Two small spherical balls of mass 10g each with charges -2mumathrmC$-2mumathrm{C}$ and 2mumathrmC$2mumathrm{C}$, are attached to two ends of very light rigid rod of length 20 cm. The arrangement is now placed near an infinite non-conducting charge sheet with uniform charge density of 100mumathrmC/m²$100mumathrm{C/m}^{2}$ such that length of rod makes an angle of 30°$30^{\circ}$ with electric field generated by charge sheet. Net torque acting on the rod is:
(Take ε*o = 8.85×10⁻¹²C²/Nm²$\epsilon*{o} = 8.85\times10^{-12}\mathrm{C}^{2}/\mathrm{Nm}^{2}$)
A.112 Nm
B.1.12 Nm
C.2.24 Nm
D.11.2 Nm
Solution & Explanation
Related Formula
Electric field due to an infinite non-conducting sheet:
E = (σ)/(2ε₀)$$E = \frac{\sigma}{2\varepsilon_0}$$
Torque on an electric dipole:
τ = pE θ$$\tau = pE\sin\theta$$
where p = qd$p = qd$ is the magnitude of the electric dipole moment.
Dipole torque vector field distribution alignment for Q19 - JEE Main 2025 Morning
Pattern Recognition
Equal and opposite charges on a rigid rod constitute an electric dipole. In a uniform field, the net translational force vanishes (Fₙₑₜ = 0$\vec{F}_{\text{net}} = \vec{0}$), leaving only a pure restoring torque τ = pE θ$\tau = pE\sin\theta$.
Evaluation Rubric / Model Answer
Option B: 1.12 Nm
Chapter Mix
Class 12 Physics: Electrostatics
More Electrostatics Previous-Year Questions — Page 2
Q42jee_main_2026_22_january_morningElectric Potential and Field
Electric field in a region is given by E = Ax i + By j$\vec{E} = Ax\hat{i} + By\hat{j}$, where A = 10~V / m²$A = 10~\mathrm{V / m^2}$ and B = 5~V / m²$B = 5~\mathrm{V / m^2}$. If the electric potential at a point (10, 20) is 500~V$500~\mathrm{V}$, then the electric potential at origin is \_\_\_\_ V.
A. 1000
B. 500
C. 2000
D. 0
Solution
Related Formula
V₂ - V₁ = -∫ E · d r$$V_2 - V_1 = -\int \vec{E} \cdot d\vec{r}$$
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
Q43jee_main_2026_22_january_morningCharged 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$\vec{E}$ 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.√(m²g² + q²E²)$\sqrt{m^2g^2 + q^2E^2}$
D.√(m²g² - q²E²)$\sqrt{m^2g^2 - q^2E^2}$
Solution
Related Formula
T = √((qE)² + (mg)²)$$T = \sqrt{(qE)^2 + (mg)^2}$$
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 = √((qE)² + (mg)²)$$T = \sqrt{(qE)^2 + (mg)^2}$$
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
Q30jee_main_2026_22_january_eveningElectric 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)$$V = \frac{kq}{r}$$Volume Conservation: N · ((4)/(3)π r³) = (4)/(3)π R³$$\text{Volume Conservation: } N \cdot \left(\frac{4}{3}\pi r^3\right) = \frac{4}{3}\pi R^3$$
Core Logic
From volume conservation of 3 coalescing droplets:
3 ((4)/(3)π r³) = (4)/(3)π R³ R = 31/3r$$3 \left(\frac{4}{3}\pi r^3\right) = \frac{4}{3}\pi R^3 \implies R = 3^{1/3}r$$
Total charge on resultant bigger bubble Q = 3q$Q = 3q$.
Calculating initial potential Vᵢ$V_i$ and final potential Vf$V_f$:
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
Q42jee_main_2026_22_january_eveningElectric 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$\vec{E}$) at the center O of the pentagon due to these five positive charges are :
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√(3)q8πε₀r² r$\vec{E} = \frac{5\sqrt{3}q}{8\pi\varepsilon_{0}r^{2}} \hat{r}$
C.V = (5q)/(4πε₀r)$V = \frac{5q}{4\pi\varepsilon_0r}$ and E = (5q)/(4πε₀r²) r$\vec{E} = \frac{5q}{4\pi\varepsilon_0r^2}\hat{r}$
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
Q48jee_main_2026_22_january_eveningSharing of Charges between Capacitors
A capacitor P with capacitance 10 × 10⁻⁶$10 \times 10^{-6}$ 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 × 10⁻⁶$20 \times 10^{-6}$ F. The charge on capacitor Q when equilibrium is established will be α × 10⁻⁵$\alpha \times 10^{-5}$ C (assume capacitor Q does not have any charge initially), the value of α$\alpha$ is ____.
Practice past-year questions one chapter at a time. Pick an exam → subject → chapter and get every PYQ for that topic — pulled together from all past papers — with the chapter's key formulas alongside.
