1. A region surrounding a stationary electric dipole has:
Correct Answer: b) Electric field only
A stationary electric dipole produces only an electric field. Magnetic fields are associated with moving charges or changing electric fields, which are not present in this case.
2. An electric dipole of moment \(\vec{p}\) is placed normal to the lines of force of electric intensity \(\vec{E}\), then the work done in deflecting it through an angle of \(\theta\) is:
Correct Answer: a) \(pE(1 - \cos\theta)\)
Work done in rotating a dipole is given by \(W = pE(1 - \cos\theta)\). When the dipole is initially perpendicular to the field (\(\theta = 90^\circ\)), this formula applies directly.
3. (Added Question) The direction of the electric field at a point on the equatorial line of an electric dipole is:
Correct Answer: b) Opposite to the dipole moment
On the equatorial line, the electric field due to a dipole is directed opposite to the dipole moment. This is because the field components from the two charges add up in the direction from +q to -q, which is opposite to the dipole moment direction (from -q to +q).
4. The electric potential at a point on the axis of an electric dipole depends on the distance \(r\) of the point from the dipole as:
Correct Answer: b) \(\frac{1}{r^2}\)
The potential \(V\) at a point on the axial line of a dipole is given by \(V = \frac{kp}{r^2}\), where \(p\) is the dipole moment and \(r\) is the distance from the dipole center. Thus, it varies as \(\frac{1}{r^2}\).
5. (Added Question) For a short electric dipole, the ratio of electric field intensities at points equidistant from the dipole on the axial and equatorial lines is:
Correct Answer: c) 2:1
For a short dipole, the electric field on the axial line is \(E_{axial} = \frac{2kp}{r^3}\) and on the equatorial line is \(E_{equatorial} = \frac{kp}{r^3}\). Therefore, the ratio \(E_{axial}:E_{equatorial} = 2:1\).
6. An electric dipole of moment \(\vec{p}\) is placed in the position of stable equilibrium in uniform electric field of intensity \(\vec{E}\). It is rotated through an angle \(\theta\) from the initial position. The potential energy of electric dipole in the final position is:
Correct Answer: a) \(-pE\cos\theta\)
The potential energy \(U\) of a dipole in an electric field is given by \(U = -\vec{p}\cdot\vec{E} = -pE\cos\theta\). In stable equilibrium, \(\theta = 0^\circ\) and \(U = -pE\) (minimum). When rotated by \(\theta\), the potential energy becomes \(-pE\cos\theta\).
7. The distance between the two charges \(+q\) and \(-q\) of a dipole is \(2a\). On the axial line at a distance \(r\) from the centre of dipole, the intensity is proportional to:
Correct Answer: d) \(\frac{2aq}{r^3}\)
The electric field on the axial line of a dipole is \(E = \frac{2kp}{r^3}\) where \(p = 2aq\) is the dipole moment. Thus, \(E \propto \frac{2aq}{r^3}\).
8. An electric dipole consisting of two opposite charges of \(2 \times 10^{-6}\) C each separated by a distance of 2 cm is placed in an electric field of \(10^5\) N/C. The maximum torque on the dipole will be:
Correct Answer: a) \(4 \times 10^{-3}\) Nm
Maximum torque \(\tau = pE = qdE = (2 \times 10^{-6} \text{ C})(0.02 \text{ m})(10^5 \text{ N/C}) = 4 \times 10^{-3}\) Nm. The maximum torque occurs when the dipole is perpendicular to the field.
9. (Added Question) The electric field due to a dipole at a large distance \(r\) from its center compared to the dipole size varies as:
Correct Answer: c) \(1/r^3\)
The electric field due to a dipole decreases with the cube of the distance (\(1/r^3\)) at large distances compared to the dipole size. This is in contrast to a point charge where the field decreases as \(1/r^2\).
10. An electric dipole when placed in a uniform electric field \(\vec{E}\) will have minimum potential energy, if the positive direction of dipole moment makes the following angle with \(\vec{E}\):
Correct Answer: c) Zero
The potential energy \(U = -\vec{p}\cdot\vec{E} = -pE\cos\theta\) is minimum when \(\cos\theta\) is maximum, i.e., when \(\theta = 0^\circ\). This occurs when the dipole moment is aligned with the electric field.
