WBCHSE Class 12 Physics Electric Field Multiple Choice Question And Answers

Electrostatics Chapter 2 Electric Field Exercise Multiple Choice Question And Answers

Question 1. The electric flux through the surface

Class 12 Physics Unit 1 Electrostatics Chapter 2 Electric Field Multiple choice question aand answer 1

  1. Is the largest
  2. Is the least
  3. (2) is the same as (3) but is smaller than (4)
  4. Is the same for all the figures

Answer: 4. Is the same for all the figures

The shape of the Gaussian surface does not determine the flux. The flux depends only on the charge enclosed by the Gaussian surface.

Question 2. Electric field lines in which an electric dipole is placed as shown. Which of the following statements is correct?

Class-12-Physics-Unit-1-Electrostatics-Chapter-2-Electric-Field-Multiple-choice-question-aand-answer-2

  1. The dipole will not experience any force.
  2. The dipole will experience a force towards the right.
  3. The dipole will experience a force towards the left.
  4. The dipole will experience a force upwards.

Answer: 3. The dipole will experience a force towards the left.

Read and Learn More Class 12 Physics Multiple Choice Questions

The electric field on the left side of the figure is greater than that on the right side. Thus the force (FL) on -q directed towards the left is more than the force (FR) on +q directed towards the right. Hence the dipole will experience a force towards the left.

Question 3. Five charges q1, q2, q3, q4, and qs are fixed at their positions as shown in Rg. 2.92. S is a Gaussian surface. The Gauss’ law is given by

⇒ \(\oint_S \vec{E} \cdot \overrightarrow{d S}=\frac{q}{\epsilon_0}\)

Which of the following statements is correct?

Class 12 Physics Unit 1 Electrostatics Chapter 2 Electric Field Multiple choice question and answer 3

  1. \(\vec{E}\) on the LHS of the equation will have a contribution from q1, q5, and q3 while q on the RHS will have a contribution from q2 and q4 only.
  2. \(\vec{E}\) on the LHS of the above equation will have a contribution from all charges while q on the RHS will have a contribution from q2 and q4 only.
  3. \(\vec{E}\) on the LHS of the above equation will have a contribution from all charges while q on the RHS will have a contribution from q2 and q4 only.
  4. Both\(\vec{E}\)E on the LHS and q on the RHS will have contributions from q2 and q4 only.

Answer: 2. \(\vec{E}\) on the LHS of the above equation will have a contribution from all charges while q on the RHS will have a contribution from q2 and q4 only.

Question 4. A hemisphere has a uniform positive charge. The electric field at a point on a diameter away from the center is directed

  1. Perpendicular to the diameter
  2. Parallel to the diameter
  3. At an angle tilted toward the diameter
  4. At an angle tilted away from the diameter

Answer: 1. Perpendicular to the diameter

The components of an electric field due to various elements parallel to the diameter neutralize one another.

Question 5. A hemisphere has a uniform positive charge. The electric field at a point on a diameter away from the center is directed

  1. Perpendicular to the plane and away from the plane
  2. Perpendicular to the plane but towards the plane
  3. Radially away from the point charge
  4. Radially towards the point charge

Answer: 1. Perpendicular to the plane and away from the plane

Question 6. If \(\oint_S \vec{E} \cdot \overrightarrow{d S}=0\) over a surface then,

  1. The electric field inside the surface and on it is zero
  2. The electric field inside the surface is necessarily uniform
  3. The number of flux lines entering the surface must be equal to the number of flux lines leaving it
  4. All charges must necessarily be outside the surface

Answer:

3. The number of flux lines entering the surface must be equal to the number of flux lines leaving it

4. All charges must necessarily be outside the surface

Question 7. The electric field at a point is

  1. Always continuous
  2. Continuous if there is no charge at that point
  3. Discontinuous only if there is a negative charge at that point
  4. Discontinuous if there is a charge at that point.

Answer:

2. Continuous if there is no charge at that point

4. Discontinuous if there is a charge at that point.

Question 8. Consider a region inside which there are various types of charges but the total charge is zero. At points outside the region.

  1. The electric field is necessarily zero
  2. The electric field is due to the dipole moment of the charge distribution only
  3. The dominant electric field is proportional to \(\frac{1}{r^3}\) for large r, where r is the distance from the origin in this region
  4. The work done to move a charged particle along a closed path, away from the region, will be zero

Answer:

3. The dominant electric field is proportional to \(\frac{1}{r^3}\) for large r, where r is the distance from the origin in this region

4. The work done to move a charged particle along a closed path, away from the region, will be zero

Question 9. The SI unit of electric field intensity is

  1. V.m-1
  2. A.m-1
  3. N.C-1
  4. J.C-1m-1

Answer:

1. V.m-1

3. N.C-1

4. J.C-1m-1

Question 10. If the net electric flux through a closed surface is zero, then

  1. There are no charges inside the surface
  2. The net charge inside the surface is zero
  3. The electric field is zero everywhere on the surface
  4. The number of electric field lines entering the surface equals the number leaving the surface

Answer:

1. There are no charges inside the surface

2. The net charge inside the surface is zero

4. The number of electric field lines entering the surface equals the number leaving the surface

Question 11. A dipole of moment \(\vec{p}\) is placed in a uniform field V \(\vec{E}\). The force on the dipole is \(\vec{P}\) and the torque is \(\vec{\tau}\) . Then

  1. \(\vec{F}=0\)
  2. \(\vec{F}=|\vec{p}| \vec{E}\)
  3. \(|\vec{\tau}|=\vec{p} \cdot \vec{E}\)
  4. \(\vec{\tau}=\vec{p} \times \vec{E}\)

Answer:

1. \(\vec{F}=0\)

4. \(|\vec{\tau}|=\vec{p} \cdot \vec{E}\)

Question 12. A proton and an electron are placed in an electric field.
The forces acting on them are F1 and F2 and their accelerations are and a2 respectively. Then

  1. F1 = F2
  2. F1 ≠ F2
  3. a1 = a2
  4. a1 ≠ a2

Answer:

1. F1 = F2

4. a1 ≠ a2

Question 13. Two identical charges +Q are kept at some fixed distance. A small particle P with charge q is placed midway between them. If P is given a small displacement Δx, it will undergo simple harmonic motion if

  1. q is positive and Δx is along the line joining the charges
  2. q is positive and Δx is perpendicular to the line joining the charges
  3. q is negative and Δx is perpendicular to the line joining the charges
  4. q is negative and Δx is along the line joining the charges

Answer:

1. q is positive and Δx is along the line joining the charges

3. q is negative and Δx is perpendicular to the line joining the charges

Question 14. A charge Q is divided into two equal parts q = \(\frac{Q}{2}\). If the charges q and q are placed at a certain distance,

  1. Coulomb force is equal in magnitude for both but opposite in direction
  2. Coulomb force is dependent on the medium in which the charges are placed
  3. Coulomb force is maximum irrespective of the medium in which the charges are placed
  4. None of the above

Answer:

