Consider two infinitely large plane parallel conducting plates as shown below. The plates are uniformly charged with a…

Year	2026	2025	2024	Total
Questions	24	39	16	79

Year

2026

2025

2024

Total

Questions

Solution & Explanation

Related Formula

E = σᵢₙₙₑᵣε₀

Core Logic

For parallel conducting plates of large area, charges redistribute on the outer and inner faces to maintain electrostatic equilibrium. Total charge per unit area on Plate 1: $q₁ = σ$ $q_1 = \sigma$ Total charge per unit area on Plate 2: $q₂ = -2σ$ $q_2 = -2\sigma$

The outer surface charge density on the far sides of both plates must be equal:

σouter = (q₁ + q₂)/(2) = (σ - 2σ)/(2) = -(σ)/(2)

Now, compute the charges on the inner facing surfaces:

Inner face of Plate 1:

σᵢₙₙₑᵣ₁ = q₁ - σouter = σ - (-(σ)/(2)) = (3σ)/(2)

Inner face of Plate 2:

σᵢₙₙₑᵣ₂ = q₂ - σouter = -2σ - (-(σ)/(2)) = -(3σ)/(2)

In the region between the plates, both inner surfaces create an electric field in the same direction (away from the positive plate 1 and towards negative plate 2):

E = σᵢₙₙₑᵣ₁2ε₀ + |σᵢₙₙₑᵣ₂|2ε₀ = (3σ/2)/(2ε₀) + (3σ/2)/(2ε₀) = (3σ)/(2ε₀)

Thus, the electrostatic force on $+q$ $+q$ is:

F = q E = (3σ q)/(2ε₀)

Step 1: Final Conclusion

The force experienced by the point charge $+q$ $+q$ is $(3σ q)/(2ε₀)$ $\frac{3\sigma q}{2\epsilon_0}$.

Pattern Recognition

For conducting plates with total charges $Q₁$ $Q_1$ and $Q₂$ $Q_2$, always calculate the outer charge first: $Qouter = (Q₁+Q₂)/(2)$ $Q_{\text{outer}} = \frac{Q_1+Q_2}{2}$. The field inside the gap is exclusively due to the inner surfaces: $E = σᵢₙₙₑᵣε₀$ .

Chapter Mix

Class 12 Physics: Electrostatics

Charge distribution and field diagram — Two parallel plates with charges +sigma and -2sigma and a point charge +q at the midpoint.

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

Solution & Explanation

Related Formula

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Step 1: Final Conclusion

Pattern Recognition

Chapter Mix

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More Electrostatics Previous-Year Questions — Page 2

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