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

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

$$\begin{gathered} 1. 1.6\times {10}^9 C, 1.4\times {10}^9 C \\ 2. 1.4\times {10}^9 C, 1.6\times {10}^9 C \\ 3. 1.34\times {10}^7 C, 1.34\times {10}^7 C \\ 4. 1.6\times {10}^8 C, 1.6\times {10}^7 C \end{gathered}$$

The ratio of the magnitude of electric force to the magnitude of gravitational force for an electron and a proton will be: $\left(m_p=1.67\times {10}^{-27} kg, m_e=9.11\times {10}^{-31} kg\right)$

$$\begin{gathered} 1. 2.4\times {10}^{39} \\ 2. 2.6\times {10}^{36} \\ 3. 1.4\times {10}^{36} \\ 4. 1.6\times {10}^{39} \end{gathered}$$

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 ${\mathrm{q}}_1, {\mathrm{q}}_{2,} {\mathrm{q}}_3$ 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?

$$\begin{gathered} 1. \frac{3}{4{\mathrm{\pi \epsilon }}_0}\frac{\mathrm{Qq}}{{\mathrm{l}}^2} \\ 2. \frac{9}{4{\mathrm{\pi \epsilon }}_0}\frac{\mathrm{Qq}}{{\mathrm{l}}^2} \\ 3. \mathrm{Zero} \\ 4. \frac{6}{4{\mathrm{\pi \epsilon }}_0}\frac{\mathrm{Qq}}{{\mathrm{l}}^2} \end{gathered}$$

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:

$$\begin{gathered} 1. \frac{{\mathrm{q}}^2}{4{\mathrm{\pi \epsilon }}_0{\mathrm{l}}^2} \\ 2. \mathrm{Zero} \\ 3. \frac{2{\mathrm{q}}^2}{4{\mathrm{\pi \epsilon }}_0{\mathrm{l}}^2} \\ 4. \frac{3{\mathrm{q}}^2}{4{\mathrm{\pi \epsilon }}_0{\mathrm{l}}^2} \end{gathered}$$

An electron falls through a distance of 1.5 cm in a uniform electric field of magnitude $2\times {10}^4 N{C}^{-1}$ [figure (a)]. The direction of the field is reversed keeping its magnitude unchanged and a proton falls through the same distance [figure (b)]. If ${\mathrm{t}}_{\mathrm{e}} \mathrm{and} {\mathrm{t}}_{\mathrm{p}}$ are the time of fall for electron and proton respectively, then:

$1. {\mathrm{t}}_{\mathrm{e}}={\mathrm{t}}_{\mathrm{p}}$
2. te>tp
3. te<tp
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: 

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

Two charges ±10 µC 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:

$$\begin{gathered} 1. 2.7\times {10}^5 {\mathrm{NC}}^{-1} \\ 2. 4.13\times {10}^6 {\mathrm{NC}}^{-1} \\ 3. 3.86\times {10}^6 {\mathrm{NC}}^{-1} \\ 4. 1.33\times {10}^5 {\mathrm{NC}}^{-1} \end{gathered}$$

Two charges ±10 µC 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:

$$\begin{gathered} 1. 2.8\times {10}^5 {\mathrm{NC}}^{-1} \\ 2. 3.9\times {10}^5 {\mathrm{NC}}^{-1} \\ 3. 1.33\times {10}^5 {\mathrm{NC}}^{-1} \\ 4. 4.1\times {10}^6 {\mathrm{NC}}^{-1} \end{gathered}$$

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)

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

An electric field is uniform, and in the positive x-direction for positive x, and uniform with the same magnitude but in the negative x-direction for negative x. It is given that E =$200 \hat{i}$ N/C
for x > 0 and E = –$200 \hat{i}$ N/C for x < 0. A right circular cylinder of length 20 cm and radius 5 cm has its centre at the origin and its axis along the x-axis so that one face is at x = +10 cm and the other is at x = –10 cm (as shown in the figure). What is the net outward flux through the cylinder?

$$\begin{gathered} 1. 0 \\ 2. 1.57 N{m}^2{C}^{-1} \\ 3. 3.14 N{m}^2{C}^{-1} \\ 4. 2.47 N{m}^2{C}^{-1} \end{gathered}$$

An electric field is uniform, and in the positive x-direction for positive x, and uniform with the same magnitude but in the negative x-direction for negative x. It is given that E =$200 \hat{i}$ N/C
for x > 0 and E = –$200 \hat{i}$ N/C for x < 0. A right circular cylinder of length 20 cm and radius 5 cm has its centre at the origin and its axis along the x-axis so that one face is at x = +10 cm and the other is along the x-axis so that one face is at x = +10 cm and the other is at x = –10 cm (as shown in the figure). What is the net charge inside the cylinder?

$$\begin{gathered} 1. 2.78\times {10}^{-11} C \\ 2 3.10\times {10}^{-12} C \\ 3. 1.37\times {10}^{-10} C \\ 4. 2.62\times {10}^{-12} C \end{gathered}$$

An early model for an atom considered it to have a positively charged point nucleus of charge Ze, surrounded by a uniform density of negative charge up to a radius R. The atom as a whole is neutral. For this model, the electric field at a distance r(r<R) from the nucleus is:

$$\begin{gathered} 1. \frac{Ze}{4\pi {\epsilon }_0}\left(\frac{1}{r^2}-\frac{r}{R^3}\right) \\ 2. \frac{Ze}{4\pi {\epsilon }_0}\frac{1}{R^2} \\ 3. \frac{Ze}{4\pi {\epsilon }_0}\frac{1}{r^2} \\ 4. \frac{Ze}{4\pi {\epsilon }_0}\frac{r}{R^3} \end{gathered}$$

The accelerations of electron and proton due to the electrical force of their mutual attraction when they are 1 Å (=${10}^{-10} m$) apart are respectively: $\left(m_p=1.67\times {10}^{-27} kg, m_e=9.11\times {10}^{-31} kg\right)$

$$\begin{gathered} 1. 2.5\times {10}^{22} m/s^2, 2.5\times {10}^{22} m/s^2 \\ 2. 2.5\times {10}^{22} m/s^2, 1.4\times {10}^{19} m/s^2 \\ 3. 1.4\times {10}^{19} m/s^2, 2.5\times {10}^{22} m/s^2 \\ 4. 1.4\times {10}^{19} m/s^2, 1.4\times {10}^{19} m/s^2 \end{gathered}$$

Two point charges +8q and -2q are located at x=0 and x=L respectively. The point on x axis at which net electric field is zero due to these charges is-

Two point charges placed in a medium of dielectric constant 5 are at a distance r between them, experience an electrostatic force ‘F’. The electrostatic force between them in vacuum at the same distance r will be-

Which statement is true for Gauss law-

A cylinder of radius r and length l is placed in an uniform electric field parallel to the axis of the cylinder. The total flux for the surface of the cylinder is given by-

Polar molecules are the molecules:

A dipole is placed in an electric field as shown. In which direction will it move?