CBSE Class 12 Physics VBQs Electric Charges and Fields

Assertions & Reasons

Question. Given below are two statements labelled as Assertion(A) and Reason(R).
Assertion(A) Insulators do not allow flow current through themselves.
Reason(R) They have no free charge carriers.
Select the most appropriate answer from the options given below.
(i) Both A and R are true, and R is the correct explanation of A.
(ii) Both A and R are true, and R is not the correct explanation of A.
(iii) A is true but R is false
(iv) A is false and R is also false.
Answer. A

Question. Given below are two statements labelled as Assertion(A) and Reason(R).
Assertion(A) During charging by rubbing, the insulating material with lower work function becomes positively charged.
Reason(R) Electrons are negatively charged.
Select the most appropriate answer from the options given below.
(i) Both A and R are true, and R is the correct explanation of A.
(ii) Both A and R are true, and R is not the correct explanation of A.
(iii) A is true but R is false
(iv) A is false and R is also false.
Answer. B

Question. Given below are two statements labelled as Assertion(A) and Reason(R).
Assertion(A) A metallic shield in the form of a hollow shell, can be built to block an electric field.
Reason(R) In a hollow spherical shell, electric field inside is not zero at every point.
Select the most appropriate answer from the options given below.
(i) Both A and R are true and, R is the correct explanation of A.
(ii) Both A and R are true and, R is not the correct explanation of A.
(iii) A is true but R is false
(iv) A is false and R is also false.
Answer. C

Question. Given below are two statements labelled as Assertion(A) and Reason(R).
Assertion (A). A charge is quantized because only integral number of electrons can be transferred.
Reason (R) There is no possibility of transfer of some fraction of electron.
Select the most appropriate answer from the options given below.
(i) Both A and R are true and, R is the correct explanation of A.
(ii) Both A and R are true and, R is not the correct explanation of A.
(iii) A is true but R is false
(iv) A is false and R is also false
Answer. A

Question. Given below are two statements labelled as Assertion(A) and Reason(R).
Assertion (A): The Coulomb force is dominating force in the universe.
Reason(R): The Coulomb force is weaker than the gravitational force.
Select the most appropriate answer from the options given below.
(i) Both A and R are true and, R is the correct explanation of A.
(ii) Both A and R are true and, R is not the correct explanation of A.
(iii) A is true but R is false
(iv) A is false and R is also false
Answer. D

Question. Given below are two statements labelled as Assertion(A) and Reason(R)
Assertion (A): Gauss theorem can used to find the electric field at any point.
Reason (R): Gauss theorem can be applied to any type of charge distribution.
(i) Both A and R are true and R is the correct explanation of A
(ii) Both A and R are true but R is not the correct explanation of A.
(iii) A is true but R is false.
(iv) A is false and R is also false.
Answer. C

Question. Given below are two statements labelled as Assertion(A) and Reason(R)
Assertion (A): If a dipole is enclosed by a surface, then according to the gauss theorem, electric flux linked with it is zero
Reason (R): The charge enclosed by the surface is zero.
(i) Both A and R are true and R is the correct explanation of A
(ii) Both A and R are true but R is not the correct explanation of A.
(iii) A is true but R is false.
(iv) A is false and R is also false.
Answer. A

Question. Given below are two statements labelled as Assertion(A) and Reason(R)
Assertion (A): positive electric flux indicates that electric lines of force are directed outwards
Reason (R): Positive electric flux is due to a positive charge
(i) Both A and R are true and R is the correct explanation of A
(ii) Both A and R are true but R is not the correct explanation of A.
(iii) A is true but R is false.
(iv) A is false and R is also false.
Answer. B

Question. Given below are two statements labelled as Assertion(A) and Reason(R)
Assertion (A): Coulombs law is useful for calculation of electric field intensity due to point charges
Reason (R): Gauss law is used for calculating electric field intensity due to symmetric charge distributions
(i) Both A and R are true and R is the correct explanation of A
(ii) Both A and R are true but R is not the correct explanation of A.
(iii) A is true but R is false.
(iv) A is false and R is also false.
Answer. B

Question. Given below are two statements labelled as Assertion(A) and Reason(R)
Assertion (A): A graph showing the variation of electric field at a point with distance due to an infinite plane sheet of charge is a straight line parallel to the distance axis
Reason (R): The electric field at a point due to an infinite plane sheet of charge is independent of the distance to the point.
(i) Both A and R are true and R is the correct explanation of A
(ii) Both A and R are true but R is not the correct explanation of A.
(iii) A is true but R is false.
(iv) A is false and R is also false
Answer. A

CASE STUDY QUESTIONS

1. Lightning is an electric current. Within a thundercloud way up in the sky, many small bits of ice (frozen raindrops) bump into each other as they move around in the air. All of those collisions create an electric charge. After a while, the whole cloud fills up with electrical charges. The positive charges or protons form at the top of the cloud and the negative charges or electrons form at the bottom of the cloud. Since opposites attract, that causes a positive charge to build up on the ground beneath the cloud. The grounds electrical charge concentrates around anything that sticks up, such as mountains, people, or single trees. The charge coming up from these points eventually connects with a charge reaching down from the clouds and lightning strikes.

Question. Charge is the property associated with matter due to which it produces and experiences
(a) electric effects only
(b) magnetic effects only
(c) both electric and magnetic effects
(d) None of these
Answer. C

Question. When some charge is transferred to ...A... it readily gets distributed over the entire surface of ... A... If some charge is put on ... B..., it stays at the same place. Here, A and B refer to
(a) insulator, conductor
(b) conductor, insulator
(c) insulator, insulator
(d) conductor, conductor
Answer. B

Question. On charging by conduction, mass of a body may
(a) increase
(b) decreases
(c) increase or decrease
(d) None of these
Answer. C

Question. If one penetrates a uniformly charged spherical cloud, electric field strength
(a) decreases directly as the distance from the centre
(b) increases directly as the distance from the centre
(c) remains constant
(d) None of these
Answer. A

Question. The law, governing the force between electric charges in the cloud is known as
(a) Ampere's law
(b) Ohm's law
(c) Faraday's law
(d) Coulomb's law
Answer. D

2.Neurons maintain different concentrations of certain ions across their cell membranes. Imagine the case of a boat with a small leak below the water line. In order to keep the boat afloat, the small amount of water entering through the leak has to be pumped out, which maintains a lower water level relative to the open sea. Neurons do the same thing, but they pump out positively charged sodium ions. In addition, they pump in positively charged potassium ions. Thus there is a high concentration of sodium ions present outside the neuron, and a high concentration of potassium ions inside. Thus sodium channels allow sodium ions through the membrane while potassium channels allow potassium ions through.

Question. When neuron pump out and in the positive sodium and positive potassium ions respectively which property of charge is to be followed
(a) Quantisation of charge
(b) Additivity of charges
(c) Conservation of charges
(d) Associativity of charge
Answer. C

Question. Coulomb’s law is true for
(a) atomic distances (= 10^–11 m)
(b) nuclear distances (= 10^–15 m)
(c) charged as well as uncharged particles
(d) all the distances
Answer. D

Question. Electric lines of force about a positive sodium or potassium ions are
(a) circular anticlockwise
(b) circular clockwise
(c) radial, inwards
(d) radial, outwards
Answer. D

Question. Electric flux produced by positive Potassium ions indicates that electric lines are directed
(a) outwards
(b) inwards
(c) either (a) or (b)
(d) None of these
Answer. A

Question. Electric flux over a surface of neuron in an electric field may be
(a) positive
(b) negative
(c) zero
(d) All of the above
Answer. D

3. Animals emit low frequency electric fields due to a process known as osmoregulation. This process allows the concentration of ions (charged atoms or molecules) to flow between the inside of our bodies and the outside. In order for our cells to stay intact, the flow of ions needs to be balanced. But balanced doesn't necessarily mean equal. The concentration of ions within a shrimp’s body is much lower than that of the sea water it swims in. Their voltage, or potential difference generated between the two concentrations across "leaky" surfaces, can then be measured.

Question. The Gaussian surface for ions in the body of animals
(a) can pass through a continuous charge distribution.
(b) cannot pass through a continuous charge distribution.
(c) can pass through any system of discrete charges.
(d) can pass through a continuous charge distribution as well as any system of discrete charges.
Answer. D

Question. Gauss's law is valid for
(a) any closed surface
(b) only regular close surfaces
(c) any open surface
(d) only irregular open surfaces
Answer. A

Question. The electric field inside a shrimp’s body of uniform charge density is
(a) zero
(b) constant different from zero
(c) proportional to the distance from the curve
(d) None of the above
Answer. A

Question. If a small piece of linear isotropic dielectric is swallowed by a shrimp and inside the body it is influenced by an electric field of strength E, then the polarization P is
(a) independent of E
(b) inversely proportional to E
(c) directly proportional to √𝐸
(d) directly proportional to E
Answer. D

Question. Field due to multiple charges/ions inside Shrimp’s body at a point is found by using
I. superposition principle.
II. Coulomb’s law.
III. law of conservation of charges.
(a) I and II
(b) II and III
(c) I and III
(d) I, II and III
Answer. A

1 MARK QUESTIONS

Question. Two identical metallic spheres of exactly equal masses are taken. One is given a positive charge q coulombs and other an equal negative charge. Are their masses after charging equal?
Answer. No. the positive charge of the body is due to deficit of electron while the negative charge is due to surplus of electrons. Hence the mass of negatively charged sphere will be slightly more than that of the positively charged sphere

Question. Force between two-point charges kept at a distant d apart in air F. If these charges are kept at the same distance in water how does the electric force between them change?
Answer. Dielectric constant for water K = 80
FWATER = FAIR / K = F/80
Thus the force in water is 1/80 times the original force in air .

Question. Why is electric field zero inside a charged conductor?
Answer. This is because charges reside on the surface of a conductor and not inside it.

Question. Why do the electrostatic field lines not form closed loops?
Answer. Electrostatic field lines start from positive charge and end on a negative charge or they fade out at infinity in case of isolated charges without forming any closed loop.

Question. Draw lines of force to represent a uniform electric field?
Answer. The lines of force of a uniform electric field are equidistant parallel lines
→
→
→
→

2 MARK QUESTIONS

Question. A force F is acting between two charges placed some distance apart in vacuum. If a brass rod is placed between these two charges how does the force change?
Answer. For any metal ,K=∞
F _brass =fvac /K = F/∞= 0
i ,e in the presence of brass rod the force between the two charges becomes zero

Question. Two dipoles made charges + q and + Q respectively have equal dipole moments. Give the (i) ratio between the separation of these two pairs of charges (ii) angle between the dipole axes of these two dipoles.
Answer. As the two dipoles have equal dipoles moments so
(i) qa = Qa’ = a’/a = q/Q
(ii) their dipoles axes must have same direction.

Question. A metallic spherical shell has an inner radius R1 and outer radius R2. A charge Q is placed at the center of the spherical cavity. What will be the surface charge density on (i) the inner surface and (ii) the outer surface?
Answer. Change -Q is induced on the inner surface and charge +Q is induced on the surface of the cavity .
Therefore surface charge density on the inner surface = -Q /4πR²
Surface charge density on the outer surface =Q /4πR²

Question. A system has two charges qA = 2.5 × 10^–7 C and qB = –2.5 × 10^–7 C located at points A: (0, 0, –15 cm) and B: (0, 0, +15 cm), respectively. What are the total charge and electric dipole moment of the system?
Answer. Total charge of electric dipole = zero coulomb
Magnitude of Dipole moment,
|p| = (Magnitude of either charge) × (Distance between 2 charges) = q.2a
Given, 2a = 30 cm, q = 2.5 × 10^‒7 C
= (2.5 × 10‒7) × (30)
= 7.5 × 10^‒8 C-m

Question. (a) An electrostatic field line is a continuous curve. That is, a field line cannot have sudden breaks. Why not?
(b) Explain why two field lines never cross each other at any point?
Answer. (a) The direction of electric field is given by tangent at each point on the curve. At sudden breaks, the field will have more than one direction which is not possible. That’s why electrostatic field line is a continuous curve.
(b) At the crossing point there will be two directions of electric field at that point given by the two tangents. This cannot happen, and so two field lines never cross each other at any point

LONG ANSWER QUESTIONS

Question. (a) State Gauss theorem in electrostatics. Using it, prove that the electric field at a point due to a uniformly charged infinite plane sheet is independent of the distance.
(b) How is the field directed if
(i) the sheet is positively charged,
(ii) negatively charged?
Answer. 1. It states, “The net electric flux through any Gaussian surface is equal to 1ε0 times the net electric charge enclosed by the surface.
Mathematically, Φ = ∫𝐸⃗. 𝑑𝑆⃗ =q/ε₀
Consider an infinite plane sheet of charge. Let a be the uniform surface charge density, i.e. the charge per unit surface area. From symmetry, we find that the electric field must be perpendicular to the plane of the sheet and that the direction of E on one side of the plane must be opposite to its direction on the other side as shown in the figure below. In such a case let us choose a Gaussian surface in the form of a cylinder with its axis perpendicular to the sheet of charge, with ends of area A.
The charged sheet passes through the middle of the cylinder’s length so that the cylinder’s ends are equidistant from the sheet. The electric field has a normal component at each end of the cylinder and no normal component along the curved surface of the cylinder. As a result, the electric flux is linked with only the ends and not the curved surface.

Therefore by the definition of electric flux, the flux Linked with the Gaussian surface is given by
Φ = ∮A𝐸⃗Δ𝐴⃗
Φ = EA + EA = 2E_A … (1)
But by Gauss’s Law
Φ = q/ε0=σA/ε0 [∵ q = σA] … (2)
From equations (1) and (2),we have
2EA = σA/ε₀ … (3)
E = σ/2ε₀ …. (4)
This gives the electric field due to an infinite plane sheet of charge which is independent of the distance from the sheet. (b) (i) directed outwards (ii) directed inwards.

Question. Use Gauss’s law to derive the expression for the electric field (E→) due to a straight uniformly charged infinite line of charge λ Cm^-1.
Answer. Consider an infinitely Long, thin wire charged positively and having uniform Linear charge density λ. The symmetry of the charge distribution shows that must be perpendicular to the tine charge and directed outwards. As a result of this symmetry, we consider a Gaussian surface in the form of a cylinder with arbitrary radius r and arbitrary Length L. with its ends perpendicular to the wire as shown in the figure. Applying Gauss’s theorem to curved surface ΔA1 and circular surface ΔA₂.
Φ EΔA₁ cos 0°+ EΔA₂ cos 90° = qε₀ = λlε₀ [∵ λ = q/l]
Or
E . 2πrl = λlε₀ ⇒ E = 1/2πε0λr
This is the expression for the electric field due to an infinitely long thin wire.