In a certain region of space, equipotential surfaces of the electric field are drawn - corresponding to \(V=10\) volt and \(V=9.9\) volts. There is no field along the \(z\text-\)direction. At a certain point \(P,\) on the \(10-\)volt surface, the distance \(PQ_1,\) to the \(9.9\) volt surface is \(2\) mm when \(\overrightarrow{P Q}_{1}\) is along the \(x\text-\)axis. On the other hand, if \(\overrightarrow{P Q}_{2}\) is taken parallel to the \(y\text-\)axis, the corresponding distance \(PQ_2=1\) mm. The electric field at \(P\) is along:
1. \(2 \hat{i}+\hat{j}\) 2. \(2 \hat{j}+\hat{i}\)
3. \(\dfrac{1}{4} \hat{i}+\hat{j}\) 4. \( \dfrac{1}{4} \hat{j}+\hat{i}\)

Subtopic:  Relation between Field & Potential |
 52%
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The capacitance of a parallel plate capacitor depends on the 'overlapping' or 'facing' area between the plates if the plates are very close together. A parallel plate capacitor is made by taking two metallic plates of the shape of equilateral triangles and placing them very close to each other at a uniform separation- facing each other. Their capacitance is now \(C_1.\) The two plates are now rotated in their planes about an axis perpendicular to their planes- passing through their centers - until their 'overlap' is least.
The capacitance is now \(C_2.\)
\(\dfrac{C_1} { C_2}\) equals:
1. \(3\) 2. \(2\)
3. \(\dfrac{3}{2}\) 4. \(\dfrac{4}{3}\)
Subtopic:  Capacitance |
 55%
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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. \(36 ~\mu \text{F}\)
Subtopic:  Capacitance |
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An arrangement consisting of two concentric spherical shells A and B has a capacitance \(C_0\) between them. If the upper 'hemispherical' space between them is filled by a dielectric of relative permittivity \(K_1\) and the lower by one of relative permittivity \(K_2,\) the new capacitance will be:
             
1. \(\left(K_{1}+K_{2}\right) C_{0}\) 2. \( \dfrac{K_{1}+K_{2}}{2} C_{0}\)
3. \(\dfrac{1}{2}\left(\dfrac{1}{K_{1}}+\dfrac{1}{K_{2}}\right) C_{0}\) 4. \( \dfrac{2 K_{1} K_{2}}{K_{1}+K_{2}} C_{0}\)
Subtopic:  Dielectrics in Capacitors |
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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
Subtopic:  Relation between Field & Potential |
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A parallel plate air capacitor having a plate separation \(d\) is charged, and the energy stored is \(U\). The force between the plates is:
1. \(\dfrac Ud\) 2. \(\dfrac {2U}d\)
3. \(\dfrac U{2d}\) 4. \(\dfrac {\sqrt2U}d\)
Subtopic:  Energy stored in Capacitor |
 60%
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Four uncharged capacitors and resistances are connected as shown and a potential difference is applied between \(P\) and \(Q\).
It is observed that \(V_A=V_B\) after a sufficiently long time.
Then:
1. \(\dfrac{C_1}{C_2}=\dfrac{C_3}{C_4}\) 2. \(\dfrac{C_1}{C_2}=\dfrac{R_3~C_3}{R_4~C_4}\)
3. \(\dfrac{C_1}{C_2}=\dfrac{R_4}{R_3}\) 4. \(\dfrac{C_1}{C_2}=\dfrac{R_4~C_3}{R_3~C_4}\)
Subtopic:  Capacitance |
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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
Subtopic:  Capacitance |
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Three metallic spheres of radii \(r_1,~ r_2,~ r_3\) are connected by very long conducting wires to form an equilateral triangle. The capacitance of the system is:

        
1. \(4 \pi \varepsilon_{0}\left(r_{1}+r_{2}+r_{3}\right)\)
2. \(4 \pi \varepsilon_{0} \dfrac{r_{1}^{2}+r_{2}^{2}+r_{3}^{2}}{r_{1}+r_{2}+r_{3}}\)
3. \(4 \pi \varepsilon_{0}\left(\dfrac{1}{r_{1}}+\dfrac{1}{r_{2}}+\dfrac{1}{r_{3}}\right)^{-1}\)
4. \(4 \pi \varepsilon_{0} \sqrt{r_{1}^{2}+r_{2}^{2}+r_{3}^{2}}\)
Subtopic:  Capacitance |
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Four identical point charges (\(q\) each) are placed at the four corners of a square of diagonal \(d.\) The potential at a point which is at a distance \(\dfrac{d}{2}\) above the centre of the square is:
\(\Big(k=\dfrac{1}{4\pi\varepsilon_0}\Big)\)
1. \(\dfrac{8~kq}{d}\) 2. \(\dfrac{4~kq}{d}\)
3. \(\dfrac{4\sqrt2~kq}{d}\) 4. \(\dfrac{\sqrt2~kq}{d}\)
Subtopic:  Electric Potential |
 69%
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