Given below are two statements: one is labelled as Assertion (A) and the other is labelled as Reason (R). Assertion (A): The outer body of an air craft is made of metal which protects persons sitting inside from lightning-strikes. Reason (R): The electric field inside the cavity enclosed by a conductor is zero. In the light of the above statements, chose the most appropriate answer from the options given below:

According to electrostatic shielding, the electric field inside a cavity of a conductor is always zero, regardless of the size and shape of the cavity and regardless of any charges located outside or on the conductor's surface. Therefore, when lightning strikes a metal aircraft, the entire charge stays on the outer metallic surface and flows down without producing an electric field inside, keeping passengers safe. * **Assertion (A):** Correct, passengers are protected from lightning because of the metallic body shield. * **Reason (R):** Correct, the field inside a cavity of a conductor is zero. * **Explanation:** Since the zero field property is precisely why passengers are protected, (R) correctly explains (A). Metallic shell / shield configuration always establishes Einside = 0. This shielding mechanism directly underpins safety features in lightning scenarios for cars and airplanes. Class 12 Physics: Electrostatics

Given below are two statements: one is labelled as Assertion (A) and the other is labelled as Reason (R). Assertion (A): The outer body of an air craft is made of metal which protects persons sitting inside from lightning-strikes. Reason (R): The electric field inside the cavity enclosed by a conductor is zero. In the light of the above statements, chose the most appropriate answer from the options given below:

According to electrostatic shielding, the electric field inside a cavity of a conductor is always zero, regardless of the size and shape of the cavity and regardless of any charges located outside or on the conductor's surface. Therefore, when lightning strikes a metal aircraft, the entire charge stays on the outer metallic surface and flows down without producing an electric field inside, keeping passengers safe. * **Assertion (A):** Correct, passengers are protected from lightning because of the metallic body shield. * **Reason (R):** Correct, the field inside a cavity of a conductor is zero. * **Explanation:** Since the zero field property is precisely why passengers are protected, (R) correctly explains (A). Metallic shell / shield configuration always establishes Einside = 0. This shielding mechanism directly underpins safety features in lightning scenarios for cars and airplanes. Class 12 Physics: Electrostatics

Given below are two statements: one is labelled as Assertion (A) and the other is labelled as Reason (R). Assertion…

Year	2026	2025	2024	Total
Questions	24	39	16	79

Year

2026

2025

2024

Total

Questions

Solution & Explanation

Core Logic

According to electrostatic shielding, the electric field inside a cavity of a conductor is always zero, regardless of the size and shape of the cavity and regardless of any charges located outside or on the conductor's surface. Therefore, when lightning strikes a metal aircraft, the entire charge stays on the outer metallic surface and flows down without producing an electric field inside, keeping passengers safe.

Step 1: Statement Evaluation

Assertion (A): Correct, passengers are protected from lightning because of the metallic body shield.
Reason (R): Correct, the field inside a cavity of a conductor is zero.
Explanation: Since the zero field property is precisely why passengers are protected, (R) correctly explains (A).

Pattern Recognition

Metallic shell / shield configuration always establishes $Einside = 0$ $E_{\text{inside}} = 0$. This shielding mechanism directly underpins safety features in lightning scenarios for cars and airplanes.

Chapter Mix

Class 12 Physics: Electrostatics

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

Core Logic

Step 1: Statement Evaluation

Pattern Recognition

Chapter Mix

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More Electrostatics Previous-Year Questions — Page 2

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Related Formula

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