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The variation of potential with distance \(x\) from a fixed point is shown in the figure. The electric field at \(x=13~\text m\) is:
     

1. \(7.5~\text{V/m}\) 2. \(-7.5~\text{V/m}\)
3. \(5~\text{V/m}\) 4. \(-5~\text{V/m}\)

Subtopic:  Relation between Field & Potential |
 52%
Level 3: 35%-60%
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Two metal spheres, one of radius \(R\) and the other of radius \(2R\) respectively have the same surface charge density \(\sigma.\) They are brought in contact and separated.  What will be the new surface charge densities on them?
1. \(\sigma_{1}=\dfrac{5}{6}\sigma ,~\sigma_{2}=\dfrac{5}{6}\sigma\)
2. \(\sigma_{1}=\dfrac{5}{2}\sigma ,~\sigma_{2}=\dfrac{5}{6}\sigma\)
3. \(\sigma_{1}=\dfrac{5}{2}\sigma ,~\sigma_{2}=\dfrac{5}{3}\sigma\)
4. \(\sigma_{1}=\dfrac{5}{3}\sigma ,~\sigma_{2}=\dfrac{5}{6}\sigma\)
Subtopic:  Electric Potential |
 58%
Level 3: 35%-60%
NEET - 2019
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Two identical capacitors \(C_{1}\) and \(C_{2}\) of equal capacitance are connected as shown in the circuit. Terminals \(a\) and \(b\) of the key \(k\) are connected to charge capacitor \(C_{1}\) using a battery of emf \(V\) volt. Now disconnecting \(a\) and \(b\) terminals, terminals \(b\) and \(c\) are connected. Due to this, what will be the percentage loss of energy?

1. \(75\%\) 2. \(0\%\)
3. \(50\%\) 4. \(25\%\)
Subtopic:  Energy stored in Capacitor |
 65%
Level 2: 60%+
NEET - 2019
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Three identical capacitors are connected as follows:
 
Which of the following shows the order of increasing capacitance (smallest first)?

1. \(\mathrm{(3), (2), (1)}\) 2. \(\mathrm{(1), (2), (3)}\)
3. \(\mathrm{(2), (1), (3)}\) 4. \(\mathrm{(2), (3), (1)}\)
Subtopic:  Combination of Capacitors |
 80%
Level 1: 80%+
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The electric potential at a point in free space due to a charge \(Q\) coulomb is \(Q\times10^{11}~\text{V}\). The electric field at that point is:
1. \(4\pi \varepsilon_0 Q\times 10^{22}~\text{V/m}\)
2. \(12\pi \varepsilon_0 Q\times 10^{20}~\text{V/m}\)
3. \(4\pi \varepsilon_0 Q\times 10^{20}~\text{V/m}\)
4. \(12\pi \varepsilon_0 Q\times 10^{22}~\text{V/m}\)

Subtopic:  Relation between Field & Potential |
 73%
Level 2: 60%+
AIPMT - 2008
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Four electric charges \(+ q,\) \(+ q,\) \(- q\) and \(- q\) are placed at the corners of a square of side \(2L\) (see figure). The electric potential at the point \(A\), mid-way between the two charges \(+ q\) and \(+ q\) is:
              
1. \(\frac{1}{4 \pi\varepsilon_{0}} \frac{2 q}{L} \left(1 + \frac{1}{\sqrt{5}}\right)\)
2. \(\frac{1}{4 \pi\varepsilon_{0}} \frac{2 q}{L} \left(1 - \frac{1}{\sqrt{5}}\right)\)
3. zero
4. \(\frac{1}{4 \pi \varepsilon_{0}} \frac{2 q}{L} \left(1 + \sqrt{5}\right)\)

Subtopic:  Electric Potential |
 75%
Level 2: 60%+
AIPMT - 2011
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Four-point charges \(-Q, -q, 2q~\text{and}~2Q\) are placed, one at each corner of the square. The relation between \(Q\) and \(q\) for which the potential at the center of the square is zero is:

1. \(Q= -q\) 2. \(Q= -2q\)
3. \(Q= q\) 4. \(Q= 2q\)
Subtopic:  Electric Potential |
 78%
Level 2: 60%+
AIPMT - 2012
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An electric dipole of moment \(p\) is placed in an electric field of intensity \(E.\) The dipole acquires a position such that the axis of the dipole makes an angle \(\theta\) with the direction of the field. Assuming that the potential energy of the dipole to be zero when \(\theta = 90^{\circ}\), the torque and the potential energy of the dipole will respectively be:
1. \(pE\text{sin}\theta, ~-pE\text{cos}\theta\)
2. \(pE\text{sin}\theta, ~-2pE\text{cos}\theta\)
3. \(pE\text{sin}\theta, ~2pE\text{cos}\theta\)
4. \(pE\text{cos}\theta, ~-pE\text{sin}\theta\)

Subtopic:  Energy of Dipole in an External Field |
 83%
Level 1: 80%+
AIPMT - 2012
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A conducting sphere of the radius \(R\) is given a charge \(Q.\) The electric potential and the electric field at the centre of the sphere respectively are:

1. zero and \(\frac{Q}{4 \pi \varepsilon_0 {R}^2}\) 2. \(\frac{Q}{4 \pi \varepsilon_0 R}\) and zero
3. \(\frac{Q}{4 \pi \varepsilon_0 R}\) and \(\frac{Q}{4 \pi \varepsilon_0{R}^2}\) 4. both are zero
Subtopic:  Electric Potential |
 85%
Level 1: 80%+
AIPMT - 2014
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A parallel plate air capacitor has capacitance \(C,\) the distance of separation between plates is \(d\) and potential difference \(V\) is applied between the plates. The force of attraction between the plates of the parallel plate air capacitor is:

1. \(\frac{C^2V^2}{2d}\) 2. \(\frac{CV^2}{2d}\)
3. \(\frac{CV^2}{d}\) 4. \(\frac{C^2V^2}{2d^2}\)
Subtopic:  Capacitance |
 71%
Level 2: 60%+
NEET - 2015
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