If the net charge inside a Gaussian surface is zero, then does it mean that electric field intensity at each point on the surface is zero?

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Solution

No, because
(i) Electric Flux $ϕ = \int_{S}^{\to} E . d \to s = \frac{q}{ε_{0}} = 0$
(ii) It is possible when $\to E ⊥ \to d s$ .

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If the net charge inside a Gaussian surface is zero, then does it mean that electric field intensity at each point on the surface is zero?

No, because(i) Electric Flux ϕ=∫→SE.d→s=qε0=0(ii) It is possible when →E⊥→ds.

No, because
(i) Electric Flux $ϕ = \int_{S}^{\to} E . d \to s = \frac{q}{ε_{0}} = 0$
(ii) It is possible when $\to E ⊥ \to d s$ .