Question

The dimensional formula for electric flux is?
A. $\left[ \text{M}{{\text{L}}^{3}}{{\text{A}}^{-1}}{{\text{T}}^{-3}} \right]$
B. $\left[ {{\text{M}}^{2}}{{\text{L}}^{2}}{{\text{A}}^{-1}}{{\text{T}}^{-2}} \right]$
C. $\left[ \text{M}{{\text{L}}^{3}}{{\text{A}}^{1}}{{\text{T}}^{-3}} \right]$
D. $\left[ \text{M}{{\text{L}}^{-3}}{{\text{A}}^{-1}}{{\text{T}}^{-3}} \right]$

Answer 1

Hint: Electric flux can be defined as the number of electric field lines through an area. Derive each quantity in terms of the fundamental quantities to find its dimensional formula.
Formula Used:
$\phi =E\cdot S\cos \theta $
$E=\dfrac{F}{Q}$
\[Q=I\cdot t\]
$F=ma$

Complete answer:
Every quantity in a physical system can be defined in terms of some fundamental quantities. These fundamental quantities are independent of each other and cannot be interlinked. These fundamental quantities are known as dimensions. Some of the fundamental quantities are Mass $\left( \text{M} \right)$, Length $\left( \text{L} \right)$, Time $\left( \text{T} \right)$ and Current $\left( \text{A} \right)$.
Now, we have to determine the dimensions of electric flux. Electric flux is defined as the number of electric field lines passing through an area. It is mathematically written as,
$\phi =E\cdot S\cos \theta $
Here, $\phi $ is electric flux, $E$ represents electric field, $S$ represents area and $\theta $ is the angle between electric field and area vector.
Now, let us determine the dimensions of the electric field. Electric field can be defined as,
$E=\dfrac{F}{Q}$
Here, $F$ is an electrostatic force and $Q$ is charged. Now, let us further derive force and charge in terms of base quantities.
The dimensions of force can be determined as,
\[\begin{align}
  & \left[ F \right]=\text{mass}\times \text{acceleration} \\
 & =\text{mass}\times \dfrac{\text{velocity}}{\text{time}} \\
 & =\text{mass}\times \dfrac{\dfrac{\text{displacement}}{\text{time}}}{\text{time}} \\
 & =\text{M}\times \dfrac{\dfrac{\text{L}}{\text{T}}}{\text{T}} \\
 & =\text{ML}{{\text{T}}^{-2}}
\end{align}\]
Now, let us determine the dimensions of electric charge. Electric charge is defined as,
\[Q=I\cdot t\]
Here, $I$ is electric current and $t$ is time. Now, current is a fundamental quantity and its dimensions are $\left[ \text{A} \right]$. Thus, the dimensions of charge are,
$\left[ Q \right]=\left[ \text{AT} \right]$
Use the dimensions of force and charge to determine the dimensions of electric flux. It can be seen as,
\[\begin{align}
  & \left[ E \right]=\dfrac{\left[ F \right]}{\left[ Q \right]} \\
 & =\dfrac{\left[ \text{ML}{{\text{T}}^{-2}} \right]}{\left[ \text{AT} \right]} \\
 & =\left[ \text{ML}{{\text{T}}^{-3}}{{\text{A}}^{-1}} \right]
\end{align}\]
Now, the dimensions of the area is $\left[ {{\text{L}}^{2}} \right]$. Now, let us use the dimensions of electric field and area to determine the dimensions of electric flux. The dimensions of electric flux are thus determined as,
$\begin{align}
  & \left[ \phi \right]=\left[ E \right]\cdot \left[ S \right] \\
 & =\left[ \text{ML}{{\text{T}}^{-3}}{{\text{A}}^{-1}} \right]\cdot \left[ {{\text{L}}^{2}} \right] \\
 & =\left[ \text{M}{{\text{L}}^{3}}{{\text{T}}^{-3}}{{\text{A}}^{-1}} \right]
\end{align}$

Thus, the correct option is (A).

Note:
Every quantity needs to be derived in terms of the base quantity to determine the dimensions. Dimensions of a particular quantity are fixed and are independent of the formula used to calculate the dimensions. Trigonometric ratios and formulae like $\sin \theta $ do not have any dimensions.