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# 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. $$1.6\times10^{39}$$

Subtopic:  Coulomb's Law |
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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.

Subtopic:  Coulomb's Law |
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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. $$\frac{6}{4\pi \epsilon _{0}} \frac{Qq}{l^2}$$

Subtopic:  Coulomb's Law |
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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. $$\frac{3}{4\pi \epsilon _{0}}\frac{q^{2}}{l^{2}}$$



Subtopic:  Coulomb's Law |
61%
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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


Subtopic:  Electric Field |
59%
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Two-point charges ${q}_{1}$ and ${q}_{2}$, of magnitude   and , respectively, are placed 0.1 m apart. The electric field at point  A (as shown in the figure) is:

Subtopic:  Electric Field |
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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. $$1.33\times10^5~\text{NC}^{-1}$$

Subtopic:  Electric Dipole |
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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:

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. $$4.1\times 10^6 ~\text{NC}^{-1}$$


Subtopic:  Electric Dipole |
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The electric field components in the shown figure are ${\mathrm{E}}_{\mathrm{x}}=\alpha {x}^{1/2}$Ey = Ez = 0, in which  . The net flux through the cube is: (Assume that a = 0.1 m)

Subtopic:  Gauss's Law |
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The electric field components in the shown figure are ${\mathrm{E}}_{\mathrm{x}}={\mathrm{ax}}^{1/2}$Ey = Ez = 0, in which  . Find the charge within the cube if net flux through the cube is : (Assume that a = 0.1 m).

Subtopic:  Gauss's Law |
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