Question

The SI unit of electric flux density is:
A. $\dfrac{N}{C}$
B. $\dfrac{N{{m}^{2}}}{C}$
C. $\dfrac{Nm}{C}$
D. $N{{m}^{2}}$

Answer 1

Hint: Electric flux density is the measurement of the intensity of the electric field generated by a source (electric charge). A charged object produces field lines around the surface. The direction of the field lines depends on the type of charge that occurred on the object.

Complete step by step solution:
The electric flux density is defined as the measure of electric flux passing through a unit area. It is denoted by D.
The electric field intensity of the field at a point is given by,
$E=\dfrac{Q}{4\pi {{\varepsilon }_{0}}{{r}^{2}}}\hat{r}$
Where,
 $r$ is the distance from the charge
$\hat{r}$ shows the direction of the field line (away from the charge)
From the above equation, we can say that E is inversely proportional to the $4\pi {{r}^{2}}$ indicating that $E$ decreases in proportion to the area of a sphere surrounding the charge. Now integrate both sides of the equation.
$\begin{align}
  & \oint{[E(r)]ds}=\oint{\left[ \hat{r}\dfrac{q}{4\pi \varepsilon {{r}^{2}}} \right]}ds \\
 & \Rightarrow q\dfrac{1}{4\pi \varepsilon {{r}^{2}}}\oint{\hat{r}ds} \\
\end{align}$
In this case $ds=\hat{r}ds$. Thus, the right-hand side simplifies to:
$q\dfrac{1}{4\pi \varepsilon \hat{r}}\oint{ds}$
So the flux density is given by,
$D=\varepsilon E$
The SI unit of electric flux density is $\dfrac{C}{{{m}^{2}}}$ it can also be written as $\dfrac{N}{C}$ .

So, the correct answer is “Option A”.

Note: When the field lines are perpendicular to the surface of the body the electric flux density of the body is zero. And the electric field density is maximum when the angle between the electric field line and surface of the cross-section is zero. The electric flux density increases with the increase in surface area and on the other hand, it decreases with the increase in the distance of the point from the charge source.