11. (Added Question) The torque experienced by an electric dipole placed in a uniform electric field is maximum when the angle between the dipole moment and the electric field is:
Correct Answer: c) 90°
The torque \(\tau = pE\sin\theta\) is maximum when \(\sin\theta\) is maximum, which occurs at \(\theta = 90^\circ\). At this angle, the dipole is perpendicular to the electric field.
12. An electron and a proton are at a distance of 1 Å. The moment of this dipole will be (C × m):
Correct Answer: a) \(1.6 \times 10^{-29}\)
Dipole moment \(p = q \times d = (1.6 \times 10^{-19} \text{ C}) \times (1 \times 10^{-10} \text{ m}) = 1.6 \times 10^{-29}\) C·m. The direction is from the negative charge (electron) to the positive charge (proton).
13. The electric field due to a dipole at a distance \(r\) on its axis is:
Correct Answer: d) Inversely proportional to \(r^3\)
The electric field on the axial line of a dipole is given by \(E = \frac{2kp}{r^3}\), which shows that it is inversely proportional to the cube of the distance (\(1/r^3\)).
14. (Added Question) The work done in rotating an electric dipole from an angle \(\theta_1\) to \(\theta_2\) in a uniform electric field is:
Correct Answer: a) \(pE(\cos\theta_1 - \cos\theta_2)\)
The work done in rotating a dipole in an electric field is equal to the change in potential energy: \(W = U_2 - U_1 = -pE\cos\theta_2 - (-pE\cos\theta_1) = pE(\cos\theta_1 - \cos\theta_2)\).
15. An electric dipole of moment \(\vec{p}\) is placed at the origin along the x-axis. The electric field at a point P, whose position vector makes an angle \(\theta\) with the x-axis, will make an angle ..... with the x-axis, where \(\tan\alpha = \frac{1}{2}\tan\theta\):
Correct Answer: a) \(\alpha\)
The electric field due to a dipole at point P has two components: radial (\(E_r \propto 2\cos\theta\)) and tangential (\(E_\theta \propto \sin\theta\)). The angle \(\alpha\) that the resultant field makes with the position vector is given by \(\tan\alpha = \frac{E_\theta}{E_r} = \frac{\sin\theta}{2\cos\theta} = \frac{1}{2}\tan\theta\).
16. (Added Question) The net force on an electric dipole placed in a uniform electric field is:
Correct Answer: a) Always zero
In a uniform electric field, the forces on the two equal and opposite charges of the dipole are equal in magnitude but opposite in direction, resulting in zero net force. The dipole experiences only torque in a uniform field.
17. A given charge is situated at a certain distance from an electric dipole in the end-on position experiences a force F. If the distance of the charge is doubled, the force acting on the charge will be:
Correct Answer: d) F/8
In the end-on (axial) position, the electric field of a dipole varies as \(1/r^3\). If distance \(r\) is doubled, the field becomes \(1/8\)th of its original value (since \(2^3 = 8\)), and hence the force \(F = qE\) also becomes \(F/8\).
18. An electric dipole is placed along the x-axis at the origin O. A point P is at a distance of r from this origin such that OP makes an angle \(\theta\) with the x-axis. If the electric field at P makes an angle \(\alpha\) with the x-axis, the value of \(\alpha\) would be:
Correct Answer: c) \(\tan^{-1}(\frac{1}{2}\tan\theta)\)
The angle \(\alpha\) is given by \(\tan\alpha = \frac{E_\theta}{E_r} = \frac{\sin\theta}{2\cos\theta} = \frac{1}{2}\tan\theta\). Therefore, \(\alpha = \tan^{-1}(\frac{1}{2}\tan\theta)\).
19. (Added Question) An electric dipole is placed in a non-uniform electric field. The dipole experiences:
Correct Answer: c) Both force and torque
In a non-uniform electric field, the forces on the two charges of the dipole are not equal and opposite, resulting in a net force. Additionally, if the dipole is not aligned with the field, it will also experience a torque.
20. The torque acting on a dipole of moment \(\vec{p}\) in an electric field \(\vec{E}\) is:
Correct Answer: a) \(\vec{p} \times \vec{E}\)
The torque \(\vec{\tau}\) acting on a dipole in an electric field is given by the cross product \(\vec{\tau} = \vec{p} \times \vec{E}\). The magnitude is \(\tau = pE\sin\theta\) where \(\theta\) is the angle between \(\vec{p}\) and \(\vec{E}\).
