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Answer 1

$\huge{\underline{\underline{\bf{Answer:-}}}}$

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Electric flux and Gauss law:

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The electric flux is defined as the dot product of the electric field and the area vector.

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• According to Gauss law, the electric flux is given as follows:

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$ϕ= \frac{Q}{ϵo}$

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Where,

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• Q is the charge enclosed in the gaussian surface.

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• ϵo is the electric permittivity of the free space.

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${\underline{\underline{\mathsf{Answer\:and\:Explanation:}}}}$

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Given Data:

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• $charge \: 1,(q1) = 3.4 \: nC$

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• $charge \: 2,(q2) = - 6.00 \: nC$

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(A)

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Now, the charge 2 will be inside the sphere of the radius $( {r}^{2} ) = 1.55 \: m,$therefore the enclosed charge would be

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$Q = q_{2}$

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$Q = - 6 \: nC$

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Therefore, the electric flux would be

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$ϕ= \frac{Q}{ϵo}$

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$= \frac{ - 6 \times 10 {}^{ - 2}C }{8.8 \times 10 {}^{ - 12} }$

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$= - 678 \: N {m}^{2} C$

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(B)

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In this sphere both charge will be inside the sphere.

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$Q = q_{1} + q_{2}$

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$Q = 3.4 - 6$

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$Q = - 2.6nC$

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Therefore, the electric flux would be

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$ϕ= \frac{Q}{ϵo}$

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$= \frac{ - 2.6 \times 10 {}^{ - 2}C }{8.5 \: \times 10 {}^{ - 12} }$

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$= - 263 \: Nm {}^{2} C$

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