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Question

A charge Q is enclosed by a Gaussian spherical surface of radius R. If the radius is doubled, then the outward electric flux will :


A
be doubled
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B
increase four times
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C
be reduced to half
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D
remain the same
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Solution

The correct option is D remain the same
$$\large \textbf{Method 1: Using Number of Field Lines}$$
$$\bullet$$  We know that, Electric flux is proportional to number of field lines.

$$\bullet$$  Form figure, we see that the number of field lines crossing both the spherical surfaces is same.

$$\bullet$$  Therefore, the flux passing both the surfaces will be same.

Hence correct Option is $$D$$

$$\large \textbf{Method 2: Using Gauss Law}$$
$$\bullet$$  Gauss Law states that, Electric flux $$\Delta \phi = \cfrac{q_{in}}{\epsilon_0}$$

$$\bullet$$  Assuming both the spherical surfaces $$S_1$$ and $$S_2$$ as gaussian surfaces. From figure, we find that the charge enclosed($$q_{in}=Q$$), same for both the surfaces irrespective of radius.

$$\bullet$$  Hence, the Flux passing through both surfaces is same.

Hence correct Option is $$D$$

$$\textbf{Note: }$$ We can observe that whatever be the size, shape and position of second surface, If it is enclosing the same charge, then the Number of field lines crossing it or charge enclosed by it remains the same as first surface. 
Hence answer will be same, even if the second surface is cube or cylinder of any size.

2108848_597493_ans_d37c36636e2d48c49a32af25f77906e7.jpg

Physics

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