21. The electric field at a point on equatorial line of a dipole and direction of the dipole moment:
Correct Answer: b) Will be in opposite direction
On the equatorial line, the electric field is directed opposite to the dipole moment. The field points from the positive to the negative charge, while the dipole moment points from the negative to the positive charge.
22. (Added Question) The potential energy of an electric dipole in a uniform electric field is zero when the angle between the dipole moment and the electric field is:
Correct Answer: c) 90°
The potential energy \(U = -pE\cos\theta\) is zero when \(\cos\theta = 0\), which occurs at \(\theta = 90^\circ\). At this angle, the dipole is perpendicular to the electric field.
23. The ratio of electric fields on the axis and at equator of an electric dipole will be:
Correct Answer: b) 2:1
The electric field on the axial line is \(E_{axial} = \frac{2kp}{r^3}\) and on the equatorial line is \(E_{equatorial} = \frac{kp}{r^3}\). Therefore, the ratio \(E_{axial}:E_{equatorial} = 2:1\).
24. Two opposite and equal charges \(4 \times 10^{-8}\) C when placed 2 cm away, form a dipole. If this dipole is placed in an external electric field \(4 \times 10^8\) N/C, the value of maximum torque and the work done in rotating it through 180° will be:
Correct Answer: a) \(3.2 \times 10^{-2}\) Nm and \(6.4 \times 10^{-2}\) J
Maximum torque \(\tau = pE = qdE = (4 \times 10^{-8} \text{ C})(0.02 \text{ m})(4 \times 10^8 \text{ N/C}) = 3.2 \times 10^{-2}\) Nm. Work done in rotating through 180° is \(W = pE(1 - \cos180°) = 2pE = 6.4 \times 10^{-2}\) J.
25. (Added Question) The electric field due to a dipole at a point on its axial line is \(E\). If the dipole is rotated by 60° about an axis perpendicular to its length, the electric field at the same point becomes:
Correct Answer: a) \(E/2\)
When the dipole is rotated by 60°, the axial component of the dipole moment becomes \(p\cos60° = p/2\). Since the electric field is proportional to the dipole moment, the new field becomes \(E/2\).
26. If \(E_1\) be the electric field strength of a short dipole at a point on its axial line and \(E_2\) that on the equatorial line at the same distance, then:
Correct Answer: b) \(E_1 = 2E_2\)
For a short dipole, \(E_1 = \frac{2kp}{r^3}\) (axial) and \(E_2 = \frac{kp}{r^3}\) (equatorial). Therefore, \(E_1 = 2E_2\) at the same distance.
27. Electric charges \(q, -2q, q\) are placed at the corners of an equilateral triangle ABC of side \(l\). The magnitude of electric dipole moment of the system is:
Correct Answer: c) \(\sqrt{3}ql\)
The system can be considered as two dipoles: one between \(q\) and \(-2q\) with moment \(ql\), and another between \(-2q\) and \(q\) with moment \(ql\). These two dipoles are at 60° to each other. The resultant dipole moment is \(\sqrt{(ql)^2 + (ql)^2 + 2(ql)(ql)\cos60°} = \sqrt{3}ql\).
28. An electric dipole is placed in an electric field generated by a point charge:
Correct Answer: d) The torque on the dipole due to the field may be zero
In the non-uniform field of a point charge, the dipole will generally experience both force and torque. However, if the dipole is aligned along the radial direction (parallel or antiparallel to the field), the torque will be zero. The net force may or may not be zero depending on the dipole's orientation and position.
29. (Added Question) The electric potential due to a dipole at a point on its axial line, at distance \(r\) from its center (where \(r\) is much larger than the dipole length) is:
Correct Answer: b) \(\frac{kp}{r^2}\)
The potential due to a dipole on its axial line is given by \(V = \frac{kp}{r^2}\), where \(p\) is the dipole moment and \(r\) is the distance from the dipole center. This is valid when \(r\) is much larger than the dipole length.
30. A point P lies on the perpendicular bisector of an electrical dipole of dipole moment \(\vec{p}\). If the distance of P from the dipole is \(r\) (much larger than the size of the dipole), then electric field at P is proportional to:
Correct Answer: b) \(p\) and \(1/r^3\)
On the equatorial line (perpendicular bisector), the electric field due to a dipole is \(E = \frac{kp}{r^3}\) for \(r\) much larger than the dipole size. Thus, \(E \propto p\) and \(E \propto 1/r^3\).
