# Why my formulary says $KE=qV$?

So according to this formulary:

$KE=\frac{1}{2}mv^{2}$

Also:

$KE=qV$

Therefore:

$\frac{1}{2}mv^{2}=&space;qV$

Now, if we substitute:

$\frac{1}{2}[kg][m^{2}s^{-2}]=[As][kgm^{2}A^{-1}s^{-3}]$

Then:

$\frac{1}{2}[kg][m^{2}s^{-2}]=[kg m^{2}s^{-2}]$

What happened to the 1/2 in the right side?

• 1/2 has no units, so $\left[\frac{1}{2}\right] =$
– d_b
Commented Nov 1, 2020 at 5:30
• This site supports MathJax so that math is readable, searchable, accessible, and consistent in appearance. Please don’t use someone else's equation editor. Commented Nov 1, 2020 at 5:52

There is nothing wrong with this. When you equate the two equations

$$\frac{1}{2} mv^2 = qV$$

you are essentially confirming what the kinetic energy of a charge $$q$$ of mass $$m$$ would gain after traversing a potential difference $$V$$.

It then appears that you are doing some dimensional analysis on both sides of the equality. While this is all good, you ask about the $$\frac{1}{2}$$ factor. This is a dimensionless number and need not be included in such an analysis.

If you are asking why $$E_k$$ should not be $$\frac{1}{2} qV$$ then the answer is that by definition, kinetic energy is $$\frac{1}{2} mv^2$$ and is equal to the potential energy $$qV$$ stored across a potential difference $$V$$, which is also equal to the kinetic energy the particle would gain once it crosses this potential difference.

Work of moving a charged particle from a point of higher potential voltage to a lower potential voltage in an ekectric or a magnet field or in the cross-field of both is done by the forces produced by those fields And work is equal negative potential energy which means that the potential energy is transferred frommtge field to the particle as kinetic energy , and the change of potential energy is equal to the product of the particle's charge and the change of potential voltage when it is moving from one point to another point in the field W= -( Uf-Ui) , and Uf-Ui = q (Vf-Vi) . Let Vf-Vi = V ---> (Uf-Ui)= qV

Since energy is conserved, Ef-Ei = Uf-Ui + KEf-KEi = 0, Uf+KEf = Ui +KEi. Since before getting in the field its initial velocity Vi = 0--> KEi = 0 After it has passed the field,the electric potential energy becomes its kinetic energy, so now its final electric potential energy Uf = 0, resulting the Eq:

Ui=KEf = qV , thus (mv^2)/2= qV