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We have a conservative force field $\vec{E}$. The conservation theorem can be expressed without mass as:

$$\frac{1}{2}v^2-\frac{1}{2}{u}^2=\int \vec{E}\cdot \vec{dS}.$$

The quantity $\frac{1}{2}v^2$ can be defined as the kinetic energy of an object moving with speed $v$. The change in this quantity would only depend on the start and end points of the object's path

What is the need to multiply both sides by $m$ in the equation?

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To make the dimension right, You have to multiphy the left-hand-side by $m$ and right-hand-side by charge $q$:

$$\frac{1}{2} m v^2-\frac{1}{2} m {u}^2= q\int \vec{E}\cdot d\vec{S}.$$

or $$\frac{1}{2} v^2-\frac{1}{2} {u}^2= \frac{q}{m} \int \vec{E}\cdot d\vec{S}.$$

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  • $\begingroup$ So would this not be needed if the force was gravity? $\endgroup$
    – Ryder Rude
    Commented Mar 24, 2021 at 5:37
  • $\begingroup$ You still need $m$ to make the righr-hand-side to be dimension of force. $\endgroup$
    – ytlu
    Commented Mar 24, 2021 at 11:04
  • $\begingroup$ But the equation $\frac{1}{2} (v^2-u^2)=\int \vec{E}\cdot \vec{dS}$ would already be dimensionally correct if $\vec{E}$ is the gravitational field $\endgroup$
    – Ryder Rude
    Commented Mar 25, 2021 at 1:57
  • $\begingroup$ Yes. If it is the gravitational filed, not gravitation force. The force and the field are dimensional different by a factor $m$. $\endgroup$
    – ytlu
    Commented Mar 25, 2021 at 5:34

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