#### exchemist

**Valued Senior Member**

How do you account for electrostatic potential energy? This is not a function of mass but of charge, of course.Obviously you have not read our work, due to the fact that we use Calculus throughoutThe Yaldon Particle Theory.

When a mass (m)is under a net force that is equal to zero, then the energy will be potential energy.

For example: When a mass is elevated and kept at a height ofymeter(s), it will have potential energy $$(E_p=mgy)$$. In the same way, when a mass travels in empty space with a constant velocity, it will have a net force that is equal to zero (Newton's First Law). Then this mass will have potential energy that is equal to $$E_p=mv^2_c$$; where $$v_c$$ is the constant velocity. When this mass (m) has a variable velocity, then it will have kinetic energy.

We will translate the above statement to Math using Newton's Laws:

For Kinetic Energy:

$$F(v)=\frac{m}{t}v$$

$$[F(v)=\frac{m}{t}v]\cdot dv$$

$$\int\limits_0^{v_{max}}F(v)\cdot dv=\frac{m}{t}\int\limits_0^{v_{max}}v\cdot dv$$

$$Fv_{max}=\frac{1}{2}\frac{m}{t}v^2_{max}$$

$$F[v_{max}\cdot t]=\frac{1}{2}mv^2_{max}$$

Where $$F\cdot s_{max}=E_k$$

$$E_k=\frac{1}{2}mv^2_{max}$$

For Potential Energy:

$$F(s)=m\frac{v}{t}$$

Where $$m\frac{v}{t}$$ is a constant value; then the velocity will also be a constant $$(v_c)$$.

After multiplying both sides of the above equation bydsand taking the limit from zero to $$s_{max}$$:

$$F(s)\int\limits_0^{s_{max}}ds=m\frac{v_c}{t}\int\limits_0^{s_{max}}ds$$

$$F\cdot s_{max}=m\frac{v_c}{t}\cdot s_{max}$$

Where $$F\cdot s_{max}=E_p$$

$$E_p=mv_c\frac{s_{max}}{t}$$

Where $$\frac{s_{max}}{t}=v_c$$, since both are properties of the same mass (m)

$$E_p=mv^2_c$$

When $$E_k=E_p$$ for the same mass (m)

$$\frac{1}{2}mv^2_{max}=mv^2_c$$

Then:

$$\frac{1}{2}v^2_{max}=v^2_c$$Or:

$$v^2_{max}=2v^2_c$$

The above mathematical proof is not needed to explainThe Yaldon Particle Theory, we just want to show the relation between the Laws of Newton and energy. Since the formulas $$E_k=\frac{1}{2}mv^2_{max}$$ and $$E_p=mgy$$ are accepted, then one must accept the Laws of Newton; due to the fact that these formulas have been derived from Newton's Laws. As one can see, we don't need a "rest mass" versus "relative mass" philosophy.

As one can see, when a mass moves with a constant velocity; it will have potential energy.The Yaldon Particle Theoryhas arrived at this conclusion; due to the correct interpretation of Newton's Laws.