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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. | \(\varepsilon_{0} \cdot \dfrac{m g}{\sqrt{2} \sigma}\) |
Subtopic: Electric Field |
63%
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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 |
Subtopic: Electric Charge |
60%
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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
Subtopic: Coulomb's Law |
73%
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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. | \(q_1<0, q_2>0~\text{and}~|q_1|<|q_2|\) |
Subtopic: Electric Field |
62%
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Three charges \(q,~q,~-q\) are placed at the three corners of an equilateral triangle \(ABC\), of side \(a.\)
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, \(E_O:E_P=\)
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. | \(\dfrac{9}{4}\) |
Subtopic: Electric Field |
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A thin uniform rod of mass \(M\) and length \(L\) is suspended from one of its ends, '\(A\)', so that it can rotate freely about it. A charge '\(q\)' is fixed to its lower end \(B\). A uniform horizontal Electric field is switched on and the rod rotates about \(A\), finally coming to equilibrium – making an angle of \(45^{\circ}\) with the vertical. If the acceleration due to gravity is '\(g\)', then,
1. \(qE =Mg\)
2. \(2qE =Mg\)
3. \(qE =2Mg\)
4. \(\sqrt{2}qE =Mg\)
1. \(qE =Mg\)
2. \(2qE =Mg\)
3. \(qE =2Mg\)
4. \(\sqrt{2}qE =Mg\)
Subtopic: Electric Field |
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Two very long insulated glass rods are charged uniformly by giving them identical charges '\(q\)', each. The rods have lengths \(L\) each and are placed parallel to each other at a distance '\(r\)' apart, where \(r\ll L\). Then, the electric force acting between the rods is proportional to:
| 1. | \(\dfrac{1}{r^2}\) | 2. | \(\dfrac{1}{r}\) |
| 3. | \(r\) | 4. | \(\dfrac{1}{r^3}\) |
Subtopic: Electric Field |
53%
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A point charge '\(q\)' is placed at the centre of a spherical cavity at the centre of a conducting sphere. The sphere is initially uncharged. The radius of the cavity is '\(a\)' and that of the sphere is '\(2a\)'. Let the charge on the outer surface of the sphere be \(Q\).
Then,
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 \(|q|>|Q|\) |
Subtopic: Gauss's Law |
50%
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A uniformly charged sphere carrying a charge \(Q\) distributed uniformly on its outer surface is placed in an isotropic medium of dielectric constant '\(K\)'.
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,
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. |
Subtopic: Electric Field |
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Identical point charges (\(q\) each), are placed at the eight corners of a cube of side \(a.\) When one of the charges is removed, the electric field at the center becomes \(E_c.\)
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,
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. | \(\dfrac{E_s}{\sqrt3}=\dfrac{E_C}{\sqrt2}\) |
Subtopic: Electric Field |
62%
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