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An electric charge q is placed at the centre of a cube of side α. The electric flux on one of its faces will be

$\frac{q}{6 ε_{0}}$

$\frac{q}{ε_{0} a^{2}}$

$\frac{q}{4 π ε_{0} a^{2}}$

$\frac{q}{ε_{0}}$

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Total electric flux coming out of a unit positive charge put in air is

$ε_{0}$

$ε_{0}^{- 1}$

${(4 p ε_{0})}^{- 1}$

$4 π ε_{0}$

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Explanation

Total flux coming out from unit charge $= \vec{E} . d \vec{s} = \frac{1}{ε_{0}} \times 1 = ε_{0}^{- 1}$

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A cube of side l is placed in a uniform field E, where $E = E \hat{i}$ . The net electric flux through the cube is

Zero

l²E

4l²E

6l²E

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Explanation

As there is no charge residing inside the cube, hence net flux is zero.

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Eight dipoles of charges of magnitude e are placed inside a cube. The total electric flux coming out of the cube will be

$\frac{8 e}{ε_{0}}$

$\frac{16 e}{ε_{0}}$

$\frac{e}{ε_{0}}$

Zero

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Explanation

The electric flux is given by the total charge enclosed by the surface multiplied by 1/ε₀.

The net charge of the dipole is zero because it consists of two equal and opposite charges. Hence, the net charge of the eight dipoles enclosed inside a cube is zero. Therefore, the total electric flux coming out of the cube will be zero.

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A charge q is placed at the centre of the open end of the cylindrical vessel. The flux of the electric field through the surface of the vessel is

Zero

$\frac{q}{ε_{0}}$

$\frac{q}{2 ε_{0}}$

$\frac{2 q}{ε_{0}}$

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According to Gauss’ Theorem, electric field of an infinitely long straight wire is proportional to

$\frac{1}{r^{2}}$

$\frac{1}{r^{3}}$

$\frac{1}{r}$

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Explanation

$e = \frac{λ}{2 π ε_{0} r}$ ⇒ $E \propto \frac{1}{r}$

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The S.I. unit of electric flux is

Weber

Newton per coulomb

Volt × metre

Joule per coulomb

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Explanation

S.I. unit of electric flux is $\frac{N \times m^{2}}{C} = \frac{J \times m}{C}$ = volt × m.

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If the electric flux entering and leaving an enclosed surface respectively is $φ_{1}$ and $φ_{2}$ the electric charge inside the surface will be

$(φ_{1} + φ_{2}) ε_{0}$

$(φ_{2} - φ_{1}) ε_{0}$

$(φ_{1} + φ_{2}) / ε_{0}$

$(φ_{2} - φ_{1}) / ε_{0}$

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Explanation

$φ_{n e t} = \frac{1}{ε_{0}} \times Q_{e n c}$ ⇒ $Q_{e n c} = (φ_{2} - φ_{1}) ε_{0}$

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An electric dipole is put in north-south direction in a sphere filled with water. Which statement is correct ?

Electric flux is coming towards sphere

Electric flux is coming out of sphere

Electric flux entering into sphere and leaving the sphere are same

Water does not permit electric flux to enter into sphere

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Explanation

In electric dipole, the flux coming out from positive charge is equal to the flux coming in at negative charge i.e. total charge on sphere = 0. From Gauss law, total flux passing through the sphere = 0.

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The electric intensity due to an infinite cylinder of radius R and having charge q per unit length at a distance r(r > R) from its axis is

Directly proportional to r²

Directly proportional to r³

Inversely proportional to r

Inversely proportional to r²

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Explanation

According to Gauss law $\oint E \cdot d s = = \frac{q l}{ε_{0}}$

$\oint d s = 2 π r l;$ (E is constant)

∴ $E \cdot 2 π r l = \frac{q l}{ε_{0}}$

⇒ $E = \frac{q}{2 π ε_{0} r}$ i.e. $E \propto \frac{1}{r}$

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