Consider Gauss-s law -E-d-s - q-0-Then- for the situation shown in Fig- at the Gaussian surface-

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Standard XII

Physics

Gauss's Law

Consider Gaus...

Question

Consider Gauss's law $\oint \to E \cdot \to d s = \frac{q}{ϵ_{0}}$ .
Then, for the situation shown in Fig. at the Gaussian surface.

\to E

due to

q_{2}

would be zero

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\to E

due to both

q_{1}

and

q_{2}

would not be zero

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ϕ

due to both

q_{1}

and

q_{2}

would not be zero

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ϕ

due to

q_{2}

would be zero

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Solution

The correct options are
B $\to E$ due to both $q_{1}$ and $q_{2}$ would not be zero
D $ϕ$ due to $q_{2}$ would be zero
In L.H.S. of Gauss's law, $\to E$ is due to all point charges present in space and $ϕ$ depends only on the enclosed charges.

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Consider Gauss's law ∮→E⋅→ds=qϵ0.Then, for the situation shown in Fig. at the Gaussian surface.

The correct options are B →E due to both q1 and q2 would not be zero D ϕ due to q2 would be zeroIn L.H.S. of Gauss's law, →E is due to all point charges present in space and ϕ depends only on the enclosed charges.

Consider Gauss's law $\oint \to E \cdot \to d s = \frac{q}{ϵ_{0}}$ .
Then, for the situation shown in Fig. at the Gaussian surface.

The correct options are
B $\to E$ due to both $q_{1}$ and $q_{2}$ would not be zero
D $ϕ$ due to $q_{2}$ would be zero
In L.H.S. of Gauss's law, $\to E$ is due to all point charges present in space and $ϕ$ depends only on the enclosed charges.