Q. No.An infinite non-conducting vertical wall carries a uniform surface charge density \(\sigma\) (positive in nature). A charged particle of mass \(m\) suspended like a pendulum-stays fixed with its string making an angle of \(45^{\circ}\) with the vertical. The charge on the particle is:
| 1. | \(\varepsilon_{0} \cdot \dfrac{2 m g}{\sigma}\) | 2. | \(\varepsilon_0\text { } \cdot \dfrac{m g}{\sigma}\) |
| 3. | \(\varepsilon_{0} \cdot \dfrac{\sqrt{2} m g}{\sigma}\) | 4. | \(\varepsilon_{0} \cdot \dfrac{m g}{\sqrt{2} \sigma}\) |
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A conductor is connected to the earth electrically; a positive point charge is brought near it while it remains earthed.
The earth connection is broken and then the positive charge is taken away. The final charge on the conductor is:
| 1. | zero | 2. | positive |
| 3. | negative | 4. | unpredictable |
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Consider electrostatic and gravitational forces among the following: electron-electron\((ee)\), electron-proton \((ep)\) & proton-proton\((pp)\). All the distances between the particles are the same. Let \(F^{gr}\) 'denote' gravitational force and \(F^{el}\) 'denote' electrostatic force and the subscripts denote the particle pairs. We consider only the magnitudes of the forces. Then:
| (a) | \(F_{e p}^{e l}=F_{p p}^{e l}=F_{e e}^{e l}\) |
| (b) | \(F_{p p}^{e l}>F_{p p}^{g r}\) |
| (c) | \(F_{e p}^{g r}<F_{e p}^{el}\) |
| (d) | \(F_{e p}^{g r}=F_{p p}^{g r}=F_{ee}^{gr}\) |
Which of the above is/are true?
1. (a) only
2. (a) and (b) only
3. (a), (b), and (c) only
4. (a) and (d) only
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A pair of field lines are drawn, connecting the charges \(q_1, q_2\) in addition to the straight field line connecting them. From the above diagram,
| 1. | \(q_1>0, q_2<0~\text{and}~|q_1|>|q_2|\) |
| 2. | \(q_1<0, q_2>0~\text{and}~|q_1|>|q_2|\) |
| 3. | \(q_1>0, q_2<0~\text{and}~|q_1|<|q_2|\) |
| 4. | \(q_1<0, q_2>0~\text{and}~|q_1|<|q_2|\) |
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The mid-point of side \(AB\) is \(P\) while the circumcenter of \(ABC\) is \(O\). Let the electric field at \(P\) be \(E_p\) and that at \(O\) be \(E_O.\)
Then, \(E_O:E_P=\)
| 1. | \(\dfrac{2}{9}\) | 2. | \(\dfrac{4}{9}\) |
| 3. | \(\dfrac{9}{2}\) | 4. | \(\dfrac{9}{4}\) |
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1. \(qE =Mg\)
2. \(2qE =Mg\)
3. \(qE =2Mg\)
4. \(\sqrt{2}qE =Mg\)
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| 1. | \(\dfrac{1}{r^2}\) | 2. | \(\dfrac{1}{r}\) |
| 3. | \(r\) | 4. | \(\dfrac{1}{r^3}\) |
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Then,
| 1. | \(q,Q\) are of the same sign and \(|q|=|Q|\) |
| 2. | \(q,Q\) are of opposite signs and \(|q|=|Q|\) |
| 3. | \(q,Q\) are of the same sign and \(|q|<|Q|\) |
| 4. | \(q,Q\) are of opposite signs and \(|q|>|Q|\) |
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The electric field within the medium due to the charge \(Q\) at some point \(P\) is \(\vec E_{Q}\). The Electric field at the same point \(P\) due to induced charge within the medium is \(\vec E_{m}\). Then,
| 1. | \(|\vec E_m|=\left|\dfrac{\vec E_Q}{K}\right|,\) and the two fields are in opposite directions. |
| 2. | \(|\vec E_Q|=\left|\dfrac{\vec E_m}{K}\right|,\) and the two fields are in the same direction. |
| 3. | \(|\vec E_Q+\vec E_m|=\left|\dfrac{\vec E_Q}{K}\right|,\) and the two fields are in opposite directions. |
| 4. | \(|\vec E_Q+\vec E_m|=\left|\dfrac{\vec E_m}{K}\right|,\) and the two fields are in the same direction. |
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Now, identical point charges (same magnitude \(q\) each), are placed at the four corners of a square of side \(a.\) When one of the charges is removed, the electric field at the center becomes \(E_s.\) Then,
| 1. | \(\dfrac{E_s}{2}=\dfrac{E_C}{3}\) | 2. | \(\dfrac{E_s}{3}=\dfrac{E_C}{2}\) |
| 3. | \(\dfrac{E_s}{\sqrt2}=\dfrac{E_C}{\sqrt3}\) | 4. | \(\dfrac{E_s}{\sqrt3}=\dfrac{E_C}{\sqrt2}\) |
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