1. Coulomb force is equal in magnitude for both but opposite in direction

2. Coulomb force is dependent on the medium in which the charges are placed

Question 15. Five-point charges each of charge +q C are placed on five vertices of a regular hexagon of side as shown in

Class 12 Physics Unit 1 Electrostatics Chapter 2 Electric Field Multiple choice question and answer Q 7

  1. The force on -q at 0 due to the charges +q at A and D are balanced
  2. The force on -q due to the charges at B and E are balanced
  3. The resultant force on -q at O is \(\frac{1}{4 \pi \epsilon_0} \cdot \frac{q^2}{h^2} \text {; along } \overrightarrow{O E}\)
  4. The resultant force on -q at O is \(\frac{1}{4 \pi \epsilon_0} \cdot \frac{q^2}{h^2}; \text { along } \overrightarrow{O C}\)

Answer:

1. The force on -q at 0 due to the charges +q at A and D are balanced

4. The resultant force on -q at O is \(\frac{1}{4 \pi \epsilon_0} \cdot \frac{q^2}{h^2}; \text { along } \overrightarrow{O C}\)

Question 16. A ring with a uniform charge Q and radius R, is placed in the yz-plane with its center at the origin. The field intensity at

  1. The origin is zero
  2. The origin is \(\frac{Q}{4 \pi \epsilon_0 R^2}\)
  3. \((x, 0,0) \text { is } \frac{Q}{4 \pi \epsilon_0 x^2}\)
  4. \((x, 0,0) \text { is } \frac{Q x}{4 \pi \epsilon_0\left(R^2+x^2\right)^{3 / 2}}\)

Answer:

1. The origin is zero

4. \((x, 0,0) \text { is } \frac{Q x}{4 \pi \epsilon_0\left(R^2+x^2\right)^{3 / 2}}\)

Question 17. A pendulum bob of mass m, carrying a charge q is at rest with its string making an angle 6 with the vertical in a uniform horizontal electric field E. The tension in the string is

  1. \(\frac{m g}{\sin \theta}\)
  2. \(\frac{m g}{\cos \theta}\)
  3. \(\frac{q E}{\sin \theta}\)
  4. \(\frac{q E}{\cos \theta}\)

Answer:

2. \(\frac{m g}{\cos \theta}\)

3. \(\frac{q E}{\sin \theta}\)

Question 18. A cubical region of side a has its center at the origin. It encloses three fixed point charges -q at \(\left(0,-\frac{a}{4}, 0\right)\), 3q at (0, 0, 0), and -q at \(\left(0, \frac{a}{4}, 0\right)\). Choose the correct options

Class 12 Physics Unit 1 Electrostatics Chapter 2 Electric Field Multiple choice question and answer Q 10

  1. The net electric flux crossing the plane x = +\(\frac{a}{2}\)is equal to the net electric flux crossing the plane x = –\(\frac{a}{2}\)
  2. The net electric flux crossing the plane y = +\(\frac{a}{2}\) is more than the net electric flux crossing the plane y = –\(\frac{a}{2}\)
  3. The net electric flux crossing the entire region is \(\frac{q}{\epsilon_0}\)
  4. The net electric flux crossing the plane z = +\(\frac{a}{2}\)is equal to the net electric flux crossing the plane x = +\(\frac{a}{2}\)

Answer:

1. The net electric flux crossing the plane x = +\(\frac{a}{2}\)is equal to the net electric flux crossing the plane x = –\(\frac{a}{2}\)

3. The net electric flux crossing the entire region is \(\frac{q}{\epsilon_0}\)

4. The net electric flux crossing the plane z = +\(\frac{a}{2}\)is equal to the net electric flux crossing the plane x = +\(\frac{a}{2}\)

Question 19. A few electric field lines for a system of two charges Q1 and Q2, fixed at two different points on the x-axis. These lines suggest that

Class 12 Physics Unit 1 Electrostatics Chapter 2 Electric Field Multiple choice question and answer Q 108

  1. \(\left|Q_1\right|>\left|Q_2\right|\)
  2. \(\left|Q_1\right|<\left|Q_2\right|\)
  3. At a finite distance to the left of Q1, the electric field is zero
  4. At a finite distance to the right of Q2, the electric field is zero

Answer:

1. \(\left|Q_1\right|>\left|Q_2\right|\)

4. At a finite distance to the right of Q2, the electric field is zero

Question 20. A spherical metal shell A of radius RA and a solid metal sphere B of radius RB(<RA) are kept far apart and each is given charge +Q. Now they are connected by a thin metal wire. Then

  1. \(E_A^{\text {inside }}=0\)
  2. QA > QB
  3. \(\frac{\sigma_A}{\sigma_B}=\frac{R_B}{R_A}\)
  4. \(E_A^{\text {on surface }}<E_B^{\text {on surface }}\)

Answer:

1. \(E_A^{\text {inside }}=0\)

2. QA > QB

3. \(\frac{\sigma_A}{\sigma_B}=\frac{R_B}{R_A}\)

4. \(E_A^{\text {on surface }}<E_B^{\text {on surface }}\)

Question 21. A non-conducting solid sphere of radius R is uniformly charged. The magnitude of electric field intensity due to the sphere at a distance r from its center

  1. Increases as r increases for r<R
  2. Decreases as r increases for 0 < r < ∞
  3. Decreases as r increases for R < r < ∞
  4. Is discontinuous at r = R

Answer:

1. Increases as r increases for r<R

3. Decreases as r increases for R < r < ∞

Question 22. Which of the following statements is not true for Coulomb’s law in electrostatics?

  1. The law is applicable only for a point charge
  2. The law is applicable to any distance
  3. According to this law force between two charges depends on the medium
  4. It is an inverse square law

Answer: 2. The law is applicable for any distance

Question 23. Select the correct statement.

  1. Both electrostatic force and gravitational force are nonconservative forces
  2. Electrostatic force is conservative but gravitational force is nonconservative
  3. Electrostatic force is nonconservative but gravitational force is conservative
  4. Both electrostatic force and gravitational force are conservative force

Answer: 4. Both electrostatic force and gravitational force are conservative force

Question 24. The number of esu charges in 1C is

  1. 3 x 1010
  2. 3 x 109
  3. 3 x 108
  4. \(\frac{1}{300}\)

Answer: 2. 3 x 109

Question 25. Charge q1 exerts force on another charge q2. A third charge q3 is brought near them. The force applied by q1 on q2

  1. Will decrease
  2. Will increase
  3. Will remain the same
  4. Will increase if the nature of q1 and q2 is the same and will decrease if their nature is opposite

Answer: 3. Will remain the same

Question 26. The number of electrons corresponding to 1 coulomb of charge is

  1. 6.25 x 1017
  2. 6.25 x 1018
  3. 6.25 x 1019
  4. 1.6 X 1019

Answer: 2. 6.25 x 1018

Question 27. Two point charges separated by a distance d repel each other with a force of 9N. If the separation between them becomes 3d, the force of repulsion will be

