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  • A charge Q is placed at a distance of 4R above the centre of a disc of radius R. The magnitude of flux through the disc is <img alt="\phi" src="https://learn.careers360.com/latex-image/?%5Cphi

A charge Q is placed at a distance of 4R above the centre of a disc of radius R. The magnitude of flux through the disc is \phi Now  a  hemispherical shell of radius R is placed over the disc such  that it forms a closed surface. The flux through the curved surface taking direction of area vector along outward normal as positive, is

Option: 1

Zero


Option: 2

\phi

 

 


Option: 3

-\phi


Option: 4

2\phi


Answers (1)

best_answer

 

Gauss's Law -

Total flux linked with a closed surface called the Gaussian surface.

Formula:

\phi = \oint \vec{E}\cdot d\vec{s}=\frac{Q_{enc}}{\epsilon _{0}}

After covering with a hemispherical shell ;\phi _{shell}+\phi _{disc}= 0\left   since there is no charge inside the hemispherical shell

\therefore \phi _{shell}= -\phi _{disc}= -\phi

Posted by

Devendra Khairwa

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