Paper Questions (Raw)
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) V=0, E = 0
2) (V=0, E \neq 0)
3) (V \neq 0, E=0)
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) (\frac{kQ^{2}}{L})
2) (\frac{\sqrt{3} kQ^{2}}{2 L})
3) (\frac{kQ^{2}}{\sqrt{3} L})
4) (\frac{\sqrt{3} kQ^{2}}{L})
Q3. P.E. of system of charges, shown in the figure, is
1) (\frac{3}{4 \pi \varepsilon_{0}} \frac{q^{2}}{r})
2) (\frac{1}{4 \pi \varepsilon_{0}} \frac{q^{2}}{r})
3) (-\frac{1}{4 \pi \varepsilon_{0}} \frac{q^{2}}{r})
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) (-\frac{q}{\sqrt{2}})
2) (-0.47 q)
3) (-\sqrt{2} q)
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
Q6. Three charges (1 \times 10^{-8} C), (2 \times 10^{-8} C) and (3 \times 10^{-8} C) are placed at (\mathrm{x}=1 cm), (\mathrm{x}=2 cm), and (\mathrm{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) 7.9 \times10-2 J
2) (8.55 \times 10^{-4} J)
3) (6.5 \times 10^{-7} J)
4) (4)
Hint: $$U}=\mp@subsup{U}{1}{}+\mp@subsup{U}{2}{}+\mp@subsup{U}{3}{ = 1 }\frac{1}{4\pi\mp@subsup{\varepsilon}{0}{}}\frac{\mp@subsup{q}{1}{}\mp@subsup{q}{2}{}}{\mp@subsup{r}{12}{}}+\frac{1}{4\pi\mp@subsup{\varepsilon}{0}{}}\frac{\mp@subsup{q}{2}{}\mp@subsup{q}{3}{}}{\mp@subsup{r}{23}{}}+\frac{1}{4\pi\mp@subsup{\varepsilon}{0}{}}\frac{\mp@subsup{q}{1}{}\mp@subsup{q}{3}{}}{\mp@subsup{r}{13}{}$$
$$= 9 \times 1 0 ^ { 9 } \times [ \frac { 2 \times 1 0 ^ { - 1 6 } } { 1 \times 1 0 ^ { - 2 } } + \frac { 6 \times 1 0 ^ { - 1 6 } } { 1 \times 1 0 ^ { - 2 } } + \frac { 3 \times 1 0 ^ { - 1 6 } } { 2 \times 1 0 ^ { - 2 } } ]$$
$$= 9 \times 1 0 ^ { 9 } \times [ ( 2 + 6 + 1 . 5 ) \times 1 0 ^ { - 1 4 } ]$$
= 9 x 10 9 x 9.5 x 10-14 = 8.55 x 10-4 J