  1. 1N
  2. 3N
  3. 6N
  4. 27 N

Answer: 1. 1N

Question 28. Charges 4Q, q, and Q are placed along the x-axis at positions 1 x = 0, x =\(\frac{l}{2}\) and x = l respectively. The value of q so that the force on charge Q is zero is

  1. Q
  2. \(\frac{Q}{2}\)
  3. – \(\frac{Q}{2}\)
  4. -Q

Answer: 4. -Q

Question 29. Mutual electrostatic force F is acting on two tiny charged spheres when they are d distance apart in air. Keeping all the external conditions fixed, if the spheres are immersed in a liquid of dielectric constant 2, then the force acting on each sphere becomes

  1. 4F
  2. 2F
  3. F
  4. \(\frac{F}{2}\)

Answer: 4. \(\frac{F}{2}\)

Question 30. When a metal plate is introduced between two charges
kept at some distance from each other, the electrostatic force between the two charges will

  1. Decrease
  2. Increase
  3. Remain the same
  4. Zero

Answer: 1. Decrease

Question 31. Two point charges +4q and +q are placed 30 cm apart.
The electric field intensity at a point on the line joining the two charges is zero. The point is situated at a distance

  1. 15 cm from 4q
  2. 20 cm from 4q
  3. 7.5 cm from q
  4. 5 cm from q

Answer: 2. 20 cm from 4q

Question 32. If E is the intensity of the electric field at a distance r(r>R) due to a uniformly charged spherical shell, then

  1. E α r
  2. \(E \propto \frac{1}{r}\)
  3. E α r2
  4. \(E \propto \frac{1}{r^2}\)

Answer: 4. \(E \propto \frac{1}{r^2}\)

Question 33. Field lines due to the charges q1 and q2 have been shown. The nature of the charges q1 and 22 is

Class 12 Physics Unit 1 Electrostatics Chapter 2 Electric Field Multiple choice question and answer Q12

  1. Both negative
  2. Both positive
  3. q1 is positive but q2 is negative
  4. q1 is negative but q2 is positive

Answer: 1. Both negative

Question 34. A hollow charged sphere of radius 2 m does not produce
any field intensity

  1. At any internal point of the sphere
  2. At any external point of the sphere
  3. At a distance greater than 2 m
  4. At a distance greater than 10 m

Answer: 1. At any internal point of the sphere

Question 35. An electron of charge -q and mass m is placed in a uniform electric field of intensity E. The value of E is such that the face on the electron due to the electric field is equal to its weight. Under this condition the value of E is

  1. \(\frac{m g}{e}\)
  2. mge
  3. \(\frac{e}{m g}\)
  4. \(\frac{e^2 g}{m^2}\)

Answer: 1. \(\frac{m g}{e}\)

Question 36. An electron enters normally in a uniform electric field of intensity 3200 V/m with a speed of 4 X 107 m/s. The electron covers a distance of 0.10 m. The deflection of the electron is

  1. 1.76 mm
  2. 17.6 mm
  3. 176 mm
  4. 0.176 mm

Answer: 1. 1.76 mm

Question 37. An electric field of intensity 9 x 104 N/C is produced at a point 2 cm away from an infinitely long straight-charged conducting wire. Electric charge per unit length becomes

  1. 2 x 10-7 C.m-1
  2. 10-7 C.m-1
  3. 9 x 104 C.m-1
  4. None of these

Answer: 2. 10-7 C.m-1

Question 38. A thin conducting ring of radius R is given a charge + Q. The electric field at the center O of the ring due to the charge on the part AKB of the ring is E. The electric field at the center due to the charge on the part ACDB of the ring is

Class 12 Physics Unit 1 Electrostatics Chapter 2 Electric Field Multiple choice question and answer Q 18

  1. E along KO
  2. 3E along OK
  3. 3E along KO
  4. E along OK

Answer: 4. E along OK

Question 39. A metal sphere is placed in a uniform electric field. The electrostatic lines of force will pass through which of the paths?

Class 12 Physics Unit 1 Electrostatics Chapter 2 Electric Field Multiple choice question and answer Q 39

  1. 1
  2. 2
  3. 3
  4. 4

Answer: 4. 4

Question 40. Due to an electric charge Q, field intensity at the position of test charge q0 is E. The test charge is replaced by -q0, then the field intensity becomes

  1. \(-q_0 \vec{E}\)
  2. \(\frac{\vec{E}}{-q_0}\)
  3. 0
  4. \(\vec{E}\)

Answer: 4. \(\vec{E}\)

Question 41. If an electric dipole of moment \(\vec{E}\) is placed in an electric field of intensity \(\vec{E}\), the torque acting on the dipole is

  1. \(\vec{\tau}=\vec{p} \cdot \vec{E}\)
  2. \(\vec{\tau}=\vec{p} \times \vec{E}\)
  3. \(\vec{\tau}=\vec{p}+\vec{E}\)
  4. \(\vec{\tau}=\vec{p}-\vec{E}\)

Answer: 2. \(\vec{\tau}=\vec{p} \times \vec{E}\)

Question 42. The direction of the intensity of the electric field at a point on the perpendicular bisector of an electric dipole (\(\vec{p}\) = electric dipole moment) is

  1. Along \(\vec{p}\)
  2. Opposite to the direction of \(\vec{p}\)
  3. Perpendicular to \(\vec{p}\)
  4. In any direction

Answer: 2. Opposite to the direction of \(\vec{p}\)

Question 43. A charge placed at a distance from an electric dipole on its axis experiences a force F. If the distance is doubled, the force will become

  1. 2F
  2. F/2
  3. F/4
  4. F/8

Answer: 4. F/8

Question 44. If E1 is the electric field strength of a short dipole at a point on the axis and E2 that on the perpendicular bisector at the same distance, then

  1. E1 = E2
  2. E2 = 2E1
  3. E2 = 2E1
  4. E1 = 3E2

Answer: 2. E2 = 2E1

Question 45. An electric dipole is placed along x-axis at the origin O . A point P is at a distance of 20 cm from this origin such that OP makes an angle with the x-axis. If the electric field at P makes an angle Η with the x-axis, then the value of 6 is

  1. \(\frac{\pi}{3}\)
  2. \(\frac{\pi}{3}+\tan ^{-1} \frac{\sqrt{3}}{2}\)
  3. \(\frac{3 \pi}{2}\)
  4. \(\tan ^{-1} \frac{\sqrt{3}}{2}\)

Answer: 2. \(\frac{\pi}{3}+\tan ^{-1} \frac{\sqrt{3}}{2}\)

Question 46. A small electric dipole is placed at the origin with its dipole moment directed along the positive X-axis. The direction of electric field at point (2, 2√2, 0) is along

  1. -Z-axis
  2. +Z-axis
  3. -Y-axis
  4. +Y-axis

Answer: 4. +Y-axis

Question 47. An electric dipole is situated in an electric field of uniform intensity E whose dipole moment is p and moment of inertial is I. If the dipole is displaced by a small angle 9 then the angular frequency of its oscillation is

