Question

How do you know that a volt per meter is the same as a newton per coulomb?

Answer 1

Yes, these are. See explanation below.

Explanation:

Let us find out the dimensions of both the given physical quantities. In terms of Mass M, Length L and Time T.

`text{Volt}/text{Meter} -> ("M"^(1/2)"L"^(1/2)"T"^-1)/"L" ->"M"^(1/2)"L"^(1/2-1)"T"^-1->"M"^(1/2)"L"^(-1/2)"T"^-1`

`text{Newton}/text{Coulomb}->("M""L""T"^-2)/("M"^(1/2)"L"^(3/2)"T"^-1) ->"M"^(1-1/2)"L"^(1-3/2)"T"^(-2-(-1))` # ->"M"^(1/2)"L"^(-1/2)"T"^-1#

We see that dimensions of both the given physical quantities are same.

Worksheet to ascertain the dimensions of quantities involved.

1. Newton is unit of Force `F=mcdota`, Its dimensions are `"M"cdot"L""T"^-2`, as acceleration is `text{Distance}/text{time}^2`
2. Coulomb is unit of electric charge.
   Coulomb Force `|vecF|prop(q_1q_2)/|vecr^2|`
   Dimensions of left hand side `"ML""T"^-2`
   Dimensions of right hand side `"Coulomb"^2 / "Distance"^2 ->"Coulomb"^2 / "L"^2`
   Equating the dimensions of two sides, taking denominator to the other side and taking square root of both sides we obtain

`sqrt("Coulomb"^2)->sqrt("ML""T"^-2cdot"L"^2)->sqrt("ML"^(1+2)"T"^-2)`

`"Coulomb"->("ML"^3"T"^-2)^(1/2)->"M"^(1/2)"L"^(3/2)"T"^-1`
3. We know that Volt `V=` Coulomb per meter. Units of Volt can be easily derived from 2. above as
`text{Volt}="Coulomb"/text{meter}->("M"^(1/2)"L"^(3/2)"T"^-1)/"L"->"M"^(1/2)"L"^(3/2-1)"T"^-1->"M"^(1/2)"L"^(1/2)"T"^-1`