Electric Charges and Fields Mock Test – Class 12 Physics

1. Two small charged bodies have charges and . They are treated as one composite system. The system is

2. A charge placed at a point experiences a force of magnitude . The electric field magnitude at the point is

3. A positive charge and another positive charge are fixed on a straight line. The electrostatic force on each charge acts

4. A diagram shows uniform electric field lines as straight, parallel, and equally spaced. This means the electric field has

5. A pair of neutral insulating bodies is rubbed together. One body finally has charge . If no charge escapes to the surroundings, the charge on the other body is

6. A uniformly charged infinite line has linear charge density . The electric field at distance from the line is directed

7. A charge is at the origin. A charge is at , and a charge is at . The electric field magnitude at the origin due to each of the two outer charges is . The magnitude of the force on a charge placed at the origin is

8. Two equal-magnitude electrostatic forces act on a charge at right angles to each other. The magnitude of the resultant force is

9. A uniformly charged thin spherical shell of radius carries total charge . For an outside point at distance from the centre, the electric field is the same as if

10. Study the charge records for four systems.

The row with an incorrect net charge is

11. In a collinear charge system, a zero net force point is more likely to occur where

12. Two electric field lines in an electrostatic field cannot intersect because

13. Two statements are made about conductors and insulators.
I. Charge can redistribute easily inside a conductor because some charge carriers are mobile.
II. An insulator contains no positive or negative charges inside its atoms.
III. A neutral conductor near a charged body may become polarized even when its net charge remains zero.
The supported statements are

14. The unit of surface charge density is

15. An electric field graph for a source is given by . If the graph of against has slope , and , the source and observation line are best identified as

16. A small pillbox Gaussian surface is drawn across the surface of a charged conductor in electrostatic equilibrium. The field just outside is , the surface charge density is , and the field inside the conductor is zero. Gauss’s law gives

17. In the relation for an infinite line charge, the absence of in the final answer occurs because

18. A negatively charged rod is held near a neutral metal sphere on an insulating stand. The sphere is then connected briefly to earth while the rod remains nearby. During earthing, electrons tend to flow

19. A closed surface has outward flux through one part and inward flux through the remaining part. According to Gauss’s law, the enclosed charge is

20. Study the table for two isolated conducting spheres after they touch and separate.

The row that does not follow equal sharing for identical spheres is

21. A mixed proportionality record is shown below. Identify the row that is not consistent with standard electrostatic results.

22. A claim says, “Induction and polarization are exactly the same because both begin with charge separation.” The best evaluation is that the claim is

23. At a far equatorial point of a dipole, where , the electric field magnitude is approximately

24. A graph is described as a straight line through the origin when force is plotted against , with the two charges kept fixed. This description supports the idea that

25. A negative charge is placed just outside a positively charged conductor surface. The conductor’s local electric field is outward normal. The force on the negative charge is

26. Two small bodies repel each other when brought close. A safe conclusion from this observation is that

27. A charge is at the centre of a spherical Gaussian surface of radius . Another charge is placed outside the sphere at distance from the centre. The electric field at the surface changes, but the net flux through the sphere is

28. If were larger while all other charge and distance values were unchanged in vacuum-form Coulomb’s law, the value of would

29. The potential energy of an electric dipole in a uniform electric field is

30. An electric dipole has charges and separated by . It is placed in a uniform electric field at an angle with the field. The torque magnitude on the dipole is

31. Read the situation and answer the question.

What is the net charge of the sphere at this stage?

32. A graph is plotted between electric field magnitude due to a fixed point charge and . The graph is a straight line through the origin. The slope of this graph is

33. The unit of can be inferred from and . The unit of is

34. In electrostatics, electric charge is treated as a scalar quantity, while electrostatic force is treated as

35. Two fixed point charges are kept at the same separation, but the experiment is repeated in a medium of dielectric constant . In a graph of versus , the slope in the medium becomes

36. Two equal positive charges are fixed at the two lower corners of an equilateral triangle. A positive charge is placed at the upper corner. If the force on due to each lower charge has magnitude , the resultant force on has magnitude

37. Read the case below.

The direction of the force on due to is

38. The pair that contains one scalar and one vector from basic electrostatics is

39. In deriving the axial field of a dipole, the subtraction of two field magnitudes occurs because

40. Study the table for a dipole in a uniform electric field and identify the row that needs correction.

41. Assertion: In vector Coulomb-law calculations, changing the definition of the unit vector can change the algebraic sign in the formula but not the physical force.
Reason: The same physical vector must be obtained when the sign convention is used consistently.

42. Assertion: The electrostatic forces between two point charges have equal magnitudes even if the charges have unequal magnitudes.
Reason: The two forces form an action-reaction pair along the line joining the charges.

43. A positively charged conductor is touched to a neutral conductor of the same material and comparable shape. After separation, the neutral conductor usually becomes

44. Assertion: Gauss’s law alone does not always allow direct calculation of electric field at a point.
Reason: The flux integral can be simplified to only when symmetry makes the field magnitude and direction suitable on the chosen surface.

45. The net charge of a system containing charges , , and is

46. A uniformly charged thin spherical shell carries total charge . At a point strictly inside the shell, the electric field is

47. Electric charge is additive. This means that the total charge of a system is found by

48. Study the table for medium effects in Coulomb force and identify the row that needs correction.

49. Near a positively charged conductor surface, the electric field just outside is directed

50. A far equatorial point is chosen at distance from the centre of a dipole. If the dipole moment is doubled while remains the same, the equatorial field magnitude becomes