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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:

$\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. \(4.1\times 10^6 ~\text{NC}^{-1}\)$\mathrm{}$

Subtopic: Electric Dipole |

68%

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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)

$1.1.05{\mathrm{Nm}}^{2}{\mathrm{C}}^{-1}$

$2.2.03{\mathrm{Nm}}^{2}{\mathrm{C}}^{-1}$

$3.3.05{\mathrm{Nm}}^{2}{\mathrm{C}}^{-1}$

$4.4.03{\mathrm{Nm}}^{2}{\mathrm{C}}^{-1}$

Subtopic: Gauss's Law |

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The electric field components in the shown figure are ${\mathrm{E}}_{\mathrm{x}}={\mathrm{ax}}^{1/2}$, 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).

$1.\mathrm{Zero}$

$2.2.7\times {10}^{-12}\mathrm{C}$

$3.7.9\times {10}^{-12}\mathrm{C}$

$4.9.2\times {10}^{-12}\mathrm{C}$

Subtopic: Gauss's Law |

54%

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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 \(\vec{E}=200\hat{i}\) N/C for \(x>0\) and \(\vec{E}=-200\hat{i}\) N/C for \(x<0\). A right circular cylinder of length \(20~\text{cm}\) and radius \(5~\text{cm}\) has its centre at the origin and its axis along the \(x\text{-}\)axis so that one face is at \(x= + 10~\text{cm}\) and the other is at \(x= -10~\text{cm}\) (as shown in the figure). What is the net outward flux through the cylinder?

1. | \(0\) | 2. | \(1.57~\text{Nm}^2\text{C}^{-1}\) |

3. | \(3.14~\text{Nm}^2\text{C}^{-1}\) | 4. | \(2.47~\text{Nm}^2\text{C}^{-1}\) |

Subtopic: Gauss's Law |

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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?

$1.2.78\times {10}^{-11}C$

$23.10\times {10}^{-12}C$

$3.1.37\times {10}^{-10}C$

$4.2.62\times {10}^{-12}C$

Subtopic: Gauss's Law |

59%

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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:

1. \(\dfrac{Z e}{4 \pi \varepsilon_{0}} \left(\dfrac{1}{r^{2}} - \dfrac{r}{R^{3}}\right)\)

2. \( \dfrac{Z e}{4 \pi \varepsilon_{0}} \dfrac{1}{R^{2}}\)

3. \( \dfrac{Z e}{4 \pi \varepsilon_{0}} \dfrac{1}{r^{2}}\)

4. \( \dfrac{Z e}{4 \pi \varepsilon_{0}} \dfrac{r}{R^{3}}\)$\frac{{\mathrm{}}^{}}{}$

Subtopic: Gauss's Law |

57%

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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:

$1.\frac{Ze}{4\pi {\epsilon}_{0}}\left(\frac{1}{{r}^{2}}-\frac{r}{{R}^{3}}\right)$

$2.0$

$3.\frac{Ze}{4\pi {\epsilon}_{0}}\frac{1}{{r}^{2}}$

$4.\frac{Ze}{4\pi {\epsilon}_{0}}\frac{r}{{R}^{3}}$

Subtopic: Gauss's Law |

55%

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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: (\(m_p=1.67\times10^{-27}~\text{kg},~m_e=9.11\times10^{-31}~\text{kg}\))

1. | \(2.5\times10^{22}\) m/s^{2}, \(2.5\times10^{22}\) m/s^{2} |

2. | \(2.5\times10^{22}\) m/s^{2}, \(1.4\times10^{19}\) m/s^{2} |

3. | \(1.4\times10^{19}\) m/s^{2}, \(2.5\times10^{22}\) m/s^{2} |

4. | \(1.4\times10^{19}\) m/s^{2}, \(1.4\times10^{19}\) m/s^{2} |

Subtopic: Coulomb's Law |

67%

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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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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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