What are the S.I units of electric flux?

Question

Accepted Answer

The correct option is (4):

Electric flux, denoted by $\Phi_E$, is a measure of the electric field passing through a given area. It is fundamentally defined by the dot product of the electric field vector ($\vec{E}$) and the area vector ($\vec{A}$), i.e., $\Phi_E = \vec{E} \cdot \vec{A}$. To determine the S.I. unit of electric flux, we need to consider the S.I. units of electric field and area.

The S.I. unit of the electric field ($\vec{E}$) is Newtons per Coulomb ($N/C$ or $N \cdot C^{-1}$). This is because the electric field is defined as the force per unit charge ($E = F/q$), where force ($F$) is measured in Newtons ($N$) and charge ($q$) in Coulombs ($C$).
The S.I. unit of area ($\vec{A}$) is square meters ($m^2$).

Therefore, the S.I. unit of electric flux ($\Phi\_E$) can be derived by multiplying the units of electric field and area:
$	ext{Unit of } \Phi\_E = 	ext{Unit of } E 	imes 	ext{Unit of } A = (N \cdot C^{-1}) 	imes (m^2) = N \cdot C^{-1} \cdot m^2$.

Alternatively, electric flux can also be expressed in Volt-meters ($V \cdot m$), since the electric field can also be expressed as Volts per meter ($V/m$ or $V \cdot m^{-1}$), and multiplying by area ($m^2$) gives $V \cdot m^{-1} \cdot m^2 = V \cdot m$. Both $N \cdot C^{-1} \cdot m^2$ and $V \cdot m$ are equivalent S.I. units for electric flux. Option (4) matches $N \cdot C^{-1} \cdot m^2$.

The incorrect option is (1): $N\~C^{-1}m^{-2}$. This option incorrectly places $m^{-2}$ instead of $m^2$ for the area unit, which would imply area is in the denominator.
The incorrect option is (2): $N\~C\~m^{-2}$. This option incorrectly has $C$ in the numerator and $m^{-2}$ in the denominator, which is not consistent with the definition of electric flux.
The incorrect option is (3): $N\~C^{-2}m^{2}$. This option incorrectly has $C^{-2}$ instead of $C^{-1}$, which would imply a division by charge squared, not just charge.