TOP 100+ NEET Questions On Electric Charges And Fields with Answers

In this post, NEET Questions On Electric Charges And Fields with Answers, we are going to discuss some most important Multiple choice questions on lectric Charges And Fields. The concept of electric field is very much important for students appearing in NEET as physics is one of the subjects and these Electric Charges And Fields Neet Questions and Answers broadly covered in it. So if you are planning to appear for NEET and want to crack it, you should go through these Electric Charges And Fields Neet Questions With Solutions as thoroughly as possible to get a clear idea about the concept of electric field.

NEET Questions On Electric Charges And Fields

1. Two 1 Coulomb charges are kept at 1m distance in air medium. Force of attraction or repulsion between them will be ________

a) 9*109 N
b) 1 dyne
c) 1 N
d) 3*103 N

Answer: a
Explanation: According to Coulomb’s Law, F=\(\frac {1}{4\pi \varepsilon_o }*\frac {q_1q_2}{r^2}\). And \(\frac {1}{4\pi \varepsilon_o}\) = 9*109N in the SI system. In this case, q1=q2=1C and r=1m. Now substituting the values we get F=9*109N. Similarly, the electric field at a distance of 1m from a 1C charge is 9*109N/C.

2. 1 emu = __________ C

a) 10
b) 3*109
c) 4.8*10-10
d) 0.1

Answer: a
Explanation: 1 emu charge is equal to 10-coulomb charge while 1 coulomb is equal to 3*109 esu charge. Emu and C both are the units of charge and hence their conversion formula is necessary for many numerical problems.

3. If the force between two charges is 9N, what will be the force if the distance between them is doubled and both the charges are increased to √2 times?

a) 9N
b) 4.5N
c) 3N
d) 3.75N

Answer: b
Explanation: The force will be 9*\(\frac {\sqrt2*\sqrt2}{2^2}\) N=4.5N because the force is proportional to the product of the two charges and inversely proportional to the square of the distance between them. Similarly, if the charges are the same but the distance is doubled, the force between them will be \(\frac {1}{4}\) times the previous value.

4. Two charges q1, q2 exert some amount of force on each other. What will happen to the force on q1 if another charge q3 is brought close to them?

a) The force will increase
b) The force will decrease
c) The force remains the same
d) The force may increase or decrease depending on whether q3 is positive or negative

Answer: d
Explanation: If both the charges q1 and q2 are positive then they will repel each other. Now if the third charge q3 is also positive, the repulsive force will increase. But if q3 is negative, the repulsive force will be reduced by some amount due to the attractive force between q1 and q3.

5. Two negative charges are kept at a certain distance in the air medium. What will happen if a dielectric slab is inserted between them?

a) The slab will get heated
b) Current will flow through the slab
c) Two charges will attract each other
d) The net force between the charges will be reduced

Answer: d
Explanation: Force between two charges is inversely proportional to the relative permittivity of the medium between them. A dielectric slab has relative permittivity more than that of air. So the force between the charges decreases. No current or heating will be noticed.

6. What will be permittivity of a medium which has dielectric constant 5.4?

a) 4.78*10-11 C2N-1m-2
b) 8.85*10-10C2N-1m-2
c) 4.5*10-10C2N-1m-2
d) 3.2*10-11C2N-1m-2

Answer: a
Explanation: εo=8.854*10-12C2N-1m-2. Therefore the permittivity will be 5.4*8.854*10-12C2N-1m-2 = 4.78*10-11C2N-1m-2. We know that permittivity of a medium is equal to its relative permittivity multiplied to the absolute permittivity of air. Air has relative permittivity =1.

7. The dimension of εo _______

a) [M-1L-3T4I2]
b) [M-1L-3T4I4]
c) [M-1L-3T2I2]
d) [M1L-3T4I2]

Answer: a
Explanation: εo is the absolute permittivity of air medium and has value 8.854*10-12C2N-1m-2. Its dimension is [M-1L-3T4I2]. The permittivity of a medium is the product of this term with the dielectric constant or relative permittivity of that medium.