  1. \(\left(\frac{p E}{I}\right)^{\frac{1}{2}}\)
  2. \(\left(\frac{p E}{I}\right)^{\frac{3}{2}}\)
  3. \(\left(\frac{I}{p E}\right)^{\frac{1}{2}}\)
  4. \(\left(\frac{p}{I E}\right)^{\frac{1}{2}}\)

Answer: 1. \(\left(\frac{p E}{I}\right)^{\frac{1}{2}}\)

Question 48. The electric field intensity at a point near a sphere of radius r and surface density of charge cr placed in a medium of dielectric constant k is

  1. \(\frac{4 \pi \sigma}{k}\)
  2. \(\frac{2 \pi \sigma}{k}\)
  3. \(\frac{\pi \sigma}{kr}\)
  4. \(\frac{\pi \sigma}{2kr}\)

Answer: 1. \(\frac{4 \pi \sigma}{k}\)

Question 49. The electric flux linked with a surface becomes maximum if the angle between the field lines and the normal to the surface is

  1. 45°
  2. 90°
  3. 180°

Answer: 1. 0°

Question 50. A circular plate of radius r is placed parallel to a uniform electric field of intensity E. The flux linked with the circular plate is

  1. 0
  2. E x πr²
  3. E x 2πr
  4. E x 4πr²

Answer: 1. 0

Question 51. If the inward and the outward electric flux through a closed surface be Φ1 and Φ2 the charge inside the closed surface is

  1. \(\left(\phi_1-\phi_2\right) \epsilon_0\)
  2. \(\left(\phi_2-\phi_1\right) \epsilon_0\)
  3. \(\left(\phi_1+\phi_2\right) / \epsilon_0\)
  4. \(\left(\phi_2+\phi_1\right) / \epsilon_0\)

Answer: 4. \(\left(\phi_2+\phi_1\right) / \epsilon_0\)

Question 52. A square of side 20 cm Is enclosed within a sphere of radius 80 cm. The centers of the sphere and the square coincide at a common point. There are four charges of 2 x 10-6 C, -5 x 10-6 C, -3 X 10-6 C, and 6 x 10-6 C at the four comers of the square. Tire total electric flux passing through the surface of the sphere in N.m2/C.

  1. 16π x 10-6
  2. zero
  3. 8π x 10-6
  4. 36π x 10-6

Answer: 2. zero

Question 53. The distribution of charges is shown in the figure. The flux of the electric field due to these charges through the surface S is

Class 12 Physics Unit 1 Electrostatics Chapter 2 Electric Field Multiple choice question and answer Q 53

  1. \(\frac{3 q}{\epsilon_0}\)
  2. \(\frac{2 q}{\epsilon_0}\)
  3. \(\frac{q}{\epsilon_0}\)
  4. 0

Answer: 4. 0

Question 54. A hollow cylinder contains a charge QC. If 888 is the electric flux in a unit of V.m associated with the curved surface B, the flux linked with the plane surface A in a unit of V.m will be (e0 = permittivity)

Class 12 Physics Unit 1 Electrostatics Chapter 2 Electric Field Multiple choice question and answer Q 54

  1. \(\frac{\phi}{3}\)
  2. \(\frac{q}{2 \epsilon_0}\)
  3. \(\frac{1}{2}\left(\frac{q}{\epsilon_0}-\phi\right)\)
  4. \(\frac{q}{\epsilon_0}-\phi\)

Answer: 3. \(\frac{1}{2}\left(\frac{q}{\epsilon_0}-\phi\right)\)

Question 55. A hemispherical bowl of radius r is kept in a uniform electric field of Intensity Total electric flux through the bowl is

Class 12 Physics Unit 1 Electrostatics Chapter 2 Electric Field Multiple choice question and answer Q 55

  1. 2πrE
  2. 4πr²E
  3. 2πr²E
  4. πr²E

Answer: 4. πr²E

Question 56. Two electric charges, + 8q and -2q, are placed at x = 0 and x = L respectively. At what point on the x-axis, does net electric field intensity due to two charges become zero?

  1. (2L, 0)
  2. \(\left(\frac{L}{4}, 0\right)\)
  3. (8L, 0)
  4. (4L, 0)

Answer: 1. (2L, 0)

Question 57. A circular copper ring of radius r, placed in a vacuum, has a charge q on it. The electric field intensity at the center of the ring is E1. The electric field intensity on the axis of the ring at a distance x from its center is E2. The value of E2 will be maximum when x = x’.

1. What is the value of E1?

  1. 0
  2. q x πr²
  3. q x 2πr²
  4. \(\frac{q}{r^2}\)

Answer: 1. 0

2. What is the value of E1?

  1. \(\frac{q}{\left(x^2+r^2\right)^{1 / 2}}\)
  2. \(\frac{q x^2}{\left(x^2+r^2\right)^{3 / 2}}\)
  3. \(\frac{q}{\left(x^2+r^2\right)^{3 / 2}}\)
  4. \(\frac{q x}{\left(x^2+r^2\right)^{3 / 2}}\)

Answer: 4. \(\frac{q x}{\left(x^2+r^2\right)^{3 / 2}}\)

3. What is the value of x’?

  1. √2r
  2. \(\frac{r}{\sqrt{2}}\)
  3. \(\frac{r}{\sqrt{3}}\)
  4. √3r

Answer: 2. \(\frac{r}{\sqrt{2}}\)

Question 58. An electron is released from rest in a uniform electric field of 106 N.C-1. The acceleration of the electron is a. The time taken by the electron in attaining a speed of 0.1c (where c = 3 x 108m.s-1 is t.

1. What is the value of a?

  1. 1.76 x 1017 m.s-2
  2. 2.56 x 1018 m s-2
  3. 1.2 x 1015 m s-2
  4. 3.45 X 1017 m.s-2

Answer: 1. 1.76 x 1017 m.s-2

2. What is the value of E?

  1. 2.8 X 10-10s
  2. 1.7 X 10-10s
  3. 3.4 X 10-10s
  4. 1.2 X 10-8s

Answer: 2. 1.7 X 10-10s

Question 59. A point mass M is attached to one end of a massless rigid non-conducting rod of length L. Another equal point mass is attached to the other end of the rod. The two particles carry charges +q and -q respectively. This arrangement is held in the region of uniform electric field E such that the rod makes a small angle 000 (say about 5° ) with field direction. The moment of inertia of the rod is I. Now answer the questions.

