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Electric Field Due to Charge Distributions - Approach

Five charges ...

Question

Five charges $q_{1}, q_{2}, q_{3}, q_{4}$ and $q_{5}$ are fixed at their positions as shown in figure. $S$ is Gaussian surface. The Gauss's law is given by $\oint_{S} \to E . \to d s = \frac{q}{ε_{0}}$ . Which of the following statements is correct?

A

\to E

on the LHS of the above equation will have a contribution from

q_{1}, q_{5}

and

q_{3}

while

q

on the RHS will have a contribution from

q_{2}

and

q_{4}

only

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B

\to E

on the LHS of the above equation will have a contribution from all charges while

q

on the RHS will have a contribution from

q_{2}

and

q_{4}

only

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C

\to E

on the LHS of the above equation will have a contribution from all charges while

q

on the RHS will have a contribution from

q_{1} q_{3}

and

q_{5}

only

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D

Both

\to E

on the LHS and

q

on the RHS will have a contributions from

q_{2}

and

q_{4}

only

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Solution

The correct option is B $\to E$ on the LHS of the above equation will have a contribution from all charges while $q$ on the RHS will have a contribution from $q_{2}$ and $q_{4}$ only
Key concept : According to Gauss' law, the term $q_{e n c l o s e d}$ on the right side of the equation $ϕ_{s} E$ . $d S = q_{e n c l o s e d / i n_{0}}$ includes the sum of all charges enclosed by the surface called (Gaussian surface).
In left side equation, the electric field is due to all the charges present both inside as well as outside the Gaussian surface.
Hence in given question, E on LHS of the above equation will have a contribution from all charges while q on the RHS will have a contribution from $q_{2}$ and $q_{4}$ only.
Hence option (b) is correct.

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Similar questions

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(A)

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