Question

Write the unit of electric flux.

Hint:

Don’t confuse {flux} \((\Phi_E)\) with {flux density}. Electric {flux density} is the field \(\vec{E}\) (units N/C or V/m), whereas electric {flux} multiplies by area, giving N·m\(^2\)/C (or V·m).

Answer 1

Step 1: Start from the definition.
Electric flux through a surface \(S\) is \[ \Phi_E = \iint_S \vec{E}. d\vec{A} \] For a uniform field making an angle \(\theta\) with the area vector, \(\Phi_E = EA\cos\theta\).
Step 2: Analyze units using \(\vec{E}\) in N/C.
\(\vec{E}\) has unit newton per coulomb (N/C) and \(A\) has unit m\(^2\). Hence, \[ [\Phi_E] = \frac{\text{N}}{\text{C}}\times \text{m}^2 = \text{N·m}^2/\text{C}. \] Step 3: Show the equivalent unit using \(\vec{E}\) in V/m.
Since \(1~\text{N/C} = 1~\text{V/m}\), an equivalent unit is \[ [\Phi_E] = \left(\frac{\text{V}}{\text{m}}\right)\times \text{m}^2 = \text{V·m}. \] Step 4: Express in SI base units (dimensional check).
\(\text{N}=\text{kg·m·s}^{-2}\) and \(\text{C}=\text{A·s}\). Therefore \[ \text{N·m}^2/\text{C} = \frac{\text{kg·m·s}^{-2}. \text{m}^2}{\text{A·s}} = \text{kg·m}^3\text{·s}^{-3}\text{·A}^{-1}. \] This confirms internal consistency.
Answer (unit of electric flux): \(\boxed{\text{N·m}^2/\text{C}}\) (equivalently, \(\boxed{\text{V·m}}\)).