1. The ratio \(\frac{\theta}{\alpha}\) is

  1. \(\frac{2 I}{2 E L}\)
  2. \(\frac{I}{q E L}\)
  3. \(\frac{I}{2 q E L}\)
  4. \(\frac{2 I}{3 q E L}\)

Answer: 2. \(\frac{I}{q E L}\)

2. The time period of SHM of the given system is

  1. \(2 \pi \sqrt{\frac{M L}{2 q E}}\)
  2. \(\frac{2 \pi}{3} \sqrt{\frac{M L}{q E}}\)
  3. \(\frac{\pi}{2} \sqrt{\frac{M L}{2 q E}}\)
  4. \(\frac{\pi}{4} \sqrt{\frac{M L}{2 q E}}\)

Answer 1. \(2 \pi \sqrt{\frac{M L}{2 q E}}\)

3. The time period for the rod to become parallel to E is

  1. \(2 \pi \sqrt{\frac{M L}{q E}}\)
  2. \(\frac{2 \pi}{3} \sqrt{\frac{M L}{q E}}\)
  3. \(\frac{\pi}{2} \sqrt{\frac{M L}{2 q E}}\)
  4. \(\frac{\pi}{4} \sqrt{\frac{M L}{2 q E}}\)

Answer 3. \(\frac{\pi}{2} \sqrt{\frac{M L}{2 q E}}\)

Question 60. An electron having charge e moves with velocity v in +x direction. An electric field acts on it along +y direction. The force on the electron acts along

  1. +z direction
  2. -z-direction
  3. +y direction
  4. -y direction

Answer: 4. -y direction

⇒ \(\vec{E}=\hat{j} E\)

Force, \(\vec{F}=q \vec{E}=-e \vec{E}=(-\hat{j}) e E\)

∴ The force acts along -y direction.

The option 4 is correct.

Question 61. A charge Q is situated inside a cube placed in air. The electric flux passing through all six faces is

  1. \(\frac{Q}{\epsilon_0}\)
  2. \(\frac{Q}{2 \epsilon_0}\)
  3. \(\frac{Q}{6 \epsilon_0}\)
  4. \(\frac{Q}{8 \epsilon_0}\)

Answer: 1.

According to Gauss’ theorem, electric flux = \(\frac{Q}{\epsilon_0}\)

Option 1 is correct.

Question 62. The line AA’ is on a charged infinite conducting plane which is perpendicular to the plane of the paper. The plane has a surface density of charge cr and B is a ball of mass m with a like charge of magnitude q.B is connected by a string from a point on line 1. The tangent of the angle (0) formed between the line AA’ and the string is

Class 12 Physics Unit 1 Electrostatics Chapter 2 Electric Field The line AA' is on a charged infinite

  1. \(\frac{q \sigma}{2 \epsilon_0 m g}\)
  2. \(\frac{q \sigma}{4 \pi \epsilon_0 m g}\)
  3. \(\frac{q \sigma}{2 \pi \epsilon_0 m g}\)
  4. \(\frac{q \sigma}{\epsilon_0 m g}\)

Answer: 1. \(\frac{q \sigma}{2 \epsilon_0 m g}\)

The outward electric field at position B due to the conducting plane AA’,

⇒ \(E=\frac{\sigma}{2 \epsilon_0}\)

∴ Outward electric force, \(F=q E=\frac{q \sigma}{2 \epsilon_0}\)

Now, downward weight = mg

At the equilibrium position of B, the string will be directed towards the resultant of F and mg.

Class 12 Physics Unit 1 Electrostatics Chapter 2 Electric Field The outward electric field at position

Then, \(\tan \theta=\frac{F}{m g}=\frac{q \sigma}{2 \epsilon_0 m g}\)

Option 1 is correct

Question 63. A charge q is placed at one corner of a cube. The electric flux through any of the three faces adjacent to the charge is zero.
The flux through any one of the other three faces is

  1. \(\frac{q}{3 \epsilon_0}\)
  2. \(\frac{q}{6 \epsilon_0}\)
  3. \(\frac{q}{12 \epsilon_0}\)
  4. \(\frac{q}{24 \epsilon_0}\)

Answer: 4. \(\frac{q}{24 \epsilon_0}\)

In three-dimensional space, if we consider a Gaussian surface surrounding the charge q, then according to Gauss’ theorem, the electric flux through that Gaussian surface

⇒ \(\frac{q}{\epsilon_0}\)

Here, the cube under consideration is occupying \(\frac{1}{8}\)th volume of the three-dimensional space surrounding charge q, therefore the electric flux through the three non-adjacent faces = \(\frac{q}{8 \epsilon_0}\)

∴ The electric flux through any one of those faces

⇒ \(\frac{1}{3} \times \frac{q}{8 \epsilon_0}=\frac{q}{24 \epsilon_0}\)

The option 4 is correct.

Question 64. Two charges +q and -q, arc placed at a distance a in a uniform electric field. The dipole moment of the combination is \(2 q a(\cos \theta \hat{i}+\sin \theta \hat{j})\), where 0 Is the angle between the direction of the field and the line joining the two charges. Which of the following statement(s) is/are correct?

  1. The torque exerted by the field on the dipole vanishes
  2. The net force on the dipole vanishes
  3. The torque is independent of the choice of coordinates
  4. The net force is independent of a

Answer:

2. The net force on the dipole vanishes

3. The torque is independent of the choice of coordinates

4. The net force is independent of a

The resultant force is applied on a dipole in a uniform electric field, but the moment of force or torque cannot be zero.

Torque, \(\vec{\tau}=\vec{p} \times \vec{E}\) [\(\vec{p}\) = dipole moment, \(\vec{E}\) = electric field]

This vector relation is independent of the choice of the coordinate system.

Again, since the resultant force is zero, hence it is independent of a.

The potions 2, 3, and 4 are correct.

Question 65. A charged particle of mass m1 and charge q1 is revolving in a circle of radius r. Another charged particle of charge q2 and mass m2 is situated at the center of the circle. If the velocity and time period of the revolving particle are v and T respectively, then

  1. \(v=\sqrt{\frac{q_1 q_2 r}{4 \pi \epsilon_0 m_1}}\)
  2. \(v=\frac{1}{m_1} \sqrt{\frac{q_1 q_2}{4 \pi \epsilon_0 r}}\)
  3. \(T=\sqrt{\frac{16 \pi^3 \epsilon_0 m_1^2 r^3}{q_1 q_2}}\)
  4. \(T=\sqrt{\frac{16 \pi^3 \epsilon_0 m_2 r^3}{q_1 q_2}}\)

Answer: 3. \(T=\sqrt{\frac{16 \pi^3 \epsilon_0 m_1^2 r^3}{q_1 q_2}}\)

Here, both q1 and q2 are either positive or negative.

Thus, \(\frac{m_1 v^2}{r}=\frac{1}{4 \pi \epsilon_0} \cdot \frac{q_1 q_2}{r^2} \quad\)

or, \(v=\sqrt{\frac{1}{4 \pi \epsilon_0 m_1} \cdot \frac{q_1 q_2}{r}}\)

Now, time period \(T=\sqrt{\frac{16 \pi^3 \epsilon_0 m_1 r^3}{q_1 q_2}}\) [∵ \(\nu=r \omega \text { and } \omega=\frac{2 \pi}{T}\)]

Hence, none of the options is correct

Question 66. A current \(I=I_0 e^{-\lambda t}\) is following in a circuit consisting of a parallel combination of resistance R and capacitance C. The total charge over the entire pulse period Is

  1. \(\frac{I_0}{\lambda}\)
  2. \(\frac{2 I_0}{\lambda}\)
  3. I0λ
  4. \(e^{I_0 \lambda}\)

Answer: 1. \(\frac{I_0}{\lambda}\)

Here, \(I=I_0 e^{-\lambda t}\)

The total charge over the entire pulse period

⇒ \(\int_0^{\infty} I d t=I_0 \int_0^{\infty} e^{-\lambda t} d t\)

= \(I_0\left[\frac{e^{-\lambda t}}{-\lambda}\right]_0^{\infty}\)

= \(\frac{I_0}{\lambda}\)

The option 1 Is correct

Question 67. A positive charge Q is situated at the center of a cube. The electric flux through any face of the cube is (in SI units)

  1. \(\frac{Q}{6 \epsilon_0}\)
  2. 4πQ
  3. \(\frac{Q}{4 \pi \epsilon_0}\)
  4. \(\frac{Q}{6 \pi \epsilon_0}\)

Answer: 1. \(\frac{Q}{6 \epsilon_0}\)

From Gauss’ theorem, the total flux linked with she surfaces of the cube is Ω, then

⇒ \(\phi=\int \vec{E} \cdot d \vec{S}=\frac{Q}{\epsilon_0}\)

∴ Flux linked with any one surface = \(=\frac{\phi}{6}=\frac{Q}{6 \epsilon_0}\)

The option 1 is correct

Question 68. A charge of 0.8C is divided into two charges Q1 and Q2. These are kept at a separation of 30cm. The force on Q1 is maximum when

  1. Q1 = Q2 = 0.4C
  2. Q1 ≈ 0.8C , Q2 negligible
  3. Q1 negligible, Q2 ≈ 0.8C
  4. Q1 = 0.2C , Q2 = 0.6C

Answer: 1. Q1 = Q2 = 0.4C

We have, Q1 + Q2 = 0.8 C

Therefore, the force of attraction between the two charges,

⇒ \(F=\frac{Q_1 Q_2}{4 \pi \epsilon_0 r^2}=\frac{Q_1\left(0.8-Q_1\right)}{4 \pi \epsilon_0 \times\left(30 \times 10^{-2}\right)^2}\)

This force will be maximum if \(\frac{d F}{d Q_1}=0\)

∴ \(\frac{d}{d Q_1}\left[\frac{Q_1\left(0.8-Q_1\right)}{4 \pi \epsilon_0 \times\left(30 \times 10^{-2}\right)^2}\right]=0\)

or, Q1 = 0.4 C and Q2 = 0.8-0.4 = 0.4 C

The option 1 is correct

Question 69. A particle with charge QC, tied at the end of an inextensible string of length R meter, revolves in a vertical plane. At the center of the circular trajectory, there is a fixed charge of magnitude QC. The mass of the moving charge M is such that \(M g=\frac{Q^2}{4 \pi \epsilon_0 R^2}\). If at the highest position of the particle, the tension of the string just vanishes, the horizontal velocity at the lowest point has to be

  1. 0
  2. \(2 \sqrt{g R}\)
  3. \(\sqrt{2 g R}\)
  4. \(\sqrt{5 g R}\)

Answer: 2. \(2 \sqrt{g R}\)

The tension on the top of the vertical circular path is zero.

Electric Field tension on the top of the vertical circular path is zero

Therefore, at that position,

⇒ \(M g-\frac{Q^2}{4 \pi \epsilon_0 R^2}=\frac{M v^2}{R}\)

But, \(M g=\frac{Q^2}{4 \pi \epsilon_0 R^2}\)

∴ v = 0

Now, from the conservation of mechanical energy,

(PE)B = (AKE)A [∵ \((\mathrm{KE})_B=\frac{1}{2} M v^2=0\)]

or, \(M g \times 2 R=\frac{1}{2} M v_0^2 \quad\)

or, \(v_0=2 \sqrt{g R}\)

The option 2 is correct.

Question 70. A particle with charge e and mass m. moving along the Y-axis with a uniform speed u enters a region where a uniform electric field E is acting along the 7-axis. The particle starts to move in a parabola. Its focal length (neglecting any effect of gravity) is

  1. \(\frac{2 m u^2}{e E}\)
  2. \(\frac{e E}{2 m u^2}\)
  3. \(\frac{m u}{2 e E}\)
  4. \(\frac{m u^2}{2 e E}\)

Answer: 4. \(\frac{m u^2}{2 e E}\)

If the particle moves through the electric field for a time t

t = \(\frac{x}{u}\) [∵t = displacement along A’-axis]

∴ Displacement along 7-axis,

⇒ \(y=\frac{1}{2} a t^2=\frac{1}{2}\left(\frac{e E}{m}\right) \cdot \frac{x^2}{u^2}\) [∵ a = acceleration of the particle = \(\)]

∴ \(x^2=\frac{2 m u^2}{e E} y\)

Now, compared with the general equation of a parabola,

x² = 4a’y

Focal length ofthe parabola \(a^{\prime}=\frac{2 m u^2}{4 e E}=\frac{m u^2}{2 e E}\)

The option 4 is correct.

Question 71. A unit negative charge with mass M resides at the midpoint of the straight line of length 2a adjoining two fixed charges of magnitude +Q each. If it is given a very small displacement x (x << a) in a direction perpendicular to the straight line, it will

  1. Come back to its original position and stay there
  2. Execute oscillations with frequency \(\frac{1}{2 \pi} \sqrt{\frac{Q}{4 \pi \epsilon_0 M a^3}}\)
  3. Fly to infinity
  4. Execute oscillations with frequency \(\frac{1}{2 \pi} \sqrt{\frac{Q}{4 \pi \epsilon_0 M a^2}}\)

Answer: 2. Execute oscillations with frequency \(\frac{1}{2 \pi} \sqrt{\frac{Q}{4 \pi \epsilon_0 M a^3}}\)

⇒ \(F_{\text {net }}=-2 F \cos \theta=-2 \times \frac{k Q \times 1}{\left(x^2+a^2\right)} \times \frac{x}{\sqrt{x^2+a^2}}\)

⇒ \(=\frac{-2 k Q}{\left(x^2+a^2\right)^{\frac{3}{2}}} x\)

∴ \(F_{\text {net }} \approx-\left(\frac{2 k Q}{a^3}\right) x\) [∵ x << a]

Electric Field A unit negative charge with mass M resides at the midpoint

Frequency of oscillation

⇒ \(\frac{1}{2 \pi} \sqrt{\frac{2 k Q}{M a^3}}\)

= \(\frac{1}{2 \pi} \sqrt{\frac{2 \times \frac{1}{4 \pi \epsilon_0} Q}{M a^3}}\)

= \(\frac{1}{2 \pi} \sqrt{\frac{Q}{2 \pi \epsilon_0 M a^3}}\)

[frequent of SHM, \(n=\frac{1}{2 \pi} \sqrt{\frac{\text { acceleration }}{\text { displacement }}}\)]

None of the options are correct.

Question 72. Consider a region in free space bounded by the surfaces of an imaginary cube having sides of length as shown in the diagram. A charge + Q is placed at the center 0 of the cube. P is such a point outside the cube that the line OP perpendicularly intersects the surface ABCD at R and also OR = RP = \(\frac{a}{2}\). A charge + Q is placed at point P also. What is the total electric flux through the five faces of the Is cube other than ABCD?

