We can charge a metal sphere positively without touching it by:

1. Conduction

2. Induction

3. Friction

4. Both (1) and (2)

If $10^9$ electrons move out of a body to another body every second, how much time approximately is required to get a total charge of $1$ C on the other body?
1. $200$ years
2. $100$ years
3. $150$ years
4. $250$ years

The amount of positive and negative charges in a cup of water ($250$ g) are respectively:

The ratio of the magnitude of electric force to the magnitude of gravitational force for an electron and a proton will be:
($m_p=1.67\times10^{-27}~\text{kg}$, $m_e=9.11\times10^{-31}~\text{kg}$)
1. $2.4\times10^{39}$
2. $2.6\times10^{36}$
3. $1.4\times10^{36}$
4. $1.6\times10^{39}$

A charged metallic sphere $A$ is suspended by a nylon thread. Another identical charged metallic sphere $B$ held by an insulating handle is brought close to $A$ such that the distance between their centres is $10$ cm, as shown in Fig.(a). The resulting repulsion of $A$ is noted. Then spheres $A$ and $B$ are touched by identical uncharged spheres $C$ and $D$ respectively, as shown in Fig.(b). $C$ and $D$ are then removed and $B$ is brought closer to $A$ to a distance of $5.0$ cm between their centres, as shown in Fig. (c). What is the expected repulsion on $A$ on the basis of Coulomb’s law?

Consider three charges $q_1,~q_2,~q_3$ each equal to $q$ at the vertices of an equilateral triangle of side $l.$ What is the force on a charge $Q$ (with the same sign as $q$) placed at the centroid of the triangle, as shown in the figure?

     
1. $\dfrac{3}{4\pi \epsilon _{0}} \dfrac{Qq}{l^2}$
2. $\dfrac{9}{4\pi \epsilon _{0}} \dfrac{Qq}{l^2}$
3. zero
4. $\dfrac{6}{4\pi \epsilon _{0}} \dfrac{Qq}{l^2}$

Consider the charges $q,~q,$ and $-q$ placed at the vertices of an equilateral triangle, as shown in the figure. Then the sum of the forces on the three charges is:

    

1. $\frac{1}{4\pi \epsilon _{0}}\frac{q^{2}}{l^{2}}$
2. zero
3. $\frac{2}{4\pi \epsilon _{0}}\frac{q^{2}}{l^{2}}$
4. $\frac{3}{4\pi \epsilon _{0}}\frac{q^{2}}{l^{2}}$

An electron falls through a distance of $1.5~\text{cm}$ in a uniform electric field of magnitude $2\times10^4~\text{N/C}$ [figure (a)]. The direction of the field is reversed keeping its magnitude unchanged and a proton falls through the same distance [figure (b)]. If $t_e$ and $t_p$ are the time of fall for electron and proton respectively, then:

   
1. $t_e=t_p$
2. $t_e>t_p$
3. $t_e<t_p$
4. none of these

Two-point charges $q_1$ and $q_2$, of magnitude $+{10}^{-8} C$  and $-{10}^{-8} C$, respectively, are placed 0.1 m apart. The electric field at point  A (as shown in the figure) is: 

$1. 3.6\times {10}^4 {\mathrm{NC}}^{-1}$
$2. 7.2\times {10}^4 {\mathrm{NC}}^{-1}$
$3. 9\times {10}^3 {\mathrm{NC}}^{-1}$
$4. 3.2\times {10}^4 {\mathrm{NC}}^{-1}$

Two charges $\pm10~\mu\text{C}$ are placed $5.0$ mm apart. The electric field at a point $P$ on the axis of the dipole $15$ cm away from its centre $O$ on the side of the positive charge, as shown in the figure is: