Question

The unit of electric field is not equivalent to:
$\begin{array}{rl}& \left(A\right){\displaystyle \frac{N}{C}}\\ & \left(B\right){\displaystyle \frac{J}{C}}\\ & \left(C\right){\displaystyle \frac{V}{m}}\\ & \left(D\right){\displaystyle \frac{J}{C-m}}\end{array}$

Answer 1

Hint: We have to compare the unit of electric field with the units of the terms given in our problem. The simplest method to do so is by writing the dimensional formula of all the terms and then just comparing them. In this way, we can find out the unit which is not equivalent to the electric field. We shall proceed in this manner only.

Complete answer:
Let us first of all find the dimensional formula of electric field and then we will find the dimensional formula of all the other terms for comparison. The dimensional formula of electric field is given by:
$\Rightarrow \left[\overrightarrow{E}\right]=\left[M^1{L}^1{I}^{-1}{T}^{-3}\right]$
We will use this as a reference for comparison.
Now, we have:
(A) Dimensional formula of ${\displaystyle \frac{N}{C}}$. Here, ‘N’ is Newton and ‘C’ is coulomb. Therefore, the dimensional formula will be given by:
$\begin{array}{rl}& \Rightarrow \left[{\displaystyle \frac{N}{C}}\right]=\left[{\displaystyle \frac{M^1{L}^1{T}^{-2}}{I^1{T}^1}}\right]\\ & \Rightarrow \left[{\displaystyle \frac{N}{C}}\right]=\left[M^1{L}^1{I}^{-1}{T}^{-3}\right]\end{array}$
This is equal to the dimensional formula of electric field. Hence, their units are the same.
(B) Dimensional formula of ${\displaystyle \frac{J}{C}}$. Here, ‘J’ is Joule and ‘C’ is coulomb. Therefore, the dimensional formula will be given by:
$\begin{array}{rl}& \Rightarrow \left[{\displaystyle \frac{J}{C}}\right]=\left[{\displaystyle \frac{M^1{L}^2{T}^{-2}}{I^1{T}^1}}\right]\\ & \Rightarrow \left[{\displaystyle \frac{J}{C}}\right]=\left[M^1{L}^2{I}^{-1}{T}^{-3}\right]\end{array}$
This is not equal to the dimensional formula of electric field. Hence, their units are not the same
(C) Dimensional formula of ${\displaystyle \frac{V}{m}}$. Here, ‘V’ is Volt (a measure of electric potential) and ‘m’ is meter. Therefore, the dimensional formula will be given by:
$\begin{array}{rl}& \Rightarrow \left[{\displaystyle \frac{V}{m}}\right]=\left[{\displaystyle \frac{M^1{L}^2{I}^{-1}{T}^{-3}}{L^1}}\right]\\ & \Rightarrow \left[{\displaystyle \frac{V}{m}}\right]=\left[M^1{L}^1{I}^{-1}{T}^{-3}\right]\end{array}$
This is equal to the dimensional formula of electric field. Hence, their units are the same.
(D) Dimensional formula of ${\displaystyle \frac{J}{C-m}}$. Here, ‘J’ is Joule, ‘C’ is coulomb and ‘m’ is meter. Therefore, the dimensional formula will be given by:
$\begin{array}{rl}& \Rightarrow \left[{\displaystyle \frac{J}{C-m}}\right]=\left[{\displaystyle \frac{M^1{L}^2{T}^{-2}}{I^1{T}^1{L}^1}}\right]\\ & \Rightarrow \left[{\displaystyle \frac{J}{C-m}}\right]=\left[M^1{L}^1{I}^{-1}{T}^{-3}\right]\end{array}$
This is equal to the dimensional formula of electric field. Hence, their units are the same.
Hence, the unit of electric field is not equivalent to ${\displaystyle \frac{J}{C}}$.

Hence, option (B) is the correct option.

Note:
Whenever calculating or comparing the units of two quantities, the method of dimensional formula is a sure shot method with no exceptions. It can be used for any quantity in Physics. One should also be careful while writing the dimensional formulas of different terms as a slight mistake can result in error in our solution.