Electric Field Consider a region in free space bounded by the surfaces

  1. \(\frac{Q}{\epsilon_0}\)
  2. \(\frac{5 Q}{6 \epsilon_0}\)
  3. \(\frac{10 Q}{6 \epsilon_0}\)
  4. 0

Answer: 1. \(\frac{Q}{\epsilon_0}\)

According to Gauss’ theorem,

⇒ \(\phi=\oint_s \vec{E} \cdot d \vec{s}=\frac{Q}{\epsilon_0}\)

The electric flux linked with each surface of the cube is equal and is given by

⇒ \(\phi=\frac{Q}{6 \epsilon_0}\)

The total electric flux linked with the five faces of the cube other than ABCD due to charge +Q at point O is \(\frac{5 Q}{6 \epsilon_0}\)

Given, OR = RP = \(\frac{a}{2}\) and the charges placed at points P and 0 are equal.

Now, if the charge +Q placed at point P is enclosed by a similar cube, then the electric flux only through surface ABCD will be \(\frac{Q}{6 \epsilon_0}\).

The same amount of flux \(\left(\frac{Q}{6 \epsilon_0}\right)\) will emerge from the remaining five surfaces of the cube centered at 0.

Hence the total electric flux linked with the five faces of other than ABCD

⇒ \(\frac{5 Q}{6 \epsilon_0}+\frac{Q}{6 \epsilon_0}\)

= \(\frac{Q}{\epsilon_0}\)

The option 1 is correct.

None of the option

Question 73. Four equal charges of value + Q are placed at any four vertices of a regular hexagon of side a . By suitably choosing the vertices, what can be the maximum possible magnitude of the electric field at the center of the hexagon?

  1. \(\frac{Q}{4 \pi \epsilon_0 a^2}\)
  2. \(\frac{\sqrt{2} Q}{4 \pi \epsilon_0 a^2}\)
  3. \(\frac{\sqrt{3} Q}{4 \pi \epsilon_0 a^2}\)
  4. \(\frac{2 Q}{4 \pi \epsilon_0 a^2}\)

Answer: 3. \(\frac{\sqrt{3} Q}{4 \pi \epsilon_0 a^2}\)

If the four +Q charges are placed at four adjacent vertices, then the net electric field at point O will be maximum and the electric field (E1) due to each charge will be equal in magnitude.

Class 12 Physics Unit 1 Electrostatics Chapter 2 Electric Field If the four +Q charges are placed at four adjacent vertices

Since ABCDEF is a regular hexagon,

So, AO = BO = CO = DO = a

The net electric field at point O due to +Q charges placed at points A and D is zero. Hence the net electric field at point O due to +Q charges placed at points B and C is the maximum value of the electric field.

Now, \(E_1=\frac{Q}{4 \pi \epsilon_0 a^2}\)

∴ \(E^{\prime}=\sqrt{E_1^2+E_1^2+2 E_1 E_1 \cos 60^{\circ}}\)

= \(\frac{\sqrt{3} Q}{4 \pi \epsilon_0 a^2}\)

The option 3 is correct.

Question 74. The bob of n provolone of muss m, Hiispeoded by no inextensible string of length l. us shown In the figure carries u small charge q. An Infinite horizontal plane conductor with uniform surface charge density σ Is placed below it. What will be the time period of the pendulum for small amplitude oscillations?

Electric Field the bob of a apendulum of mass

  1. \(2 \pi \sqrt{\frac{L}{\left(g-\frac{m q}{c_0 \sigma}\right)}}\)
  2. \(\sqrt{\frac{L}{\left(g-\frac{m q \sigma}{\epsilon_0}\right)}}\)
  3. \(\frac{1}{2 \pi} \sqrt{\frac{L}{\left(g-\frac{q \sigma}{\epsilon_0 m}\right)}}\)
  4. \(2 \pi \sqrt{\frac{L}{\left(g-\frac{q \sigma}{\epsilon_0 m}\right)}}\)

Answer: 4. \(2 \pi \sqrt{\frac{L}{\left(g-\frac{q \sigma}{\epsilon_0 m}\right)}}\)

Electric field due to a conducting plate, \(E=\frac{\sigma}{\epsilon_0}\)

The electric force on the bob of the pendulum, \(F_e=q E=\frac{q \sigma}{\epsilon_0}\)

The force Fe is directed opposite to gravity

Electric Field the bob of a apendulum of mass

The net force on the bob is directed vertically downward,

⇒ \(F^{\prime}=m g^{\prime}=m g-F_e=m\left(g-\frac{q \sigma}{c_0 m}\right)\)

or, \(g^{\prime}=g-\frac{q \sigma}{\epsilon_0 m}\)

∴The time period for small amplitude oscillations,

⇒ \(T=2 \pi \sqrt{\frac{L}{g^{\prime}}}=2 \pi \sqrt{\frac{L}{\left(g-\frac{q \sigma}{\epsilon_0 m}\right)}}\)

Option 4 Is correct

Question 75. A long cylindrical shell carries a positive surface charge in the upper half and a negative surface charge -tr In the lower half. The electric field lines around the cylinder will look like the figure given (figures are schematic and not drawn to scale)

Electric Field A long cylindrical shell carries positive surface charge

Answer: 1.

Actually, the cylindrical shell Is an electric dipole.

Option 1 Is correct.

Question 76. The region between two concentric spheres of radii a and b, respectively, has volume charge density \(\frac{A}{r}\), where A is a constant and r is the distance from the center. At the center of the spheres is a point charge Q. The value of A such that the electric field in the region between the spheres will be constant, is:

Electric Field The region between two concentric spheres of radii

  1. \(\frac{Q}{2 \pi a^2}\)
  2. \(\frac{Q}{2 \pi\left(b^2-a^2\right)}\)
  3. \(\frac{2Q}{\pi\left(a^2-b^2\right)}\)
  4. \(\frac{2 Q}{\pi a^2}\)

Answer: 1. \(\frac{Q}{2 \pi a^2}\)

Let us consider having external radius r and internal radius a.

Amount of charge on the shell,

⇒ \(Q_1=\int_a^r 4 \pi r^2 d r \cdot \frac{A}{r}=2 \pi A\left(r^2-a^2\right)\)

Class 12 Physics Unit 1 Electrostatics Chapter 2 Electric Field sphericalshell

Thus, the field intensity at any point at a distance r from the center,

⇒ \(E(r)=\frac{Q+Q_1}{4 \pi \epsilon_0 r^2}\)

= \(\frac{1}{4 \pi \epsilon_0 r^2} \quad\left[Q+2 \pi A\left(r^2-a^2\right)\right]\)

⇒ \(\frac{Q}{4 \pi \epsilon_0 r^2}+\frac{2 \pi A}{4 \pi \epsilon_0}-\frac{2 \pi A a^2}{4 \pi \epsilon_0 r^2}\)

For uniform electric field, \(\)

[since the magnitude of the uniform electric field does not depend on the value of r ]

∴ \(A=\frac{Q}{2 \pi a^2}\)

Option 1 is correct.

