Chapter 21.2, Problem 43SSC

To determine

To Show: A volt per meter is the same as a newton per coulomb.

Explanation of Solution

Introduction:

The electric field strength on a charge $q\text{'}$ is given by,

  $E=\frac{F_{\text{on }q\text{'}}}{q\text{'}}\mathrm{...................}\left(1\right)$

Where,

  $F_{\text{on }q\text{'}}=\text{Electric force on }q\text{'}$

In terms of potential difference $\left(\Delta V\right)$ , Electric field strength $\left(E\right)$ can be given as,

  $E=\frac{\Delta V}{d}\left(2\right)$

Here, d is the distance moved by the charge.

Now it is known that SI unit of force is Newton (N), of charge is Coulomb (C), of potential difference is Volts (V), and of distance is meter (m).

Thus, SI unit of Electric field using equation (1) will be,

  $\text{SI}\text{}\text{ unit of }E=\frac{N}{C}$

And using equation (2), SI unit of Electric field will be,

  $\text{SI}\text{}\text{ unit of }E=\frac{V}{m}$

Conclusion:

Hence, equating above two equations, it can be shown that,

  $\frac{N}{C}=\frac{V}{m}$