Electric Charges and Fields

Electric Charges and Fields

Easy Questions

Electrostatics

Electric Charges and Fields
Electrostatic Potential and Capacitance
  1. A charge Q is enclosed by a Gaussian spherical surface of radius R. If the radius is doubled, then  the outward electric flux will








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    By Gauss’s theorem,

    φ =
    qin
    ε0

    Thus, the net flux depends only on the charge enclosed by the surface. Hence, there will be no effect on the net flux if the radius of the surface is doubled.

    Correct Option: C

    By Gauss’s theorem,

    φ =
    qin
    ε0

    Thus, the net flux depends only on the charge enclosed by the surface. Hence, there will be no effect on the net flux if the radius of the surface is doubled.

  1. What is the flux through a cube of side 'a' if a point charge of q is at one of its corner :








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    ​Eight identical cubes are required to arrange so that this charge is at centre of the cube formed so flux.

    φ =
    q
    8ε0


    Correct Option: B

    ​Eight identical cubes are required to arrange so that this charge is at centre of the cube formed so flux.

    φ =
    q
    8ε0


  1. A point charge +q is placed at mid point of a cube of side ‘L’. The electric flux emerging from the cube is








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    By Gauss theorem

    Total electric flux =
    Total charge inside cube
    ε0

    ⇒ φ =
    q
    ε0

    Correct Option: A

    By Gauss theorem

    Total electric flux =
    Total charge inside cube
    ε0

    ⇒ φ =
    q
    ε0

  1. There is an electric field E in x-direction. If the work done on moving a charge of 0.2 C through a distance of 2 m along a line making an angle 60° with x-axis is 4J, then what is the value of E?​​








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    Charge (q) = 0.2 C ; Distance (d) = 2m ;
    Angle θ = 60° and work done (W) = 4J. ​
    Work done in moving the charge (W) ​= F.d cos θ = qEd cos θ

    or , E =
    W
    		 =
    4
    		 =
    4
    		 = 20 N / C
    qd cos θ0.2 × 2 × cos 60°0.4 × 0.5

    Correct Option: D

    Charge (q) = 0.2 C ; Distance (d) = 2m ;
    Angle θ = 60° and work done (W) = 4J. ​
    Work done in moving the charge (W) ​= F.d cos θ = qEd cos θ

    or , E =
    W
    		 =
    4
    		 =
    4
    		 = 20 N / C
    qd cos θ0.2 × 2 × cos 60°0.4 × 0.5

  1. An electric dipole, consisting of two opposite charges of 2 × 10–6 C each separated by a distance 3 cm is placed in an electric field of 2 × 105 N/C. Torque acting on the dipole is​​








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    Charges (q) = 2 × 10–6 C, Distance (d) = 3 cm = 3 × 10–2 m and electric field (E) = 2 × 105 N/C.
    Torque (τ) = q.d. ​
    E = (2 × 10–6) × (3 × 10–2) × (2 × 105) ​
    = 12 × 10–3 N–m .

    Correct Option: C

    Charges (q) = 2 × 10–6 C, Distance (d) = 3 cm = 3 × 10–2 m and electric field (E) = 2 × 105 N/C.
    Torque (τ) = q.d. ​
    E = (2 × 10–6) × (3 × 10–2) × (2 × 105) ​
    = 12 × 10–3 N–m .