Question 77. An electric dipole has a fixed dipole moment \(\vec{p}\), which makes angle d with respect to x-axis. When subjected to an electric field \(\vec{E}_1=E \hat{i}\), it experiences a torque \(\vec{T}_1=\tau \hat{k}\). When subjected to another electric field \(\vec{E}_2=\sqrt{3} E_1 \hat{j}\) it experiences a torque \(\vec{T}_2=-\vec{T}_1\). The angle d is

  1. 30°
  2. 45°
  3. 60°
  4. 90°

Answer: 3. 60°

Since the dipole makes an angle d with the x-axis,

∴ \(\vec{p}=p \cos \theta \hat{i}+p \sin \theta \hat{j}\)

Class 12 Physics Unit 1 Electrostatics Chapter 2 Electric Field An electric dipole has a fixed dipole moment

∴ \(\vec{T}_1=\vec{p} \times \vec{E}_1=(p \cos \theta \hat{i}+p \sin \theta \hat{j}) \times(E \hat{i})\)

or, \(\tau \hat{k}=-p E \sin \theta \hat{k}\)…(1)

and \(\vec{T}_2=\vec{p} \times \vec{E}_2\)

or, \(-\vec{T}_1=(p \cos \theta \hat{i}+p \sin \theta \hat{j}) \times \sqrt{3} E_1 \hat{j}\)

or, \(-\tau \hat{k}=(p \cos \theta \hat{i}+p \sin \theta \hat{j}) \times \sqrt{3} E \hat{j}\) [∵ E1 = E]

⇒ \(\sqrt{3} p E \cos \theta \hat{k}\)

or, \(\tau \hat{k}=-\sqrt{3} p E \cos \theta \hat{k}\)….(2)

Comparing (1) and (2),

⇒ \(p E \sin \theta=\sqrt{3} p E \cos \theta\)

or, 0 = 60°

The option 3 is correct

Question 78. The electric field in a certain region is acting radially outward and is given by E = Ar. A charge contained in a sphere of radius V centered at the origin of the field, will be given by

  1. 4π∈0Aa²
  2. A∈0
  3. 4π∈0Aa³
  4. 0Aa³

Answer: 3. 4π∈0Aa³

If q is the amount of charge inside the sphere, by considering the upper surface of the sphere to be the Gaussian surface,

⇒ \(E S=\frac{q}{\epsilon_0}\)

or, \(A a \cdot 4 \pi a^2=\frac{q}{\epsilon_0} \quad\)

or, \(q=4 \pi \epsilon_0 A a^3\)

The option 3 is correct.

Question 79. Two identical charged spheres suspended from a common point by two massless strings of length / are initially at a distance d (d << l ) apart because of their mutual repulsion. The charge begins to leak from both spheres at a constant rate. As a result, the charges approach each other with a velocity of v. Then as a function of distance x between them

  1. v ∝ x-1
  2. v ∝ x1/2
  3. v ∝ x
  4. v ∝ x -1/2

Answer: 4. v ∝ x -1/2

The repulsive force between the charged spheres,

F = Tsinθ

Weight of each sphere = mg = Tcosθ

∴ \(\tan \theta=\frac{F}{m g} \quad\)

or, \(\frac{x}{2 l}=\frac{\frac{1}{4 \pi \epsilon_0} x^2}{m g} \quad\)

or, \(\frac{q^2}{2 l}=\frac{1}{4 \pi \epsilon_0} \cdot \frac{q^2}{m g}\)

Electric Field The repulsive force between the charged spheres

Differentiating with respect to t,

⇒ \(\frac{d q}{d t} \propto \frac{3}{2} x^{\frac{1}{2}} \frac{d x}{d t} \quad \text { or, } v \propto x^{-\frac{1}{2}}\) [∵ \(\frac{dp}{dt}\)]

The option 4 is correct.

Question 80. A wheel having mass m has charges +q and -q on diametrically opposite points. It remains in equilibrium on a rough inclined plane in the presence of a vertical electric field E. Then the value of E is

Electric Field A wheel having mass m has charges

  1. \(\frac{m g \tan \theta}{q}\)
  2. \(\frac{m g}{q}\)
  3. \(\frac{m g}{2 q}\)
  4. \(\frac{m g \tan \theta}{2 q}\)

Answer: 3. \(\frac{m g}{2 q}\)

The magnitude of torque due to the electric field

⇒ \(|\vec{p} \times \vec{E}|=p E \sin \theta=(q \cdot 2 R) E \sin \theta\)

= 2qERsinθ [ R = radius of the wheel]

Component of weight along the surface = mgsinθ

Torque produced due to it = (mgsinθ)K

In equilibrium, 2qEsinθ = mgsinθ.R

or, \(E=\frac{m g}{2 q}[latex]

The option 3 is correct.

Question 81. A toy car with large q mows on a frictionless horizontal plane surface under the influence of a uniform electric field [latex]\vec{E}\). Due to the force q\(\vec{E}\), its velocity increases from 0 to 6 m/s in one-second duration. At that instant, the direction of the field is reversed. The car continues to move for two more seconds under the influence of this field. The average velocity and the average speed of the toy car between 0 to 3 seconds are respectively

  1. 1 m/s, 3.5 m/s
  2. 1 m/s, 3 m/s
  3. 2 m/s, 4 m/s
  4. 1.5 m/s, 3 m/s

Answer: 2. 1 m/s, 3 m/s

Electric Field A toy car witli cliarge q mows on a frictionless horizontal

Total displacement = area under the v-t curve

= \(\frac{1}{2}\) x 2 x 6 – \(\frac{1}{2}\) x 1 X 6

∴ Average velocity \(\frac{\text { total displacement }}{\text { total time }}=\frac{3}{3}=1 \mathrm{~m} / \mathrm{s}\)

Total path travelled = \(\frac{1}{2} \times 2 \times 6+\frac{1}{2} \times 1 \times 6=9 \mathrm{~m}\)

∴ Average speed = \(\frac{\text { total path travelled }}{\text { total time }}=\frac{9}{3}=3 \mathrm{~m} / \mathrm{s}\)

The option 2 is correct

Question 82. An electron falls from rest through a vertical distance h in a uniform and vertically upward-directed electric field E. The direction of the electric field is now reversed, keeping its magnitude the same. A proton is allowed to fall from rest in it through the same vertical distance h. The time off of the electron, in comparison to the time of fall of the proton is

  1. 10 times greater
  2. 5 times greater
  3. smaller
  4. equal

Answer: 3.

Acceleration of the electron,

⇒ \(a_e=g+\frac{e E}{m_e}\) [e = charge of electron, me = mass of electron]

Acceleration of the proton,

⇒ \(a_p=g+\frac{e E}{m_p}\)

where mp = mass of a proton

Since, mp > me

∴ ap < ae

So the time of fall will be less for the electron, i.e., te < tp

The option 3 is correct.

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