Please refer to below Exam Question for Class 12 Physics Chapter 1 Electric Charges and Fields. These questions and answers have been prepared by expert Class 12 Physics teachers based on the latest NCERT Book for Class 12 Physics and examination guidelines issued by CBSE, NCERT, and KVS. We have provided Class 12 Physics exam questions for all chapters in your textbooks. You will be able to easily learn problems and solutions which are expected to come in the upcoming class tests and exams for standard 12th.

Chapter 1 Electric Charges and Fields Class 12 Physics Exam Question

Question. Sketch the electric field lines for two point charges q₁ and q₂ for q₁ = q₂ and q₁ > q₂ separated by a distance d.
Answer. When the charges are equal, the neutral point N lies at the centre of the line joining the charges.
However, when the charges are unequal, the point N is closer to the smaller charge.

Question. Draw the pattern of electric field lines, when a point charge –Q is kept near an uncharged conducting plate. 

Answer. As –Q charge is kept near an uncharged conducting plate, positive charge is induced on the plate due to electrostatic induction. The field lines will be perpendicular to the metal surface.

Question. Why do the electrostatic field lines not form closed loops?
Answer. Electric field lines start from positive charge and terminate at negative charge. If there is a single positive charge, the field lines start from the charge and terminate at infinity. So, the electric field lines do not form closed loops.

Question. Does the charge given to a metallic sphere depend on whether it is hollow or solid? Give reason for your answer.
Answer. No, Reason: This is because the charge resides only on the surface of the conductor.

Question. Two identical conducting balls A and B have charges –Q and +3Q respectively. They are brought in contact with each other and then separated by a distance d apart. Find the nature of the Coulomb force between them.
Answer.

The nature of the coulomb force between them is repulsive.

Question. Two insulated charged copper spheres A and B of identical size have charges q_A and q_B respectively. A third sphere C of the same size but uncharged is brought in contact with the first and then in contact with the second and finally removed from both. What are the new charges on A and B?
Answer.

Question. Fig. shows three point charges +2q, – q and +3q. The charges +2q and –q are enclosed within a surface ‘S’. What is the electric flux due to this configuration through the surface ‘S’?

Answer.

Question. What is the electric flux through a cube of side 1 cm which encloses an electric dipole?
Answer. Net electric flux is zero.
Reason : (i) Independent to the shape and size.
(ii) Net charge of the electric dipole is zero.

Question. Two metallic spheres A and B kept on insulating stands are in contact with each other. A positively charged rod P is brought near the sphere A as shown in the figure. The two spheres are separated from each other, and the rod P is removed. What will be the nature of charges on spheres A and B?

Answer. • Sphere A will be negatively charged.
• Sphere B will be positively charged.
Explanation: If positively charged rod P is brought near metallic sphere A due to induction negative charge starts building up at the left surface of A and positive charge on the right surface of B.

If the two spheres are separated from each other, the two spheres are found to be oppositely charged. If rod P is removed, the charges on spheres rearrange themselves and get uniformly distributed over them.

Question. Two charges of magnitudes – 2Q and + Q are located at points (a, 0) and (4a, 0) respectively.
What is the electric flux due to these charges through a sphere of radius ‘3a’ with its centre at the origin?
Answer.

Concept: Imagine a sphere of radius 3a about the origin and observe that only charge –2Q is inside the sphere.

Question. A metal sphere is kept on an insulating stand. A negatively charged rod is brought near it, then the sphere is earthed as shown. On removing the earthing, and taking the negatively charged rod away, what will be the nature of charge on the sphere? Give reason for your answer.

Answer. The sphere will be positively charged due to electrostatic induction.
Explanation: When a negatively charged rod is brought near a metal sphere, the electrons will flow to the ground while the positive charges at the near end will remain held there due to the attractive force of the negative charge on the rod. On disconnecting the sphere from the ground, the positive charge continues to be held at the near end. On removing the electrified rod, the positive charge will spread uniformly over the sphere.

Question. How does the electric flux due to a point charge enclosed by a spherical Gaussian surface get affected when its radius is increased?
Answer. Electric flux through a Gaussian surface, enclosing the charge

This is independent of radius of Gaussian surface, so if radius is increased, the electric flux through the surface will remain unchanged.

Question. A charge Q μC is placed at the centre of a cube. What would be the flux through one face?
Answer.

Question. A charge q is placed at the centre of a cube of side l. What is the electric flux passing through two opposite faces of the cube?
Answer. By symmetry, the flux through each of the six faces of the cube will be same when charge q is placed at its centre.

Thus, electric flux passing through two opposite faces of the cube

Question. What orientation of an electric dipole in a uniform electric field corresponds to its (i) stable and (ii) unstable equilibrium?
Answer. (i) In stable equilibrium the dipole moment is parallel to the direction of electric field (i.e., θ = 0).
(ii) In unstable equilibrium PE is maximum, so θ = π , i.e., dipole moment is antiparallel to electric field.

Question. What is the nature of electrostatic force between two point electric charges q₁ and q₂ if 
(a) q₁ + q₂>0? (b) q₁ + q₂<0?
Answer. (a) If both q₁ and q₂ are positive, the electrostatic force between these will be repulsive.
However, if one of these charges is positive and is greater than the other negative charge, the electrostatic force between them will be attractive.
Thus, the nature of force between them can be repulsive or attractive.
(b) If both q₁ and q₂ are –ve, the force between these will be repulsive.
However, if one of them is –ve and it is greater in magnitude than the second+ve charge, the force between them will be attractive.
Thus, the nature of force between them can be repulsive or attractive.

Question. Figure shows a point charge +Q, located at a distance R/2 from the centre of a spherical metal shell. Draw the electric field lines for the given system.

Answer.

Question. Sketch the electric field lines for a uniformly charged hollow cylinder shown in figure.

Answer.

Question. The dimensions of an atom are of the order of an Angstrom. Thus there must be large electric fields between the protons and electrons. Why, then is the electrostatic field inside a conductor zero?
Answer. The electric fields bind the atoms to neutral entity. Fields are caused by excess charges. There can be no excess charge on the inner surface of an isolated conductor. So, the electrostatic field inside a conductor is zero.

Question. An arbitrary surface encloses a dipole. What is the electric flux through this surface?
Answer. Net charge on a dipole = – q + q = 0. According to Gauss’s theorem, electric flux through the surface,

Short Answer Questions–I

Question. (a) An electrostatic field line is a continuous curve. That is, a field line cannot have sudden breaks. Why is it so?
(b) Explain why two field lines never cross each other at any point. 
Answer. (a) An electrostatic field line is the path of movement of a positive test charge (q₀ → 0) A moving charge experiences a continuous force in an electrostatic field, so an electrostatic field line is always a continuous curve.
(b) Two electric lines of force can never cross each other because if they cross, there will be two directions of electric field at the point of intersection (say A); which is impossible.

Question. Define electric dipole moment. Is it a scalar or a vector quantity? What are its SI unit?
Answer. The electric dipole moment is defined as the product of either charge and the distance between the two charges. Its direction is from negative to positive charge.
i.e., |p|=q(2l)
Electric dipole moment is a vector quantity.
Its SI unit is coulomb-metre.

Question. Depict the orientation of the dipole in (a) stable, (b) unstable equilibrium in a uniform electric field.
Answer. (a) Stable equilibrium,

Question. Two equal balls having equal positive charge ‘q’ coulombs are suspended by two insulating strings of equal length. What would be the effect on the force when a plastic sheet is inserted between the two?
Answer. Force will decrease.
Reason: Force between two charges each ‘q’ in vacuum is

Question. Plot a graph showing the variation of coulomb force (F) versus (1/r²) d n, where r is the distance between the two charges of each pair of charges: (1 μC, 2 μC) and (2 μC, – 3 μC). Interpret the graphs obtained.
Answer.

Since, magnitude of the slope is more for attraction, therefore, attractive force is greater than repulsive force.

Question. An electric dipole is held in a uniform electric field.
(i) Show that the net force acting on it is zero.
(ii) The dipole is aligned parallel to the field. Find the work done in rotating it through the angle of 180°.

Question. (a) Define electric flux. Write its SI unit.
(b) A spherical rubber balloon carries a charge that is uniformly distributed over its surface. As the balloon is blown up and increases in size, how does the total electric flux coming out of the surface change? Give reason.
Answer. (a) Total number of electric field lines crossing a surface normally is called electric flux.
Its SI unit is Nm²C^–1 or Vm.
(b) Total electric flux through the surface q/∈0
As charge remains unchanged when size of balloon increases, electric flux through the surface remains unchanged

Question. Two concentric metallic spherical shells of radii R and 2R are given charges Q₁ and Q₂ respectively. The surface charge densities on the outer surfaces of the shells are equal.
Determine the ratio Q₁ : Q₂.
Answer. Surface charge density σ is same.

Question. The sum of two point charges is 7 μC. They repel each other with a force of 1 N when kept 30cm apart in free space. Calculate the value of each charge. 
Answer. q₁ + q₂ = 7 × 10 ^–6 C

(q₁ – q₂)² = (q₁ + q2)² – 4q₁q₂
= (7×10^–6)² – 4×10^–11
= 49 × 10^–12 – 40 × 10^–12 = 9 × 10^–12
q₁ – q₂ = 3×10^–6 C …(iii)
Solving (i) and (iii), we get
q₁ =5×10^–6 C, q₂=2×10^–6 C
⇒ q₁= 5 μC, q₂ = 2 μC

Question. Two identical point charges, q each, are kept 2 m apart in air. A third point charge Q of unknown
magnitude and sign is placed on the line joining the charges such that the system remains in equilibrium. Find the position and nature of Q.
Ans. System is in equilibrium therefore net force on each charge of system will be zero.
For the total force on ‘Q’ to be zero

For the equilibrium of charge “q” the nature of charge Q must be opposite to the nature of charge q.

Question. Figure shows two large metal plates P₁ and P₂, tightly held against each other and placed between two equal and unlike point charges perpendicular to the line joining them.
(i) What will happen to the plates when they are released?
(ii) Draw the pattern of the electric field lines for the system. 
Answer. (i) Charges induced on outer surfaces of P₁ and P₂ are – Q and + Q respectively.
When plates are released, they will tend to move away from one another; plate P₁ moving towards +Q and P₂ towards –Q due to attraction.
(ii) The field pattern is shown in fig.

Question. Calculate the amount of work done in rotating a dipole, of dipole moment 3 × 10^–8 Cm, from its position of stable equilibrium to the position of unstable equilibrium, in a uniform electric field of intensity 10⁴ N/C.
Answer. P = 3×10^–8 Cm; E = 104 N/C
At stable equilibrium (θ₁) = 0°
At unstable equilibrium (θ₂)=180°
Work done in a rotating dipole is given by:
W = PE (cos θ₁ – cos θ₂) = (3 × 10–8) (104) [cos 0° – cos 180°] = 3 × 10^–4 [1 – (–1)]
W = 6 × 10^–4 J