How does the electric flux due to a point charge enclosed by a spherical gaussian surface get affected when its radius is increased?

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Solution

Electric flux, $ϕ = \frac{q_{e n c l o s e d}}{ϵ_{o}}$
As electric dipole consists of equal positive and negative charges separated by small distance. Thus net charge on a dipole is Zero.
$∴$ Charge enclosed by the cube, $q_{e n c l o s e d} = 0$
$⟹$ $ϕ = \frac{0}{ϵ_{o}} = 0$

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How does the electric flux due to a point charge enclosed by a spherical gaussian surface get affected when its radius is increased?

Electric flux, ϕ=qenclosedϵoAs electric dipole consists of equal positive and negative charges separated by small distance. Thus net charge on a dipole is Zero.∴ Charge enclosed by the cube, qenclosed=0⟹ ϕ=0ϵo=0

Electric flux, $ϕ = \frac{q_{e n c l o s e d}}{ϵ_{o}}$
As electric dipole consists of equal positive and negative charges separated by small distance. Thus net charge on a dipole is Zero.
$∴$ Charge enclosed by the cube, $q_{e n c l o s e d} = 0$
$⟹$ $ϕ = \frac{0}{ϵ_{o}} = 0$