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Paper Questions (Raw)

Subject: Physics
Topic: 21. Electrostatics
Subtopic: 21.7 P.E of System of Charges
Test Type: NEET
Created At: 2025-12-14 09:55:19
Record ID: 693e89871ca1a96b64116aeb
Raw snapshot: /physics_outputs/693e89871ca1a96b64116aeb_raw.html
Q1. Three charges, (+q),(+q),(-q), are placed at the three corners of an equilateral triangle, of side (L). The potential energy of the system is
1) (1) (V=0, E = 0)
2) (2) (V=0, E \neq 0)
3) (3) (V \neq 0, E=0)
4) (4) (V \neq 0, E \neq 0)
Q2. Two similar charged small spheres, each having a charge (Q) are suspended from a point by a thread of length (L) . At equilibrium, the angle made by the threads with the vertical is (30^{\circ}). The electrostatic PE of the system at equilibrium, is
1) (1) (\frac{kQ^2}{L})
2) (2) (\frac{\sqrt{3} kQ^2}{2 L})
3) (3) (\frac{kQ^2}{\sqrt{3} L})
4) (4) (\frac{\sqrt{3} kQ^2}{L})
Q3. P.E. of system of charges, shown in the figure, is
1) (1) (\frac{3}{4 \pi \varepsilon_0} \frac{q^2}{r})
2) (2) (\frac{1}{4 \pi \varepsilon_0} \frac{q^2}{r})
3) (3) (-\frac{1}{4 \pi \varepsilon_0} \frac{q^2}{r})
4) (4) (-\frac{3}{4 \pi \varepsilon_0} \frac{q^2}{r})
Q4. The value of (Q) , for which the PE of system is zero, is
1) (1) (-\frac{q}{\sqrt{2}})
2) (2) (-0.47 q)
3) (3) (-\sqrt{2} q)
4) (4) (\frac{q}{\sqrt{2}})
Q5. Three positive charges are placed at the vertices of an equilateral triangle. A negative charge is placed at the centre of the triangle. The P.E of the system will
1) (1) Increase
Q6. Three charges (1 \times 10^{-8} C), (2 \times 10^{-8} C) and (3 \times 10^{-8} C) are placed at (x=1 cm), (x=2 cm), and (x=3 cm), respectively, on x -axis from the origin. The potential energy of this arrangement relative to the potential energy for infinite separation is
1) (1) (7.9 \times 10^{-2} J)
2) (2) (8.55 \times 10^{-4} J)
3) (3) (6.5 \times 10^{-7} J)
4) (4)
Hint: (U = U_1 + U_2 + U_3) (= \frac{1}{4\pi\varepsilon_0}(\frac{q_1 q_2}{r_{12}}+\frac{q_2 q_3}{r_{23}}+\frac{q_1 q_3}{r_{13}})) (= 9 \times 10^9 \times [ \frac{2 \times 10^{-16}}{1 \times 10^{-2}} + \frac{6 \times 10^{-16}}{1 \times 10^{-2}} + \frac{3 \times 10^{-16}}{2 \times 10^{-2}} ]) (= 9 \times 10^9 \times [ (2 + 6 + 1.5) \times 10^{-14} ]) (= 9 \times 10^9 \times 9.5 \times 10^{-14} = 8.55 \times 10^{-4} J)