8. What is the C.G.S. unit of charge?

a) Stat Coulomb
b) Coulomb
c) EMU
d) Pascal

Answer: a
Explanation: The C.G.S. unit of charge is Stat Coulomb or ESU but in the SI system the unit is Coulomb. EMU is another unit of charge in the SI system. Pascal is the unit of pressure.

9. Three equal positive charges are kept at the corners of an equilateral triangle. What will be the vector sum of the forces acting on the particles?

a) Directed towards the center
b) Directed radially outside
c) Acts along one of the sides of the triangle
d) Zero

Answer: d
Explanation: The magnitude of the force acting on each of the charges will be the same and they will differ from each other by 120 degrees. Therefore their resultant force will always be zero.

10. A 1C charge is placed at the origin. Other infinite numbers of unit charges are placed at √2, √4, √8, √16,….(up to infinite) distances from the origin in a straight line. What will be the total force acting on the 1st charge?

a) 9*109 N
b) 18*109 N
c) Infinite
d) 1 N

Answer: a
Explanation: The force on 1st charge will be = 9*10-9*(\(\frac {1}{\sqrt{2}^2} + \frac {1}{\sqrt{4}^2} + \frac {1}{\sqrt{8}^2}\) +⋯) = 9*10-9*\(\frac {\frac {1}{2}}{1-\frac {1}{2}}\) = 9*10-9N [infinite GP formula].

11. Four charges are kept at the corner points of a square. The net force on a charge kept at the center of the square is _________

a) Along diagonal
b) Zero
c) Along one side
d) Depends on the nature of the charges

Answer: d
Explanation: Depending on the nature and quantity of charges, the net force on the central charge may vary its magnitude and direction. If all the charges are the same the net force will be zero.

12. Which among the following is false?

a) Coulomb force is a central force
b) The force between two charges depend on the medium between them
c) Coulomb force is a weak force
d) The net force on a charge is the vector sum of the forces acting on it due to several other charges

Answer: c
Explanation: Coulomb force is a strong force. It can be shown that the Coulomb force between two electrons is 1043 times than gravitational force. All the other options are basic properties of Coulomb’s Law.

13. The amount of force exerted on a unit positive charge in an electric field is known as _____

a) Electric field intensity
b) Electric flux
c) Electric potential
d) Electric lines of force

Answer: a
Explanation: The zone near a charge where its attraction or repulsion force works, is known as the electric field of that charge. Theoretically, it is up to infinite but practically it has limitations. If a unit positive charge is kept in that field, it will undergo some force which is known as electric field intensity at that point.

14. The direction of electric field created by a negative charge is ___________

a) Directed outwards
b) Directed towards the charge
c) Maybe outwards or towards the charge
d) Circular in shape

Answer: b
Explanation: If a unit positive charge is kept near a negative charge, the unit positive charge will be attracted towards the negative charge. That means the electric field is towards the negative charge. But in case of positive charge, the field is directed away from the charge.

15. Electric field inside a hollow conducting sphere ______

a) Increases with distance from the center of the sphere
b) Decreases with distance from the center of the sphere
c) Is zero
d) May increase or decrease with distance from the center

Answer: c
Explanation: According to Gauss’s law, if there is no charge inside a closed surface, the field inside the closed surface will always be zero. We know the charge is distributed on the outer surface of a conducting hollow sphere because the charges want to maintain maximum distance among them due to repulsion. So there is no charge inside the sphere and hence no electric field.

16. Electric field due to a uniformly charged hollow sphere at a distance of r (where r is greater than the radius of the sphere) is __________

a) Proportional to r
b) Inversely proportional to r
c) Proportional to r2
d) Inversely proportional to r2

Answer: d
Explanation: If the total charge of the sphere is q then the electric field at a distance of r is equal to \(\frac {q}{4\pi\varepsilon_o r^2}\).Therefore the electric field is proportional is \(\frac {1}{r^2}\) (if r > radius of the sphere). But if r < radius of the sphere the electric field will be zero i.e. electric field inside a hollow sphere is always zero.

