
What is the electric flux through the surface \[S\] in a vacuum?
Answer
492.6k+ views
Hint: The electric flux is one of the properties of an electric field and that means the number of lines of electric force intersects a given area. Generally, the positive flux and the negative flux always become equal. So that the net electric flux is zero. The electric flux does not depend on the size and shape of the surface. We are going to find the electric flux through the surface in a vacuum.
Formula used:
The electric flux over the closed surface in a vacuum, ${\phi _E} = \dfrac{{{Q_{net}}}}{{{\varepsilon _ \circ }}}$
Where \[{Q_{net}}\] - total charge
${\varepsilon _ \circ }$- permittivity of the medium
Complete answer:
According to Gauss law, the electric flux over the closed surface is defined as the ratio between the total net charge enclosed inside the surface and the permittivity of the medium. The charges are not enclosed inside the surface and cannot be taken into account of the net charge.
Given, the charges are over the surface in a vacuum. The electric flux can be calculated by the above formula,
${\phi _E} = \dfrac{{{Q_{net}}}}{{{\varepsilon _ \circ }}}$
${Q_{net}} = {q_1} + {q_2} + {q_3} + .....$
The total net charge is equal to the sum of all individual charges enclosed inside the surface.
${Q_{net}} = \left( { + 2q - q} \right)$
${Q_{net}} = q$
Therefore the electric flux is given by
${\phi _E} = \dfrac{q}{{{\varepsilon _ \circ }}}$
It found that the electric flux through the surface s in a vacuum is ${\phi _E} = \dfrac{q}{{{\varepsilon _ \circ }}}$
Note: The surface imagined in Gauss law is called Gaussian’s surface. Application of Gauss law is the electric field due to infinitely long wire, infinite charged plane sheet, two parallel charged sheets, and uniformly charged spherical shell. The permittivity of the medium in a vacuum is $8.85 \times {10^{ - 12}}\dfrac{F}{m}$(Faraday per meter). In day-to-day life, the electric flux is used in photocopying machines, electrostatic precipitators, and air purifiers.
Formula used:
The electric flux over the closed surface in a vacuum, ${\phi _E} = \dfrac{{{Q_{net}}}}{{{\varepsilon _ \circ }}}$
Where \[{Q_{net}}\] - total charge
${\varepsilon _ \circ }$- permittivity of the medium
Complete answer:
According to Gauss law, the electric flux over the closed surface is defined as the ratio between the total net charge enclosed inside the surface and the permittivity of the medium. The charges are not enclosed inside the surface and cannot be taken into account of the net charge.
Given, the charges are over the surface in a vacuum. The electric flux can be calculated by the above formula,
${\phi _E} = \dfrac{{{Q_{net}}}}{{{\varepsilon _ \circ }}}$
${Q_{net}} = {q_1} + {q_2} + {q_3} + .....$
The total net charge is equal to the sum of all individual charges enclosed inside the surface.
${Q_{net}} = \left( { + 2q - q} \right)$
${Q_{net}} = q$
Therefore the electric flux is given by
${\phi _E} = \dfrac{q}{{{\varepsilon _ \circ }}}$
It found that the electric flux through the surface s in a vacuum is ${\phi _E} = \dfrac{q}{{{\varepsilon _ \circ }}}$
Note: The surface imagined in Gauss law is called Gaussian’s surface. Application of Gauss law is the electric field due to infinitely long wire, infinite charged plane sheet, two parallel charged sheets, and uniformly charged spherical shell. The permittivity of the medium in a vacuum is $8.85 \times {10^{ - 12}}\dfrac{F}{m}$(Faraday per meter). In day-to-day life, the electric flux is used in photocopying machines, electrostatic precipitators, and air purifiers.
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