(ID 15793)

In the case of a spherical Gaussian surface, the electric field ($\vec{E}$) is constant in the direction of the versor normal to the section ($\hat{n}$). Therefore, using the charge ($Q$), the electric field constant ($\epsilon_0$), and the dielectric constant ($\epsilon$), it can be calculated by integrating over the surface where the electric field is constant ($dS$):

$\displaystyle\int_S\vec{E}\cdot\hat{n}\,dS=\displaystyle\frac{Q}{\epsilon_0\epsilon}$

with the surface ($S$) for a sphere of radius a distance between charges ($r$):

$ S = 4 \pi r ^2$

Thus, the electric field of a point charge ($E_p$) results in:

$ E_p =\displaystyle\frac{1}{4 \pi \epsilon_0 \epsilon }\displaystyle\frac{Q}{ r ^2}$

(ID 11835)

The electric potential, point charge ($\varphi_p$) is calculated from the radial integration of the electric field of a point charge ($E_p$) from the radius ($r$) to infinity, which results in

$ \varphi_p = -\displaystyle\int_r^{\infty} du\,E_p$

On the other hand, for the charge ($Q$), the dielectric constant ($\epsilon$), and the electric field constant ($\epsilon_0$), the value of the electric field of a point charge ($E_p$) is

$ E_p =\displaystyle\frac{1}{4 \pi \epsilon_0 \epsilon }\displaystyle\frac{Q}{ r ^2}$

This implies that by integrating

$\varphi_p = -\displaystyle\int_{r}^{\infty} du \displaystyle\frac{ Q }{4 \pi \epsilon_0 \epsilon u^2 }= -\displaystyle\frac{ Q }{ 4 \pi \epsilon_0 \epsilon }\displaystyle\frac{1}{ r }$

we obtain

$ \varphi_p = -\displaystyle\frac{ Q }{4 \pi \epsilon_0 \epsilon }\displaystyle\frac{1}{ r } $

As illustrated in the following graph:

the field at two points must have the same energy. Therefore, the variables the charge ($Q$), the particle mass ($m$), the speed 1 ($v_1$), the speed 2 ($v_2$), and the electric potential 1 ($\varphi_1$) according to the equation:

$ \varphi_1 = -\displaystyle\frac{ Q }{4 \pi \epsilon_0 \epsilon }\displaystyle\frac{1}{ r_1 } $

and the electric potential 2 ($\varphi_2$), according to the equation:

$ \varphi_2 = -\displaystyle\frac{ Q }{4 \pi \epsilon_0 \epsilon }\displaystyle\frac{1}{ r_2 } $

must satisfy the following relationship:

$ \displaystyle\frac{1}{2} m v_1 ^2 + q \varphi_1 = \displaystyle\frac{1}{2} m v_2 ^2 + q \varphi_2 $

(ID 11842)

(ID 15803)

The electric field of a point charge ($E_p$) is a function of the charge ($Q$), the electric field constant ($\epsilon_0$), the dielectric constant ($\epsilon$), and the distance between charges ($r$) and is calculated as follows:

$ E_p =\displaystyle\frac{1}{4 \pi \epsilon_0 \epsilon }\displaystyle\frac{Q}{ r ^2}$

(ID 11442)

The electric field of a point charge ($E_p$) is a function of the charge ($Q$), the electric field constant ($\epsilon_0$), the dielectric constant ($\epsilon$), and the distance between charges ($r$) and is calculated as follows:

$ E_p =\displaystyle\frac{1}{4 \pi \epsilon_0 \epsilon }\displaystyle\frac{Q}{ r ^2}$

(ID 11442)

The electric potential, point charge ($\varphi_p$) is with the charge ($Q$), the distance between charges ($r$), the dielectric constant ($\epsilon$) and the electric field constant ($\epsilon_0$) equal to:

$ \varphi_p = -\displaystyle\frac{ Q }{4 \pi \epsilon_0 \epsilon }\displaystyle\frac{1}{ r } $

(ID 11576)

The electric potential, point charge ($\varphi_p$) is with the charge ($Q$), the distance between charges ($r$), the dielectric constant ($\epsilon$) and the electric field constant ($\epsilon_0$) equal to:

$ \varphi_p = -\displaystyle\frac{ Q }{4 \pi \epsilon_0 \epsilon }\displaystyle\frac{1}{ r } $

(ID 11576)

Electric potentials, which represent potential energy per unit of charge, influence how the velocity of a particle varies. Consequently, due to the conservation of energy between two points, it follows that in the presence of variables the charge ($q$), the particle mass ($m$), the speed 1 ($v_1$), the speed 2 ($v_2$), the electric potential 1 ($\varphi_1$), and the electric potential 2 ($\varphi_2$), the following relationship must be satisfied:

$ \displaystyle\frac{1}{2} m v_1 ^2 + q \varphi_1 = \displaystyle\frac{1}{2} m v_2 ^2 + q \varphi_2 $

(ID 11596)