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Q. No.The left half of a parallel plate capacitor is filled with a dielectric of relative permittivity \(K\) while the right half is filled with air. The capacitor is charged by connecting its plates to a battery. The electric field within the dielectric is \(E_K\) and that within the air is \(E.\) Which, of the following, is true?
| 1. | \(E_K=E\) | 2. | \({\Large\frac{E_K}{E}}=K\) |
| 3. | \({\Large\frac{E_K}{E}}={\large\frac{1}{K}}\) | 4. | \({\Large\frac{E_K}{E}}=\sqrt K\) |
Subtopic: Dielectrics in Capacitors |
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Two charges \(q,\) \(-q\) are placed at the two ends of the hypotenuse of an isosceles right-angled triangle, the smaller sides being of length, \(a.\) A dipole of dipole moment \(p\) is placed at the right-angled vertex with its axis pointing towards the positive charge, \(q.\) The torque acting on the dipole is:
1. \(\dfrac{kpq}{a^2}\)
2. zero
3. \(\dfrac{2kpq}{a^2}\)
4. \(\dfrac{\sqrt{5}kpq}{a^2}\)
Subtopic: Energy of Dipole in an External Field |
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Two identical capacitors, each of capacitance \(C\), are connected in series and are charged by means of an ideal battery of emf \(E\). They are disconnected and reconnected in parallel and connected to the same battery. During this reconnection, the positive terminals of the capacitors are connected to the positive terminal of the battery and their negative terminals are similarly connected together. Let, the work done by the battery during the first connection be \(W_1\), and during the second be \(W_2\). Then,
1. \(W_1=W_2\)
2. \(2W_1 =W_2\)
3. \(W_1 = 2W_2\)
4. \(4W_1 = W_2\)
1. \(W_1=W_2\)
2. \(2W_1 =W_2\)
3. \(W_1 = 2W_2\)
4. \(4W_1 = W_2\)
Subtopic: Energy stored in Capacitor |
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A uniform electric field exists in a certain region of space. The potential at the following points are given (all units are in SI):
• \(A \left ( 1, 0, 0 \right )\) \(V_{A}=2\) volt
• \(B \left ( 0, 2, 0 \right )\) \(V_{B}=4\) volt
• \(C \left ( 0, 0, 2 \right )\) \(V_{C}=6\) volt
• \(D \left ( 1, 1, 0 \right )\) \(V_{D}=-1\) volt
The component of the electric field along the \(x\text-\)axis is:
1. \(2\) V/m
2. \(8\) V/m
3. \(3\) V/m
4. \(-6\) V/m
• \(A \left ( 1, 0, 0 \right )\) \(V_{A}=2\) volt
• \(B \left ( 0, 2, 0 \right )\) \(V_{B}=4\) volt
• \(C \left ( 0, 0, 2 \right )\) \(V_{C}=6\) volt
• \(D \left ( 1, 1, 0 \right )\) \(V_{D}=-1\) volt
The component of the electric field along the \(x\text-\)axis is:
1. \(2\) V/m
2. \(8\) V/m
3. \(3\) V/m
4. \(-6\) V/m
Subtopic: Relation between Field & Potential |
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A uniformly charged thin rod of length \(L\) carries a total charge \(q.\) The potential at a point \(A,\) on the perpendicular bisector of the rod, and at a distance \(L\) from its centre is:
\(\left(\text{take}~ k=\dfrac{1}{4\pi\varepsilon_0}\right) \)
\(\left(\text{take}~ k=\dfrac{1}{4\pi\varepsilon_0}\right) \)
| 1. | \(\dfrac{kq}{L}\) | 2. | less than \(\dfrac{kq}{L}\) |
| 3. | greater than \(\dfrac{kq}{L}\) | 4. | zero |
Subtopic: Electric Potential |
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A capacitance is formed by connecting two metallic balls of radius \(r\) by a conducting wire, and two oppositely charged identical metallic hemispheres \((A,B)\) slightly larger than the balls. The separation between the hemispheres and the respective balls is \(d.\) The capacitance between \(A,B\) is:
| 1. | \(\dfrac{4\pi\varepsilon_0r^2}{d}\) | 2. | \(\dfrac{2\pi\varepsilon_0r^2}{d}\) |
| 3. | \(\dfrac{\pi\varepsilon_0r^2}{d}\) | 4. | \(\dfrac{\pi\varepsilon_0r^2}{2d}\) |
Subtopic: Capacitance |
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A charge is uniformly distributed on the circumference of a disc, and the potential at its centre is \(5\) volt. If the charge was uniformly distributed on the surface of this disc, the potential at a point \(P\) on its axis, at a distance equal to the disc's radius from its centre, equals:
1. \(10\) V
2. \(5 \sqrt 2\) V
3. \(10 \sqrt 2\) V
4. \(10 (\sqrt {2} -1)\) V
1. \(10\) V
2. \(5 \sqrt 2\) V
3. \(10 \sqrt 2\) V
4. \(10 (\sqrt {2} -1)\) V
Subtopic: Electric Potential |
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The arrangement shown in the figure is set up with capacitors initially uncharged, and the circuit is completed. A potential difference is imposed across \(AB\) so that the charge on the upper capacitor is doubled without changing its sign.
Then, \(V_{A}-V_{B}=\)
1. \(E_0\)
2. \(2E_0\)
3. \(-E_0\)
4. zero
Then, \(V_{A}-V_{B}=\)
1. \(E_0\)
2. \(2E_0\)
3. \(-E_0\)
4. zero
Subtopic: Capacitance |
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A \(12 ~\mu \text{F}\) capacitor is charged by means of a \(6~\text{V}\) battery and the charged capacitor and the battery are connected in series so that their combined potential difference is twice as much. When a second unknown capacitor (initially uncharged) is connected across this combination, the first capacitor is observed to lose half of its initial charge.
The capacitance of the unknown capacitor is:
1. \(4 ~\mu \text{F}\)
2. \(6~ \mu \text{F}\)
3. \(24 ~\mu \text{F}\)
4. \(36 ~\mu \text{F}\)
The capacitance of the unknown capacitor is:
1. \(4 ~\mu \text{F}\)
2. \(6~ \mu \text{F}\)
3. \(24 ~\mu \text{F}\)
4. \(36 ~\mu \text{F}\)
Subtopic: Capacitance |
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An electron (charge: \(e\) & mass: \(m\)) emitted from the positive plate \(A\) of the capacitor \(AB\), just manages to reach the negative plate \(B\). If the potential difference between the plates is \(V_0\), then the speed of the electron when it is emitted is:
| 1. | \(\sqrt{\frac{e V_{0}}{m}} \quad\) | 2. | \(\sqrt{\frac{e V_{0}}{2 m}} \quad\) |
| 3. | \(\sqrt{\frac{2 e V_{0}}{m}}\) | 4. | \(\sqrt{\frac{2 e V_{0} d}{m}}\) |
Subtopic: Electric Potential Energy |
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