17. Two point charges q1 and q2 are situated at a distance d. There is no such point in between them where the electric field is zero. What can we deduce?

a) There is no such point
b) The charges are of the same polarity
c) The charges are of opposite polarity
d) The charges must be unequal

Answer: c
Explanation: If both the charges are of the same polarity (maybe of unequal magnitude), there must be a point in between them where the electric field intensities of the charges are of equal magnitude and in opposite direction. Hence they balance each other and the net field intensity must be zero. But if the charges are of opposite polarities their field intensities aid each other and net field intensity can never be zero.

18. Two point charges of the same polarities are hung with the help of two threads and kept close. The angle between the threads will be _________ if the system is taken to space.

a) 180 degree
b) 90 degree
c) 45 degree
d) 60 degree

Answer: a
Explanation: There is gravitational field on earth, so if we hang the two same charges there will be an interaction of vertical gravitational field and horizontal electric field. The system will achieve equilibrium by creating a certain angle between the threads and hence the vertical and horizontal components of forces will balance. But in space, there is no gravity. So the charges will be at 180-degree separation.

19. Electric field is a _______

a) Scalar quantity
b) Vector quantity
c) Tensor quantity
d) Quantity that has properties of both scalar and vector

Answer: b
Explanation: A scalar quantity is a quantity with magnitude only but no direction. But a vector quantity possesses both magnitude and direction. An electric field has a very specific direction (away from a positive charge or towards a negative charge). Hence electric field is a vector quantity. Moreover, we have to use a vector addition for adding two electric fields.

20. Two point charges +4q and +q are kept at a distance of 30 cm from each other. At which point between them, the field intensity will be equal to zero?

a) 15cm away from the +4q charge
b) 20cm away from the +4q charge
c) 7.5cm away from the +q charge
d) 5cm away from the +q charge

Answer: b
Explanation: The electric field at a distance of r from a charge q is equal to \(\frac {q}{4\pi\varepsilon_or^2}\). Let the electric field intensity will be zero at a distance of x cm from +4q charge, so the fields due to the two charges will balance each other at that point. Therefore \(\frac {4q}{4\pi\varepsilon_ox^2}=\frac {q}{4\pi\varepsilon_o(30-x)^2}\). Solving this we get x=20cm. Therefore the point will be 20cm away from the +4q charge.

21. What is the dimension of electric field intensity?

a) [M L T-2 I-1]
b) [M L T-3 I-1]
c) [M L T-2 I-2]
d) [M L T-3 I]

Answer: b
Explanation: Electric field intensity is defined as the force on a unit positive charge kept in an electric field. Hence we can simply consider its dimension as\(\frac {the \, dimension \, of \, force}{the \, dimension \, of \, charge}\). The dimension of force is [MLT-2] and the dimension of charge is [IT]. Therefore the dimension of field intensity is [M L T-3 I-1].

22. V/m is the unit of ______

a) Electric field intensity
b) Electric flux
c) Electric potential
d) Charge

Answer: a
Explanation: E=-\(\frac {dV}{dx}\) where E is the field intensity, V is potential and x is distance. Therefore unit of electric field intensity will be \(\frac {unit \, of \, potential}{unit \, of \, distance} = \frac {V}{m}\). Electric flux has unit V*m, V is the unit of electric potential whereas charge has a unit of Coulomb or esu.

23. Electric field intensity at the center of a square is _____ if +20 esu charges are placed at each corner of the square having side-length as 10 cm.

a) 0
b) 0.4 dyne/esu
c) 2 dyne/esu
d) 1.6 dyne/esu

Answer: a
Explanation: Distance of center from each corner point of the square is = \(\frac {10\sqrt2}{2}\) = 5√2. Therefore field intensity at the center due to a single charge is = \(\frac {20}{(5\sqrt2)^2}\) dyne/esu. But the fields due to the four charges are equal and are at perpendicular to each other. So the fields balance each other and the net electric field at the center will be equal to zero.

