If electric flux entering and leaving an enclosed surface is b

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 Multiple Choice QuestionsMultiple Choice Questions

171.

The unit of permittivity (ε0) of space is

  • newton-metre/ coulomb2

  • coulomb / newton-metre

  • coulomb / newton-metre2

  • coulomb/ newton-metre2


172.

Two charges each of equal magnitude 3.2 x 10-19 coulomb but of opposite sign form an electric dipole. The distance between the two charges is 2.4 A. If the dipole is placed in an electric field of 5 x 105 volt/metre, then in equilibrium its potential energy will be

  • 3 × 15-23 joule

  • -3.84 × 10-23 joule

  • -6 × 10-23 joule 

  • -2 × 10-26 joule 


173.

1 newton/coulomb is equivalent to

  • 1 C/V

  • 1 J

  • 1 V/M

  • 1 J/C


174.

An electron is moving with velocity v on a circular path of radius r in a transverse electric field B. The specific charge (e/m) of the electron is

  • Bvr

  • νBr

  • Bvr2

  • B


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175.

Number of extra electrons in a body of charge -80μC, is

  • 80 × 1015

  • 80 × 10−15

  • 5 × 1014

  • 1.28 × 10−17


176.

If the electric fluxes entering and leaving an enclosed surface respectively are  Φ1 and Φ2, the electric charge inside the surface will be

  • 2 − Φ1) ε0

  • 1 + Φ2) / ε0

  • 2 − Φ1) / ε0

  • 1 + Φ2) / ε0


177.

Shown below is a distribution of charges. The flux of electric field due to these charges through the surface is

  • 3qε0

  • zero

  • 2qε0

  • qε0


178.

Two point charges + q and + 4q are located at x = O and x = L respectively. The location of a point on the x-axis at which the net electric field due to these two points charges is zero, is

  • L / 3

  • 2 L

  • 4 L

  • 8 L


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179.

If electric flux entering and leaving an enclosed surface is ϕ1 and ϕ2 respectively, the electric charge inside the enclosed surface will be

  • ϕ1 + ϕ2 ε0

  • ϕ2 - ϕ1 ε0

  • ϕ1 + ϕ2 / ε0

  • ϕ2 - ϕ1 / ε0


B.

ϕ2 - ϕ1 ε0

Applying Gauss theorem, ϕ = qε0                                        q =ε0 ϕSince, ϕ = ϕ2 - ϕ1           q = ϕ2 - ϕ1 ε0


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180.

The electric field in a region is given by E = 4 i + 3 jV/m. The net flux passing through a square area of side 4 m parallel to y-z plane is

  • 32 V-m

  • 16 V-m

  • 12 V-m

  • 64 V-m


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