Q

Need clarity, kindly explain! - Electrostatics - JEE Main-3

If the electric flux entering and leaving an enclosed surface respectively is and the electric charge inside the surface will be

• Option 1)

• Option 2)

• Option 3)

• Option 4)

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N

As we learnt in

Gauss's Law -

Total flux linked with a closed surface called Gaussian surface.

Formula:

$\dpi{100} \phi = \oint \vec{E}\cdot d\vec{s}=\frac{Q_{enc}}{\epsilon _{0}}$

- wherein

No need to be a real physical surface.

Qenc - charge enclosed by closed surface.

According to Gauss theorem,

$\left ( \phi _{2} -\phi _{1}\right )= \frac{Q}{\varepsilon _{0}}\Rightarrow Q= \left ( \phi _{2} -\phi _{1}\right )\varepsilon _{0}$

The flux enters the enclosure if one has a negative charge $\left ( -q _{2} \right )$ and flux goes out if one has a +ve charge $\left ( +q _{1} \right )$

As one does not know whether $\phi _{1} > \phi _{2}\or \, \, \, \phi _{2} > \phi _{1}, Q=\left ( \phi _{2} -\phi _{1}\right )\varepsilon _{0}$

Option 1)

Correct

Option 2)

Incorrect

Option 3)

Incorrect

Option 4)

Incorrect

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