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

1. Conduction

2. Induction

3. Friction

4. Both (1) and (2)

Subtopic: Electric Charge |

70%

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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

Subtopic: Electric Charge |

57%

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

1. | \(1.6 \times10^9\) C, \(1.4 \times10^9\) C |

2. | \(1.4 \times10^9\) C, \(1.6 \times10^9\) C |

3. | \(1.34 \times10^7\) C, \(1.34 \times10^7\) C |

4. | \(1.6 \times10^8\) C, \(1.6 \times10^7\) C |

Subtopic: Electric Charge |

68%

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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 |

58%

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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 |

68%

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

Subtopic: Coulomb's Law |

87%

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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. \(\frac{3}{4\pi \epsilon _{0}}\frac{q^{2}}{l^{2}}\)

$\mathrm{}$

Subtopic: Coulomb's Law |

62%

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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$\mathrm{}$

Subtopic: Electric Field |

59%

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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:

$1.3.6\times {10}^{4}{\mathrm{NC}}^{-1}$

$2.7.2\times {10}^{4}{\mathrm{NC}}^{-1}$

$3.9\times {10}^{3}{\mathrm{NC}}^{-1}$

$4.3.2\times {10}^{4}{\mathrm{NC}}^{-1}$

Subtopic: Electric Field |

54%

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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.

Subtopic: Electric Dipole |

56%

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