If electric flux entering and leaving an enclosed surface respectively is {phi ₁} and &

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1. Electric Charges and Fields

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If the electric flux entering and leaving an enclosed surface respectively is ${\phi _1}$ and ${\phi _2}$ the electric charge inside the surface will be

A

$\left( {{\phi _2} - {\phi _1}} \right){\varepsilon _0}$

B

$\frac{{\left( {{\phi _1} + {\phi _2}} \right)}}{{{\varepsilon _0}}}$

C

$\frac{{\left( {{\phi _2} - {\phi _1}} \right)}}{{{\varepsilon _0}}}$

D

$\left( {{\phi _1} + {\phi _2}} \right){\varepsilon _0}$

Solution

Net electric flux coming out from the closed surface

$\phi_{2}-\phi_{1}=\frac{q}{\varepsilon_{0}}$ (Gauss's theorem )

$\therefore $ ${q=\varepsilon_{0}\left(\phi_{2}-\phi_{1}\right)}$

Standard 12

Physics

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