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Electric Charges and Fields Test 4

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Electric Charges and Fields Test 4
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  • Question 1
    1 / -0

    If the electric field is given by 26i – k. Calculate the electric flux through a surface of area 20 units lying in x-y plane.

    Solution

    ► Electric field (E) = 26i - k
    ► A = 20k
    Electric flux = electric field.area
    = E.A
    = (26i - k)*(20k)
    = (26*0 -1)*20
    = 20 units

  • Question 2
    1 / -0

    A point charge of 2.0 μC is at the center of a cubic Gaussian surface 9.0 cm on edge. What is the net electric flux through the surface?

    Solution

    Net electric flux (ΦNet) through the cubic surface is given by,

    Where, ∈0 = Permittivity of free space

    = 8.854 × 10−12 N−1C2 m−2

    q = Net charge contained inside the cube = 2.0 μC = 2 × 10−6 C

    = 2.2 × 105 N m2 C−1

    The net electric flux through the surface is 2.2 ×105 N m2C−1.

  • Question 3
    1 / -0

    Electric flux through an area dA for an electric field E is given by​:

    Solution

    For constant electric field

  • Question 4
    1 / -0

    A point charge causes an electric flux of −1.0×103 Nm2/C to pass through a spherical Gaussian surface of 10.0 cm radius centered on the charge.
    (a) If the radius of the Gaussian surface were doubled, how much flux would pass through the surface?
    (b) What is the value of the point charge?

    Solution

    Electric flux is given by  since amount of charge not depends on size and shape so by making radius <  double the amount of charge remain same so electric flux remain same.

  • Question 5
    1 / -0

    Choose the correct statement:

    Solution

    By definition of Solid Angle,

    Where, Ω is represented in Steradians.

  • Question 6
    1 / -0

    Electric flux is a _______ quantity.

    Solution

    Electric flux is a scalar quantityElectric Flux is dot/scalar product of magnetic field vector and area vector. And the result of a dot/scalar quantity is always a scalar quantity. So electric flux is also a scalar quantity 

  • Question 7
    1 / -0

    If the electric field is given by 6i+3j+4k calculate the electric flux through a surface of area 20 units lying in y-z plane.

    Solution

    ► Area = 20i since it is in the Y-Z plane
    ► E = 6i + 3j + 4k
    ► E.S = Flux
    = 20*6 = 120 units

  • Question 8
    1 / -0

    A uniform line charge with linear density λ lies along the y-axis. What flux crosses a spherical surface centred at the origin with r = R?

    Solution

    Line charge density = λ
    Radius, r = R
    Now,  λdl=dθ
    For total charge enclosed by radius R, 0R Q = 0R λdl
    Qinc = λR
    Now, ϕ = Qinco    {from Gauss law}
    = λR/ε0

  • Question 9
    1 / -0

    Dimensional formula of ØE is​:

    Solution

    Flux = E. A
    ⇒ F = Eq
    ⇒ E= F/q= N/C
    ⇒ Flux= N/C × m²
    = Nm2C-1

  • Question 10
    1 / -0

    Electric flux density is a function of _______.

    Solution

    Electric flux density is the charge per unit area. Hence it is a function of charge and not any of the other values.

  • Question 11
    1 / -0

    Which of the following correctly states Gauss law?

    Solution

    The electric flux passing through any closed surface is equal to the total charge enclosed by that surface. In other words, electric flux per unit volume leaving a point (vanishing small volume), is equal to the volume charge density

  • Question 12
    1 / -0

    Electric flux of a surface is maximum, when

    Solution

    By definition,

    For maximum, E.dA must be maximum. This is possible when E || dA

  • Question 13
    1 / -0

    The Gaussian surface for a point charge will be

    Solution

    A point charge is single dimensional. The three dimensional imaginary enclosed surface of a point charge will be sphere.

  • Question 14
    1 / -0

    At what point is the electric field intensity due to a uniformly charged spherical shell is maximum?

    Solution

    By Gauss Law,

    Electric Field due to spherical shell shows the following behavior:

  • Question 15
    1 / -0

    A charge q is placed at the centre of the open end of cylindrical vessel whose height is equal to its radius. The electric flux of electric field of charge q through the surface of the vessel is

    Solution

    We can assume a similar vessel, such that the charge lies at the interfaces of both the vessels. Now, by Gauss law, all the electric flux must pass through these two vessels. As they are kept symmetrically with respect to the charge equal amount of flux would pass through them

    From both vessels,

    For each vessel,

  • Question 16
    1 / -0

    The charge enclosed by a spherical Gaussian surface is 8.85 X 10-8 C. What will be the electric flux through the Surface?

    Solution

    We know that the flux linked with a Gaussian surface is given as:

    Φ = E.ds = q/ε0

    Here,

    q = 8.85 x 10-8 C

    ε0 = 8.85 x 10-12 F/m

    So,

    Φ = 8.85x10-8/ 8.85x10-12

    Thus, flux will be

    Φ = 10-4 Nm2/C

  • Question 17
    1 / -0

    If the electric field is given by  calculate the electric flux through as surface of area 10 units lying in y-z plane.

    Solution



    Hence,

  • Question 18
    1 / -0

    The flux associated with a spherical surface is 104 N m2C-1.  What will be the flux, if the radius of the spherical shell is doubled?

    Solution

    If the radius of the spherical shell is double then also flux associated with spherical surface remain unchanged because the charge enclosed in system always remains unchanged.

  • Question 19
    1 / -0

    The Gaussian surface for a line charge will be

    Solution

    A line charge can be visualized as a rod of electric charges. The three dimensional imaginary enclosed surface of a rod can be a cylinder.

  • Question 20
    1 / -0

    The electric field intensity due to a sphere (solid or hollow) at an external point varies as:

    Solution

    We calculate this using spherical symmtery in Gauss Law

    For spherical Shell,

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