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Electric Charges and Fields Test - 9

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Electric Charges and Fields Test - 9
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  • Question 1
    4 / -1

    Two parallel infinite line charges + λ and -λ are placed with a separation distance R in free space. Then etelectric field exactly mid - way between the two line charges is 

    Solution

    Electric field at point P due to line charge distribution +λ,
    E =  away from +λ

    Electric field at point P due to line charge distribution -λ,

    E and E have same direction,

  • Question 2
    4 / -1

    An electric dipole consists of charges ±2.0 x 10-8C separated by a distance of 2.0 x 10-3 m. It is placed near a long line charge of linear charge density 4.0 x 10-4C m-1 as shown in the figure, such that the negative charge is at a distance of 2.0 cm from the line charge. The force acting on the dipole will be

    Solution

    The electric field at a distance r from the line charge of linear density λ is given by
    E = λ/2πε0r
    Hence, the field at the negative charge,
    E
    The force on the negative charge, F1 = (3.6 x 108) (2.0 x 10-8) = 7.2 N towards the line charge
    Similarly, the field at the positive charge,
    i.e., at r = 0.022 m is E= 3.3 x108 N -1
    The force on the positive charge,
    F2 = (3.3 x 108) x (2.0 x 10-8) = 6.6 N away from the line charge.
    Hence, the net force on the dipole = 7.2N - 6.6N = 0.6 N towards the line charge.

  • Question 3
    4 / -1

    Two infinite plane parallel sheets, separated by a distance d have equal and opposite uniform charge desities σ. Electric field at a point between the sheets is

    Solution

    Electric field, E = σ/ε0

  • Question 4
    4 / -1

    Two large, thin metal places are parallel and close to each other. On their faces, the plates have surface charge densities of opposite signs and of magnitude 16 x 10-22 C m-2. The electric field between the plates is

    Solution

    Here 

  • Question 5
    4 / -1

    Two large thin metal plates are parallel and close to each other. On their inner faces, the plates have surface charge densities of opposite signs and magnitude 27 x 10-22C m-2. The electric field  in region II in between the plates is

    Solution

    The value of  in the region II, in between the plates = 3.05 x 10-10N C-1

  • Question 6
    4 / -1

    A charged ball B hangs from a silk thread S, which makes an angle θ with a large charged conducting sheet P as shown in the figure. The surface charge density of the sheet is proportional to

    Solution


    T sinθ = σq/ε0
    T cosθ = mg
    ∴ tanθ = σq/ε0mg
    ∴ σ is proportional to tanθ.

  • Question 7
    4 / -1

    Consider a thin spherical shell of radius R consisting of uniform surface charge density σ. The electric field at a point of distance x from its centre and outside the shell is

    Solution

    For a thin uniformly charged spherical shell, the field points outside the shell at a distance x from the centre is

    If the radius of the sphere is R, Q = σ4πR2

    This is inversely proportional to square of the distance from the centre. It is as if the whole charge is concentrated at the centre.

  • Question 8
    4 / -1

     

    A non conducting sphere of radius a has a net charge +q uniformly distributed throughout its volume. A spherical conducting shell having inner and outer radii b and c and net charge −q is concentric with the sphere (see the figure).

    Read the following statements

    (i) The electric field at a distance r from the center of the sphere for r<a is 

    (ii) The electric field at distance r for A<r<b is 0
    (iii) The electric field at distance r for b<r<c is 0
    (iv) The charge on the inner surface of the spherical shell is −q
    (v) The charge on the outer surface of the spherical shell is +q
    Which of the above statements are true?

  • Question 9
    4 / -1

    There is a solid sphere of radius R having uniformly distributed charge throughout it. What is the relation between electric field E and distance r from the centre (r < R)?

  • Question 10
    4 / -1

    An early model for an atom considered it to have a positively charged point nucleus of charge Ze, surrounded by a uniform density of negative charge upto a radius R. The atom as a whole is neutral. The electric field at a distance r from the nucleus is (r < R).

    Solution

    Charge on nucleus is =+Ze
    total negative charge =−Ze(∵ atoms is electrical neutral)
    Negative charge density, ρ= charge/volume = 

    Consider a Gaussian surface with radius r
    By Gauss's theorem

    Charge enclosed by Gaussian surface
     Using (i)
    From (ii),

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