31. An electric dipole is kept in non-uniform electric field. It experiences:
Correct Answer: a) A force and a torque
In a non-uniform electric field, the dipole experiences both force and torque. The force arises because the field is different at the two ends of the dipole, and the torque occurs unless the dipole is aligned with the field direction.
32. An electric dipole in a uniform electric field experiences (When it is placed at an angle \(\theta\) with the field):
Correct Answer: c) Torque but no force
In a uniform electric field, the dipole experiences no net force (as forces on the two charges are equal and opposite) but does experience a torque unless it is aligned with the field (\(\theta = 0\)).
33. (Added Question) The potential due to an electric dipole at a point on its equatorial plane is:
Correct Answer: c) Zero
The potential due to a dipole at any point on its equatorial plane is zero because the potentials due to the two equal and opposite charges cancel each other out.
34. The electric intensity due to a dipole of length 10 cm and having a charge of \(500 \mu C\), at a point on the axis at a distance 20 cm from one of the charges in air, is:
Correct Answer: a) \(6.25 \times 10^7\) N/C
Distance from center = 20 cm - 5 cm = 15 cm = 0.15 m. Dipole moment \(p = qd = (500 \times 10^{-6} \text{ C})(0.1 \text{ m}) = 5 \times 10^{-5}\) C·m. Electric field on axis \(E = \frac{2kp}{r^3} = \frac{2(9 \times 10^9)(5 \times 10^{-5})}{(0.15)^3} \approx 6.25 \times 10^7\) N/C.
35. If the magnitude of intensity of electric field at a distance \(x\) on axial line and at a distance \(y\) on equatorial line on a given dipole are equal, then \(x:y\) is:
Correct Answer: c) \(2^{1/3}:1\)
Equating the fields: \(\frac{2kp}{x^3} = \frac{kp}{y^3}\) ⇒ \(2/x^3 = 1/y^3\) ⇒ \(x^3 = 2y^3\) ⇒ \(x/y = 2^{1/3}\). Thus, \(x:y = 2^{1/3}:1\).
36. The potential at a point due to an electric dipole will be maximum and minimum when the angles between the axis of the dipole and the line joining the point to the dipole are respectively:
Correct Answer: a) 0 and \(\pi\)
The potential \(V = \frac{kp\cos\theta}{r^2}\) is maximum when \(\cos\theta = 1\) (\(\theta = 0\)) and minimum when \(\cos\theta = -1\) (\(\theta = \pi\)).
37. (Added Question) The force between two electric dipoles whose centers are separated by a distance \(r\) (much larger than their sizes) varies as:
Correct Answer: d) \(1/r^4\)
The force between two dipoles varies as \(1/r^4\) when the distance \(r\) between their centers is much larger than their sizes. This is because the field of one dipole varies as \(1/r^3\), and the force on the other dipole is proportional to this field and its own dipole moment.
38. The distance between \(H^+\) and \(Cl^-\) ions in HCl molecule is 1.28 Å. What will be the potential due to this dipole at a distance of 12 Å on the axis of dipole:
Correct Answer: a) 0.13 V
Dipole moment \(p = qd = (1.6 \times 10^{-19} \text{ C})(1.28 \times 10^{-10} \text{ m}) = 2.048 \times 10^{-29}\) C·m. Potential on axis \(V = \frac{kp}{r^2} = \frac{(9 \times 10^9)(2.048 \times 10^{-29})}{(12 \times 10^{-10})^2} \approx 0.13\) V.
39. Electric potential at an equatorial point of a small dipole with dipole moment \(p\) (r, distance from the dipole) is:
Correct Answer: a) Zero
The potential due to a dipole at any point on its equatorial plane is zero because the potentials due to the positive and negative charges cancel each other exactly.
40. (Added Question) The energy required to turn an electric dipole end for end (180°) in a uniform electric field \(E\) from its most stable position is:
Correct Answer: b) \(2pE\)
The work done to rotate a dipole from \(\theta = 0^\circ\) (stable equilibrium) to \(\theta = 180^\circ\) is \(W = pE(1 - \cos180^\circ) = 2pE\). This is the energy required to flip the dipole.
41. (Added Question) An electric dipole is placed at the center of a spherical surface. The electric flux through the surface is:
Correct Answer: a) Zero
A dipole consists of two equal and opposite charges. According to Gauss's law, the net charge enclosed by the surface is zero (\(+q - q = 0\)), so the total electric flux through the surface is zero.
