The units of the quantity are:
[tex] \frac{[kg] [m]^2}{[s]^2 [C]} [/tex]
We can isolate one [tex] \frac{[m]}{[s]^2} [/tex], which corresponds to an acceleration, a:
[tex]= \frac{[kg][m] a}{[C]} =[/tex]
[kg] corresponds to a mass, m; [m] corresponds to a length, L; C corresponds to a charge, Q:
[tex]= \frac{maL}{Q} [/tex]
But the product ma is a force, F:
[tex]= \frac{FL}{Q} [/tex]
and the product FL is a work, so an energy, U:
[tex]= \frac{U}{Q} [/tex]
and this ratio corresponds to an electrical potential. So, the quantity is an electrical potential.
