J J Science Academy - Electrostatics Questions

1) 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) $V \neq 0, E \neq 0$

2) 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) $\frac{\sqrt{3} kQ^{2}}{L}$

3) 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) $-\frac{3}{4 \pi \varepsilon_{0}} \frac{q^{2}}{r}$

4) 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) $\frac{q}{\sqrt{2}}$

5) 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) Increase

6) 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) 7.9 \times10-2 J

2) $8.55 \times 10^{-4} J$

3) $6.5 \times 10^{-7} J$

4) 4

$$U=U_1+U_2+U_3 = \frac{1}{4\pi\varepsilon_0}\frac{q_1 q_2}{r_{12}}+\frac{1}{4\pi\varepsilon_0}\frac{q_2 q_3}{r_{23}}+\frac{1}{4\pi\varepsilon_0}\frac{q_1 q_3}{r_{13}}$$ $$= 9 \times 1 0 ^ { 9 } \times [ \frac { 2 \times 1 0 ^ { - 1 6 } } { 1 \times 1 0 ^ { - 2 } } + \frac { 6 \times 1 0 ^ { - 1 6 } } { 1 \times 1 0 ^ { - 2 } } + \frac { 3 \times 1 0 ^ { - 1 6 } } { 2 \times 1 0 ^ { - 2 } } ]$$ $$= 9 \times 1 0 ^ { 9 } \times [ ( 2 + 6 + 1 . 5 ) \times 1 0 ^ { - 1 4 } ]$$ = 9 x 10 9 x 9.5 x 10-14 = 8.55 x 10-4 J