Given below are two statements: one is labelled as Assertion A and the other is labelled as Reason R. Assertion A: Work done in moving a test charge between two points inside a uniformly charged spherical shell is zero, no matter which path is chosen. Reason R: Electrostatic potential inside a uniformly charged spherical shell is constant and is same as that on the surface of the shell. In the light of the above statements, choose the correct answer from the options given below:

where, WA arrow B = work done in moving a test charge q from point A to B VA, VB = electrostatic potentials at points A and B For a uniformly charged spherical shell of radius R and charge Q, the electric field inside the shell is zero (E = 0). Consequently, the electric potential V remains constant throughout the interior of the shell and equals its value on the surface: Since the potential is identical at all interior points (VA = VB), the potential difference is zero: Thus, the work done in moving any test charge inside is strictly zero: 1. **Assertion A**: "Work done in moving a test charge inside is zero..." - This is **True**. 2. **Reason R**: "Electrostatic potential inside is constant and same as on the surface..." - This is **True** and directly explains why the potential difference Delta V = 0, making the work done zero. Therefore, both statements are true and R is the correct explanation of A. Sees: "Uniformly charged spherical shell" + "Work done inside" → Potential difference Delta V = 0 W = 0. Shortcut: Since Einside = 0, potential inside is flat/constant. No potential difference means zero work. Both statements are true and connected. ✓ 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: Work done in moving a test charge between two points inside a uniformly charged spher

where, WA arrow B = work done in moving a test charge q from point A to B VA, VB = electrostatic potentials at points A and B For a uniformly charged spherical shell of radius R and charge Q, the electric field inside the shell is zero (E = 0). Consequently, the electric potential V remains constant throughout the interior of the shell and equals its value on the surface: Since the potential is identical at all interior points (VA = VB), the potential difference is zero: Thus, the work done in m

Given below are two statements: one is labelled as Assertion A and the other is labelled as Reason R. Assertion A: Work done in moving a test charge between two points inside a uniformly charged spherical shell is zero, no matter which path is chosen. Reason R: Electrostatic potential inside a uniformly charged spherical shell is constant and is same as that on the surface of the shell. In the light of the above statements, choose the correct answer from the options given below:

where, WA arrow B = work done in moving a test charge q from point A to B VA, VB = electrostatic potentials at points A and B For a uniformly charged spherical shell of radius R and charge Q, the electric field inside the shell is zero (E = 0). Consequently, the electric potential V remains constant throughout the interior of the shell and equals its value on the surface: Since the potential is identical at all interior points (VA = VB), the potential difference is zero: Thus, the work done in moving any test charge inside is strictly zero: 1. **Assertion A**: "Work done in moving a test charge inside is zero..." - This is **True**. 2. **Reason R**: "Electrostatic potential inside is constant and same as on the surface..." - This is **True** and directly explains why the potential difference Delta V = 0, making the work done zero. Therefore, both statements are true and R is the correct explanation of A. Sees: "Uniformly charged spherical shell" + "Work done inside" → Potential difference Delta V = 0 W = 0. Shortcut: Since Einside = 0, potential inside is flat/constant. No potential difference means zero work. Both statements are true and connected. ✓ 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: Work done in moving a test charge between two points inside a uniformly charged spher

where, WA arrow B = work done in moving a test charge q from point A to B VA, VB = electrostatic potentials at points A and B For a uniformly charged spherical shell of radius R and charge Q, the electric field inside the shell is zero (E = 0). Consequently, the electric potential V remains constant throughout the interior of the shell and equals its value on the surface: Since the potential is identical at all interior points (VA = VB), the potential difference is zero: Thus, the work done in m

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

Year	2026	2025	2024	Total
Questions	24	39	16	79

Year

2026

2025

2024

Total

Questions

Solution & Explanation

Related Formula

WA arrow B = q(VB - VA)

where, $WA arrow B$ $W_{A \rightarrow B}$ = work done in moving a test charge $q$ $q$ from point $A$ $A$ to $B$ $B$ $VA, VB$ $V_A, V_B$ = electrostatic potentials at points $A$ $A$ and $B$ $B$

Core Logic

For a uniformly charged spherical shell of radius $R$ $R$ and charge $Q$ $Q$, the electric field inside the shell is zero ( $E = 0$ $E = 0$).

Consequently, the electric potential $V$ $V$ remains constant throughout the interior of the shell and equals its value on the surface:

Vinside = Vsurface = (1)/(4πε₀) (Q)/(R)

Since the potential is identical at all interior points ( $VA = VB$ $V_A = V_B$), the potential difference is zero:

Δ V = VB - VA = 0

Thus, the work done in moving any test charge inside is strictly zero:

W = q Δ V = 0

Step 1: Analyzing the Statements

Assertion A: "Work done in moving a test charge inside is zero..." - This is True.
Reason R: "Electrostatic potential inside is constant and same as on the surface..." - This is True and directly explains why the potential difference $Δ V = 0$ $\Delta V = 0$, making the work done zero.

Therefore, both statements are true and $R$ $R$ is the correct explanation of $A$ $A$.

Pattern Recognition

Sees: "Uniformly charged spherical shell" + "Work done inside" → Potential difference $Δ V = 0 W = 0$ $\Delta V = 0 \implies W = 0$. Shortcut: Since $Einside = 0$ $E_{\text{inside}} = 0$, potential inside is flat/constant. No potential difference means zero work. Both statements are true and connected. ✓

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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 $1 / 32/3$ $1 / 3^{2/3}$.

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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

Core Logic

Step 1: Analyzing the Statements

Pattern Recognition

Chapter Mix

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