The ratio of the magnitude of electric force to the magnitude of gravitational force for an electron and a proton will be: (\(m_p=1.67\times10^{-27}~\mathrm{kg}\), \(m_e=9.11\times10^{-31}~\mathrm{kg}\))
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?

1. Electrostatic force on A due to B remains unaltered.

2. Electrostatic force on A due to B becomes double.

3. Electrostatic force on A due to B becomes half.

4. Can't say.

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. \(\frac{3}{4\pi \epsilon _{0}} \frac{Qq}{l^2}\)
2. \(\frac{9}{4\pi \epsilon _{0}} \frac{Qq}{l^2}\)
3. zero
4. \(\frac{6}{4\pi \epsilon _{0}} \frac{Qq}{l^2}\)$\mathrm{}$

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 \(1.5\) cm in a uniform electric field of magnitude \(2\times10^4\) 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$\mathrm{}$

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:

        
1. \(2.7\times10^5~\text{NC}^{-1}\)
2. \(4.13\times10^6~\text{NC}^{-1}\)
3. \(3.86\times10^6~\text{NC}^{-1}\)
4. \(1.33\times10^5~\text{NC}^{-1}\)

Two charges \(\pm 10~ \mu \text{C}\) are placed \(5.0\) mm apart. The electric field at a point \(Q\), \(15\) cm away from O on a line passing through \(O\) and normal to the axis of the dipole, as shown in the figure is:

$\mathrm{}$1. \(2.8 \times 10^5~\text{NC}^{-1}\)
2. \(3.9\times 10^5 ~\text{NC}^{-1}\)
3. \(1.33\times 10^5 ~\text{NC}^{-1}\)
4. \(4.1\times 10^6 ~\text{NC}^{-1}\)$\mathrm{}$

The electric field components in the shown figure are ${\mathrm{E}}_{\mathrm{x}}=\alpha x^{1/2}$, E_y = E_z = 0, in which  $\mathrm{\alpha }=800 \mathrm{N}/\mathrm{C} {\mathrm{m}}^{1/2}$. The net flux through the cube is: (Assume that a = 0.1 m)

$1. 1.05 {\mathrm{Nm}}^2{\mathrm{C}}^{-1}$
$2. 2.03 {\mathrm{Nm}}^2{\mathrm{C}}^{-1}$
$3. 3.05 {\mathrm{Nm}}^2{\mathrm{C}}^{-1}$
$4. 4.03 {\mathrm{Nm}}^2{\mathrm{C}}^{-1}$

The electric field components in the shown figure are ${\mathrm{E}}_{\mathrm{x}}={\mathrm{ax}}^{1/2}$, E_y = E_z = 0, in which  $\mathrm{\alpha }=800 \mathrm{N}/\mathrm{C} {\mathrm{m}}^{1/2}$. Find the charge within the cube if net flux through the cube is $1.05 \mathrm{N} {\mathrm{m}}^2 {\mathrm{C}}^{-1}$: (Assume that a = 0.1 m). 

$1. \mathrm{Zero}$
$2. 2.7\times {10}^{-12} \mathrm{C}$
$3. 7.9\times {10}^{-12} \mathrm{C}$
$4. 9.2\times {10}^{-12} \mathrm{C}$