42. An electric dipole has the magnitude of its charge as \(q\) and its dipole moment is \(p\). It is placed in a uniform electric field \(E\). If its dipole moment is along the direction of the field, the force on it and its potential energy are respectively:
Correct Answer: c) Zero and minimum
In a uniform electric field, the net force on a dipole is zero. When the dipole is aligned with the field (\(\theta = 0\)), the potential energy \(U = -pE\) is minimum (most stable configuration).
43. For a dipole \(q = 2 \times 10^{-6}\) C and \(d = 0.02\) m. Calculate the maximum torque for this dipole if \(E = 10^5\) N/C:
Correct Answer: a) \(4 \times 10^{-3}\) Nm
Maximum torque \(\tau = pE = qdE = (2 \times 10^{-6} \text{ C})(0.02 \text{ m})(10^5 \text{ N/C}) = 4 \times 10^{-3}\) Nm. Maximum torque occurs when \(\theta = 90^\circ\).
44. A molecule with a dipole moment \(p\) is placed in an electric field of strength \(E\). Initially the dipole is aligned parallel to the field. If the dipole is to be rotated to be anti-parallel to the field, the work required to be done by an external agency is:
Correct Answer: d) \(2pE\)
Work done \(W = \Delta U = U_f - U_i = pE - (-pE) = 2pE\). Here, initial potential energy \(U_i = -pE\) (parallel) and final \(U_f = pE\) (anti-parallel).
45. What is the angle between the electric dipole moment and the electric field strength due to it on the equatorial line:
Correct Answer: c) 180°
On the equatorial line, the electric field due to a dipole is directed opposite to the dipole moment (from +q to -q, while dipole moment points from -q to +q). Hence, the angle is 180°.
46. An electric dipole of moment \(\vec{p}\) placed in a uniform electric field \(\vec{E}\) has minimum potential energy when the angle between \(\vec{p}\) and \(\vec{E}\) is:
Correct Answer: a) Zero
Potential energy \(U = -\vec{p}\cdot\vec{E} = -pE\cos\theta\) is minimum when \(\cos\theta\) is maximum (1), which occurs at \(\theta = 0\). This is the most stable configuration.
47. Two electric dipoles of moment \(P\) and \(64P\) are placed in opposite direction on a line at a distance of 25 cm. The electric field will be zero at point between the dipoles whose distance from the dipole of moment \(P\) is:
Correct Answer: a) 5 cm
Let \(x\) be the distance from \(P\) dipole. Fields cancel when \(\frac{2kP}{x^3} = \frac{2k(64P)}{(25-x)^3}\) ⇒ \(\frac{1}{x^3} = \frac{64}{(25-x)^3}\) ⇒ \(\frac{1}{x} = \frac{4}{25-x}\) ⇒ \(25-x = 4x\) ⇒ \(x = 5\) cm.
48. When an electric dipole \(\vec{p}\) is placed in a uniform electric field \(\vec{E}\) then at what angle between \(\vec{p}\) and \(\vec{E}\) the value of torque will be maximum:
Correct Answer: b) \(\pi/2\)
Torque \(\tau = pE\sin\theta\) is maximum when \(\sin\theta = 1\), which occurs at \(\theta = \pi/2\) (90°). At this angle, the dipole is perpendicular to the field.
49. Two charges \(+4 \times 10^{-6}\) C and \(-4 \times 10^{-6}\) C kept 2.4 Å apart forms a dipole. If it is kept in uniform electric field of intensity \(4 \times 10^8\) N/C then what will be its electrical energy in equilibrium:
Correct Answer: a) \(-3.84 \times 10^{-24}\) J
In equilibrium, the dipole aligns with the field (\(\theta = 0\)). Potential energy \(U = -pE = -qdE = -(4 \times 10^{-6} \text{ C})(2.4 \times 10^{-10} \text{ m})(4 \times 10^8 \text{ N/C}) = -3.84 \times 10^{-24}\) J.
50. The electric field due to an electric dipole at a distance \(r\) from its centre in axial position is \(E\). If the dipole is rotated through an angle of 90° about its perpendicular axis, the electric field at the same point will be:
Correct Answer: c) \(E/2\)
After 90° rotation, the point becomes an equatorial point. The ratio of axial to equatorial field is 2:1, so the new field will be \(E/2\).