE$
B. $mg + qE$ $mg + qE$
C. $\sqrt(m²g² + q²E²)$ $\sqrt{m^2g^2 + q^2E^2}$
D. $\sqrt(m²g² - q²E²)$ $\sqrt{m^2g^2 - q^2E^2}$

Solution

Related Formula

T = \sqrt((qE)² + (mg)²)

Core Logic

Solution pendulum diagram for Q43 - JEE Main 2026 Morning

At equilibrium, the effective forces acting on the bob are vertical gravitational force $mg$ $mg$ and horizontal electric force $qE$ $qE$. The string tension balances the resultant of these orthogonal forces:

T = \sqrt((qE)² + (mg)²)

Pattern Recognition

Sees: Charged pendulum in horizontal electric field. Shortcut: Combine orthogonal forces ( $mg$ $mg$ downwards and $qE$ $qE$ horizontally) via Pythagorean vector addition. Check: Matches option (3). ✓

Chapter Mix

Class 12 Physics: Electrostatics

Q30 jee_main_2026_22_january_evening Electric Potential of Coalescing Bubbles

Three small identical bubbles of water having same charge on each coalesce to form a bigger bubble. Then the ratio of the potentials on one initial bubble and that on the resultant bigger bubble is :

A. $1:31/3$ $1:3^{1/3}$
B. $1:22/3$ $1:2^{2/3}$
C. $32/3:1$ $3^{2/3}:1$
D. $1:32/3$ $1:3^{2/3}$

Solution

Related Formula

V = (kq)/(r)

Volume Conservation: N \cdot ((4)/(3)π r³) = (4)/(3)π R³

Core Logic

From volume conservation of 3 coalescing droplets:

3 ((4)/(3)π r³) = (4)/(3)π R³ R = 31/3r

Total charge on resultant bigger bubble $Q = 3q$ $Q = 3q$.

Calculating initial potential $Vᵢ$ $V_i$ and final potential $Vf$ $V_f$:

Vᵢ = (kq)/(r)

Vf = (k(3q))/(R) = 3kq31/3r = 32/3 (kq)/(r)

Ratio of initial to final potential:

(Vᵢ)/(Vf) = 132/3 = 1 : 32/3

Coalescing charged bubbles diagram for Q30 - JEE Main 2026 Evening

Step 1: Final Conclusion

The ratio of potentials is $1 : 32/3$ $1 : 3^{2/3}$.

Pattern Recognition

Coalescing droplets rule: For $N$ $N$ identical drops, $R = N1/3r$ $R = N^{1/3}r$ and $Q = Nq$ $Q = Nq$. Potential ratio $Vᵢ / Vf = 1 / N2/3$ $V_i / V_f = 1 / N^{2/3}$. For $N=3$ $N=3$, ratio is $1 / 32/3$ $1 / 3^{2/3}$.

Chapter Mix

Class 12 Physics: Electrostatic Potential and Capacitance

Q42 jee_main_2026_22_january_evening Electric Field and Potential of Polygon of Charges

Five positive charges each having charge q are placed at the vertices of a pentagon as shown in the figure. The electric potential (V) and the electric field (

E

) at the center O of the pentagon due to these five positive charges are :

Regular pentagon charged vertices diagram for Q42 - JEE Main 2026 Evening — The figure illustrates a regular pentagon with five equal positive charges q placed at each vertex at distance r from center O.

A. $V = (5q)/(4πε₀r)$ $V = \frac{5q}{4\pi\varepsilon_0r}$ and $E = 0$ $\vec{E} = 0$
B. $V = 5q4πε₀r$ $V = \frac{5q}{4\pi\varepsilon_{0}r}$ and $E = 5\sqrt(3)q8πε₀r² r$
C. $V = (5q)/(4πε₀r)$ $V = \frac{5q}{4\pi\varepsilon_0r}$ and $E = (5q)/(4πε₀r²) r$
D. $V = 0$ $V = 0$ and $E = 0$ $\vec{E} = 0$

Solution

Related Formula

V = Σ (k qᵢ)/(r)

Ecenter = Σ Eᵢ = 0 (Symmetric Polygon)

Core Logic

Due to spatial symmetry of identical charges at the 5 vertices of a regular pentagon, vector sum of electric fields at center O cancels out:

$E = 0$ $\vec{E} = 0$

Electric potential is a scalar sum:

V = 5 \times ((q)/(4πε₀ r)) = (5q)/(4πε₀ r)

Step 1: Final Conclusion

Option (1) gives the correct values $V = (5q)/(4πε₀r)$ $V = \frac{5q}{4\pi\varepsilon_0r}$ and $E = 0$ $\vec{E} = 0$.

Pattern Recognition

Symmetry rule: Identical charges at vertices of any regular polygon $Ecenter = 0$ $\implies \vec{E}_{center} = 0$. Potential is scalar addition $V = N (kq)/(r)$ $V = N \frac{kq}{r}$.

Chapter Mix

Class 12 Physics: Electrostatics

Q48 jee_main_2026_22_january_evening Sharing of Charges between Capacitors

A capacitor P with capacitance

10 \times 10⁻⁶

F is fully charged with a potential difference of 6.0 V and disconnected from the battery. The charged capacitor P is connected across another capacitor Q with capacitance

20 \times 10⁻⁶

F. The charge on capacitor Q when equilibrium is established will be

α \times 10⁻⁵

C (assume capacitor Q does not have any charge initially), the value of

α

is ____.

Numerical Answer. Answer: 4 to 4

Solution

Related Formula

Vcommon = (C₁ V₁ + C₂ V₂)/(C₁ + C₂)

Q₂ = C₂ Vcommon

Core Logic

Given $C₁ = 10 \times 10⁻⁶ ~F, V₁ = 6.0 ~V$ and $C₂ = 20 \times 10⁻⁶ ~F, V₂ = 0 ~V$ :

Vcommon = 10⁻⁵ \times 6 + 010⁻⁵ + 2 \times 10⁻⁵ = 6 \times 10⁻⁵3 \times 10⁻⁵ = 2 ~V

Calculating final charge on capacitor Q ( $C₂$ $C_2$):

Q₂ = C₂ Vcommon = (20 \times 10⁻⁶ ~F) \times 2 ~V = 40 \times 10⁻⁶ ~C = 4 \times 10⁻⁵ ~C

Comparing with $α \times 10⁻⁵ ~C α = 4$ .

Step 1: Final Conclusion

The value of $α$ $\alpha$ is 4.

Pattern Recognition

Charge distribution rule: Total initial charge $Qtotal = C₁ V₁ = 60$ . Final charge splits in proportion to capacitance ratio $C₂ / (C₁+C₂) = 2/3$ $C_2 / (C_1+C_2) = 2/3$. $Q₂ = (2/3) \times 60 = 40 = 4 \times 10⁻⁵~C$ .

Chapter Mix

Class 12 Physics: Electrostatic Potential and Capacitance

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 2

Solution

Related Formula

Core Logic

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Final Conclusion

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Final Conclusion

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Final Conclusion

Pattern Recognition

Chapter Mix

More Electrostatics Questions — jee_main_2025_02_april_morning

Select Your Target Exam

Full Mock Test Hub

Topic-wise Practice Hub

Exam Analysis Tools

Chapter Weightage & Trends

Topic-wise Practice Hub

JEE Main Exam Papers

Chapter Weightage Board

Formula Recognition

Exam Questions

Select an Exam

Welcome back

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: