Questions - Answers

Very Short Answer type Questions

1. What is meant by the statement "charge is quantized?"
A: The minimum charge that can be transferred from one body to another is equal to charge of an electron. So, the charge always exists as an integral multiple of charge of electron i.e. Q = ± ne. Then charge is said to be quantized.
  Where        n = 1, 2, 3, .....;  e = 1.6 × 10^-19 C .
 

2. Repulsion is sure test of charging than attraction. Why?
A: A charged body can attract opposite charged body and a uncharged body (or neutral body). But repulsion exists only between two like charges. Hence repulsion is sure test than attraction.
 

3. How many electrons constitute 1 C of charge?
A:  Charge   q = ± ne
                      1 C = ± n(1.6 × 10⁻¹⁹)

                        

4. What happens to the weight of a body when it is charged positively?
A:  When a body is positively charged it loses electrons, hence its weight decreases.
 

5. What happens to the force between two charges, if the distance between them is (a) halved (b) doubled.

(ii)         If d₂ = 2d₁

        
i.e. distance is doubled force is decreased by four times.
 

6. The electric line of force do not intersect. Why?
A: The electric line of force do not intersect. If they intersect, the intersecting point should have field in two directions, which is not possible.
 

7. Consider two charges +q and -q placed at B and C of an equilateral triangle ABC. For this system total charge is zero. But electric field at A, which is equidistant from B and C is not zero. Why?
A: Charge is a scalar, total charge Q = -q + q = 0. But electric field intensity is a vector. So, they must be add up vectorially at given point. Thus intensity is non zero.

8. Electrostatic field lines do not form closed loops. If they form closed loops then work done in moving charge along a closed path will not be zero. In above cases can you guess the nature of electrostatic force?
A: The electrostatic force is a conservative force.
 

9. State Gauss Law.
A:  Total electric flux through any closed surface is equal to  

 times the charge enclosed by the surface.

           
 

10. When is electric flux negative and when it is positive?

11. Write the expression for electric intensity due to infinite long charge wire at radial distance 'r' from the wire.

 

12. Write the expression for electric intensity due to an infinite plane sheet of charge?

13. Write the expression for electric intensity due to a charged conducting spherical shell at points outside and inside the shell.

Short Answer type Questions (SAQ)

1. State and explain Coulomb's inverse square law in electricity.

A:  The force of attraction of repulsion between two stationary charges is directly proportional to the product of the magnitude of the charges and inversely proportional to the square of the distance between them. Force acts along the line joining the two charges.

The electrostatic force between two charges depends on the nature of the medium between them.

     

2. Define intensity of electric field at a point. Derive an expression for the intensity due to a point charge.
A: The force experienced by a unit positive charge placed at a point in the electric field is called intensity of electric field.

                             
Consider a point charge 'q' is placed at a point 'O',  be the electric field strength at point P, is at a distance of 'r' from 'O'. A test charge q₀ is placed at point 'P', then force acting on charge q₀ due to q is

      

Where   is unit vector along . Due to positive charge field is away from it. Due to negative charge field is towards it, field is non-uniform.
 

3. Derive an equation for the couple acting in a uniform electric field.
A: Consider a dipole of moment 'p' in a uniform electric field E, situated at an angle 'θ' with the field. A positive charge 'q' experiences a force qE in the direction of field and negative charge experiences equal force qE, but in opposite direction.
Thus net force on dipole is zero. These forces constitute a couple. They will produce torque on the dipole.

Torque τ = Force × Perpendicular distance
                 = (qE) (2a sin θ)
                 = (q × 2a) E sin θ
                 = pE sin θ  (   p = q × 2a)

             
Direction of torque is normal to the plane of the paper.             
        τ_max = pE   (   θ = 90°)

4. Derive an expression for the intensity of the electric field at a point on the axial line of an electric dipole.
A: Consider an electric dipole with charges q, - q with separation 2a between them. Let P be a point on the axial line at a distance 'r' from the centre of dipole.
The intensity of electric field at point 'P' is E_axial = E_+q + E_−q .

         
Electric field intensity due to +q, -q at point P is

     

              
5. Derive an expression for the intensity of electric field at a point on the equitorial plane of the electric dipole.
A: Consider an electric dipole with charges -q, +q with a separation '2a' between them. Consider a point 'P' at a distance 'r' from centre. Electric field due to +q at P is

         
         Electric field due to -q at P is

6. State Gauss's Law in electric field and explain its importance.
A: Gauss's Law: Total electric flux through any closed surface is equal to  times net charge enclosed by the surface.

 
Importance:

* Using Gauss's law we can find field due to a distribution of charge.
* It is used to calculate electrostatic field when the system has symmetry.
 

Long Answer type Questions

1. Define electric flux. Applying Gauss's Law and derive the expression for electric intensity due to an infinite long straight charged wire. (Assume that electric field is every where radial and depends only on the radial distance r of the point from the wire).
A: Electric flux: The number of electric line of force passing through the given area and perpendicular to the surface is called electric flux (φ).
Let us consider an infinitely long thin straight wire, having uniform linear charge density (λ).
Consider a cylindrical Gaussian surface ABCD of length 'l' and radial distance 'r' with its axis coinciding with the charged wire.

 

Its direction will be radially outwards if λ is positive.
Its direction will be radially inwards if λ is negative.

 

         
                     

Let us consider Gaussian surface PQRS in the form of a horizontal cylinder.         

3. Applying Gauss's Law derive the expression for electric intensity due to a charged conducting spherical shell at (i) a point outside the shell (ii) a point on the surface of the shell (iii) a point inside the shell.
A: Consider a charged spherical shell of radius R and of uniform charge density σ.

 (i) Field outside the shell:
Consider a spherical Gaussian surface around the conductor of a distance of 'r' from the centre. Let 'P' be a point on the Gaussian surface. Flux through the

  
(ii) Field at a point on the shell:
      If the point 'P' lies on the spherical shell r = R

 

(iii) Field at a point inside the shell:
If point 'P' lies inside the shell with centre as '0'. Flux through this surface is   

The field inside a uniformly charged shell at all points inside is zero.