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    Questions Based on NCERT/Part - 1 [Physics] 

    Electric Charge & Fields

    1. (a) Explain the meaning of the statement 'Electric charge of a body is quantized'.

    (b) Why can one ignore quantization of electric charge when dealing with macrocopic, i.e. large scale

    charges?

    SOLUTION :

    (a) The electric charge of a body is quantized means that the charge on a body can occur in some integral values

    only. Charge on any body is the integral multiple of charge on an electron because the charge of an electron

    is the elementary charge in nature. The on any body can be expressed by the formula  q ne  where,

    n = number of electrons transferred and  e = charge on one electron

    The cause of quantization is that only integral number of electrons can be transferred from one body to other.

    (b) We can ignore the quantization of electric charge when dealing with macroscopic charges because the

    charge on one electron is 191 6 10. C     in magnitude, which is very small as compared to the large scale

    change.

    2. When a glass rod is rubbed with a silk cloth, charges appear on both. A similar phenomenon is observed with

    many other pair of bodies. Explain how this observation is consistent with the law of conservation of charge.

    SOLUTION :

    According to the law of conservation of charge, "charge can neither be created nor be destroyed but can be transferred

    from one body to another body".

    Before rubbing the two bodies they both are neutral i.e, the total charge of the system is zero. When the glass rod is

    rubbed with a silk cloth, the charge appears on both glass rod and the silk cloth. Some electrons from glass rod attain

     positive charge (due to loss of electrons) and silk cloth attains same negative charge (due to gain of electrons).

    Again the total charge of the system is zero i.e., the charge before rubbing is same as the charge after rubbing. This is

    consistent with the law of conservation of charge. Here, we can also say that changes can be created only in equal and

    unlike pairs.

    3. (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?

    SOLUTION :

    (a)  An electrostatic field line represents that actual path travelled by a unit positive charge in an electric field. If

    the line have sudden breaks it means the unit positive test charge jumps from one place to another which is

    not possible. It also means that electric field becomes zero suddenly at the breaks which is not possible. So,

    the field line cannot have any sudden breaks.

    (b)  If two field lines cross each other, then we can draw two tangents at the point of

    intersection which indicates that (as tangent drawn at any point on electric line of

    force gives the direction of electric field point) there are two directions of electric

    field at a particular point, which is not possible at the same instant. Thus, two field

    lines never cross each other at any point.

    4. The given figure shows tracks of three charged

    particles in a uniform electrostatic field. Given the

    signs of the three charges. Which particle has the

    highest charge to mass ratio ?  

    SOLUTION :

    We know that a positively charged particle is attracted towards the negatively charged plate and a negatively charged

     particle is attracted towards the positively charged plate.

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      Here, particle 1 and particle 2 are attracted towards positive plate that means particle 1 and particle 2 are negatively

    charged. Particle 3 is attracted towards negatively charged plate so it is positively charged. As the deflection in the

     path of a charged particle is directly proportional to the charge/mass ratio.

    q  y

    m   

    Here, the deflection in particle 3 is maximum, so the charge to mass ratio of particle 3 is maximum.

    5. (a) A conductor A with a cavity as shown in Fig. (a) is given a charge Q. Show that the entire charge

    must appear on the outer surface of the conductor.

    (b) Another conductor B with charge q  is inserted into

    the cavity keeping  B  insulated from  A. Shown that

    the total charge on the outside surface of  A is Q + q 

    [Fig. (b)]

    (c) A sensitive instrument is to be shielded from the

    strong electrostatic field in its environment.

    Suggest a possible way. 

    SOLUTION :

    (a) As we know the property of conductor that the net electric field inside a charged conductor is zero,

    i.e., E  = 0

     Now let us choose a Gaussian surface lying completely inside the conductor enclosing the cavity. So, from

    Gauss's theorem

    0

    q  E. dS 

        

    As  E  = 0

    0

    0 q

        

    q = 0

    That means the charge inside the cavity is zero. Thus, the entire charge Q on the conductor must appear on

    the outer surface of the conductor.

    (b) As the conductor  B  carrying a charge +q  inserted in the cavity, the charge q   is induced on the metal

    surface of the cavity and then charge + q induced on the outside surface of the conductor  A. Initially the

    outer surface of A has a charge Q  and now it has a charge + q  induced, so the total charge on the outer

    surface of A is Q + q.

    (c)  To protect any sensitive instrument from electrostatic field, the sensitive instrument must be put in the

    metallic cover. This is known as electrostatic shielding.

    6. A hollow charged conductor has a tiny hole cut into its surface. Show that the electric field in the hole is

    02

         

     n  

    , where  n  is the unit vector in the outward normal direction and   is the surface charge density near

    the hole.

    SOLUTION : 

    Surface charge density near the hole =   . Unit vector =   n  (normal directed outwards).

    Let P be the point on the hole. The electric field at Point P closed to the surface of conductor, according to Gauss's theorem,

    0

    q  E . dS 

        

    where, area q

    q ds dS   ds

              

      

    where, q is the charge near the hole.

    0

    dS   EdS cos

         

        

      Angle between electric field and area vector is 0  

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         0

    dS   Eds

       

         ( q / dS q dS       where, dS  = area)

    0

     E     

        

    0

     E n  

        

    This electric field is due to the filled up hole and the field due to the rest of the charged conductor. The two fields

    inside the conductor are equal and opposite. So, there is no electric field inside the conductor. Outside the conductor,

    the electric field at P due to each part = 

    0

    1

    2 2

     E n  

     

      

    7. (a) Consider an arbitrary electrostatic field configuration. A small test charge is placed at a null point

    (i.e., where, E = 0) of the configuration. Show that the equilibrium of the test charge is necessarily

    unstable.

    (b) Verify this result for the simple configuration of two charges of the same magnitude and sign placed a

    certain distance apart.

    SOLUTION : 

    (a)  Let us consider that initially the test charge is in the stable equilibrium. When the test charge is displaced

    from the null point (where,  E  = 0) in any direction, it must experience a restoring force towards the null

     point.

    This means that there is a net inward flux through a closed surface around the null point. According to the Gauss's theorem, the net electric flux through a surface net enclosing any charge must be zero. Hence, the

    equilibrium is not stable.

    (b)  The middle point of the line joining two like charges is a null point. If

    we displace at test charge slightly along the line, the restoring force

    try to bring the test charge back to the centre. If we displace the test

    charge normal to the line, the net force on the test charge takes it

    further away from the null point. Hence, the equilibrium is not stable.

    8. A polythene piece rubbed with wool is found to have a negative charge of 73 10 C  .

    (a) Estimate the number of electrons transferred (from which to which)?

    (b) Is there a transfer of mass from wool to polythene?

    SOLUTION : 

    Given charge on polythene = 73 10 C     

    (a)  The charge on an object is given by q ne   

    The number of electrons transferred n  Total charge q( )

    Charge of electron e( )  

    7 12

    19

    3 10 1 875 10

    1 6 10 n .

    .

      

        

     

    Thus, the number of electrons transferred is 121 875 10.    . Electrons will be transferred from wool to

     polythene because polythene attains the negative charge that means it gains the electrons.

    (b)  As the electrons are transferred from wool to polythene, the mass is