24. The electric field lines diverge from _______

a) Positive charge
b) Negative charge
c) Dipole
d) Zero potential point

Answer: a
Explanation: Electric field lines always converge at a negative charge and diverge from the positive charge. If a unit positive charge is kept in an electric field, the path of motion of that charge is known as field lines. Now we know that a positive charge is always pushed away by another positive charge. Thus we can find the direction of field lines near a positive charge.

25. Two electric field lines ______

a) Always intersect each other
b) Never intersect
c) May intersect sometimes
d) Are always perpendicular to each other

Answer: b
Explanation: We know that the electric field at a point is directed towards the tangent of electric field lines at that point. Now if two field lines intersect at a point, two tangents can be drawn at that point. That means at a particular point there would be two separate directions of the electric field which is impossible. So, we can conclude that two electric field lines never intersect each other.

26. Consider the lines of force as shown in the figure. Two unit positive charges are kept at points A and B.

a) Charge at A will suffer greater force
b) Charge at B will suffer greater force
c) Force at both points will be same but non-zero
d) Force at both points will be the same

Answer: b
Explanation: More number of field lines cross unit cross-section area at point B. Therefore electric field intensity will be more at point B than that of point A. Thus a unit positive charge will face greater force at point B than point A. So we can conclude that field intensity will be greater where the field lines are dense.

27. Two plates are oppositely charged uniformly and kept parallel to each other at a certain distance. What will be the nature of electric field lines in between them?

a) Circular
b) Parallel to each other throughout the cross section
c) Not uniformly distributed
d) Parallel and uniform in central part but fringes out at the extreme ends

Answer: d
Explanation: The electric field between two uniformly charged plates is uniform. Electric field lines in the uniform electric field are parallel to each other and uniformly distributed. This phenomenon is observed in most of the parts between the plates. But in the two extreme ends, the field line fringes out due to the non-uniform electric field.

28. Which among the following is false about electric field lines?

a) They are continuous
b) They attract each other
c) They remain parallel in a uniform electric field
d) They diverge from positive charge

Answer: b
Explanation: Electric lines of force are always diverging from positive charge and they converge at a negative charge. Besides, they repel each other because of repulsion between two similar charges. They remain parallel to each other in the uniform field but may divert in case of a non-uniform field. These lines are continuously stretched between two opposite charges and always try to maintain minimum length.

29. At ‘Electrical Neutral Point’ the lines of force _______

a) Are absent
b) Coincide with each other
c) Always converge
d) Always diverge

Answer: a
Explanation: Electrical neutral point is a point in between two positive and negative charges, where the electric field intensity is zero. It means that no lines of force pass through that point because electric field lines are the measure of field intensity i.e. more lines mean more field intensity. The lines never converge or diverge neither they coincide at that point.

30. What is the dimension of electric flux?

a) [M L3 T-3 I-1]
b) [M L2 T-3 I-1]
c) [M L3 T-3 I1]
d) [M L3 T3 I-1]

Answer: a
Explanation: Electric flux=electric field intensity* area. The dimension of field intensity is [M L T-3 I-1] and the dimension of the area is [L2]. Therefore, the dimension of flux = [M L3 T-3 I-1]. This can also be justified that flux=potential*length. By putting the dimensions of potential and length, we can get the same result.

NEET Questions On Electric Charges And Fields and Answers

Electric charges and fields questions asked in the National Eligibility cum Entrance Test (NEET) were intended to check whether students could take care of basic electric charges and fields concepts.

A charged body produces a force on another charged body brought near it. This force is usually experienced as the attractive or repulsive electrostatic force between two bodies. It causes such substances to move, but the magnitude of the force cannot be determined without knowing the nature of the charge and its magnitude on both the bodies. The way in which a charge influences another charge, is through an electric field, this is surrounded by an electrically charged body.

Electric charges are described by charge (measured in coulombs), which can be either positive or negative, and location. The quantum of electric charge is the electron, and a single charge will repel another unless they are paired (in atomic structure). More confusingly, you must take into consideration how the pair is aligned: sometimes two positive charges will repel, other times they attract. In addition, unlike gravity, there is no attractive force between two bodies with an overall neutral charge.

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