
The SI unit of electric flux is
(A) Voltmeter
(B) Joule/meter
(C) Newton
(D) None
Answer
135k+ views
Hint: Electric flux is given by the number of electric field lines passing through a given area. It is a measure of the electric field intensity in a given area. Formulate the equation for electric flux and substitute the units of the related terms in the equation to find the SI unit of electric flux.
Complete step by step solution:
Let $E$ denote the electric field in an area represented by the variable $A$ . Let the electric flux be represented by $\Phi $ . Electric flux is defined as the measure of the electric field in a given area. So, the formulated equation from the definition looks like
$\Phi =\overset{\to }{\mathop{E}}\,\cdot \overset{\to }{\mathop{dA}}\,$
The electric field is defined as the electric force per unit charge. SI unit of force is Newton, $N$and the SI unit of charge is $C$(coulomb). Therefore, the SI unit of the electric field will be Newton per coulomb, $N{{C}^{-1}}$ . And the SI unit of area is ${{m}^{2}}$ (meter square).
Therefore, the SI unit of electric flux is $N{{m}^{2}}{{C}^{-1}}$ (Newton meter square per coulomb)
Newton is the SI unit of force and meter is the SI unit of displacement. As force multiplied by displacement is work which is expressed in terms of joules, we can write the SI unit of electric flux as $Jm{{C}^{-1}}$ (Joule meter per coulomb).
But, the units of potential is $J{{C}^{-1}}$ (joules per coulomb).
Therefore, the SI unit of electric flux becomes $Vm$ (voltmeter).
Option (A) is the correct answer.
Note:
All units can be written in different ways that involve different terms if we break the units into the fundamental units, or by using dimensional analysis. And errors will be reduced if we use dimensional analysis.
Complete step by step solution:
Let $E$ denote the electric field in an area represented by the variable $A$ . Let the electric flux be represented by $\Phi $ . Electric flux is defined as the measure of the electric field in a given area. So, the formulated equation from the definition looks like
$\Phi =\overset{\to }{\mathop{E}}\,\cdot \overset{\to }{\mathop{dA}}\,$
The electric field is defined as the electric force per unit charge. SI unit of force is Newton, $N$and the SI unit of charge is $C$(coulomb). Therefore, the SI unit of the electric field will be Newton per coulomb, $N{{C}^{-1}}$ . And the SI unit of area is ${{m}^{2}}$ (meter square).
Therefore, the SI unit of electric flux is $N{{m}^{2}}{{C}^{-1}}$ (Newton meter square per coulomb)
Newton is the SI unit of force and meter is the SI unit of displacement. As force multiplied by displacement is work which is expressed in terms of joules, we can write the SI unit of electric flux as $Jm{{C}^{-1}}$ (Joule meter per coulomb).
But, the units of potential is $J{{C}^{-1}}$ (joules per coulomb).
Therefore, the SI unit of electric flux becomes $Vm$ (voltmeter).
Option (A) is the correct answer.
Note:
All units can be written in different ways that involve different terms if we break the units into the fundamental units, or by using dimensional analysis. And errors will be reduced if we use dimensional analysis.
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