(ID 15792)

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 cylinder of the axle distance ($r$) and the conductor length ($L$):

$ S =2 \pi r h $

what is shown in the graph

and the linear charge density ($\lambda$) calculated with the charge ($Q$):

$ \lambda = \displaystyle\frac{ Q }{ L }$

Thus,

$ E_w =\displaystyle\frac{1}{2 \pi \epsilon_0 \epsilon }\displaystyle\frac{ \lambda }{ r }$

(ID 11837)

The electric potential, infinite wire ($\varphi_w$) is derived from the radial integration of the electric field of an infinite wire ($E_w$) from the reference radius ($r_0$) to the axle distance ($r$), resulting in the following equation:

$ \varphi_w = -\displaystyle\int_{r_0}^r du E_w$

Furthermore, for the variables the charge ($Q$), the dielectric constant ($\epsilon$), and the electric field constant ($\epsilon_0$), the value of the electric field of an infinite wire ($E_w$) is given as:

$ E_w =\displaystyle\frac{1}{2 \pi \epsilon_0 \epsilon }\displaystyle\frac{ \lambda }{ r }$

This implies that by performing the integration

$\varphi_w = -\displaystyle\int_{r_0}^r du \displaystyle\frac{ \lambda }{ 2 \pi \epsilon_0 \epsilon u }= -\displaystyle\frac{ \lambda }{ 2 \pi \epsilon_0 \epsilon } \ln\left(\displaystyle\frac{ r }{ r_0 }\right)$

the following equation is obtained:

$ \varphi_w = -\displaystyle\frac{ \lambda }{ 2 \pi \epsilon_0 \epsilon }\ln\left(\displaystyle\frac{ r }{ r_0 }\right)$

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{ \lambda }{ 2 \pi \epsilon_0 \epsilon }\ln\left(\displaystyle\frac{ r_1 }{ r_0 }\right)$

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

$ \varphi_2 = -\displaystyle\frac{ \lambda }{ 2 \pi \epsilon_0 \epsilon }\ln\left(\displaystyle\frac{ r_2 }{ r_0 }\right)$

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 11844)

(ID 15802)

The electric field of an infinite wire ($E_w$) is a function of the linear charge density ($\lambda$), the axle distance ($r$), the dielectric constant ($\epsilon$) and the electric field constant ($\epsilon_0$) and is calculated through:

$ E_w =\displaystyle\frac{1}{2 \pi \epsilon_0 \epsilon }\displaystyle\frac{ \lambda }{ r }$

(ID 11444)

The linear charge density ($\lambda$) is calculated as the charge ($Q$) divided by the conductor length ($L$):

$ \lambda = \displaystyle\frac{ Q }{ L }$

(ID 11459)

The electric field of an infinite wire ($E_w$) is a function of the linear charge density ($\lambda$), the axle distance ($r$), the dielectric constant ($\epsilon$) and the electric field constant ($\epsilon_0$) and is calculated through:

$ E_w =\displaystyle\frac{1}{2 \pi \epsilon_0 \epsilon }\displaystyle\frac{ \lambda }{ r }$

(ID 11444)

The electric potential, infinite wire ($\varphi_w$) is with the pi ($\pi$), the electric field constant ($\epsilon_0$), the dielectric constant ($\epsilon$), the linear charge density ($\lambda$), the axle distance ($r$) and the reference radius ($r_0$) is equal to:

$ \varphi_w = -\displaystyle\frac{ \lambda }{ 2 \pi \epsilon_0 \epsilon }\ln\left(\displaystyle\frac{ r }{ r_0 }\right)$

(ID 15813)

The electric potential, infinite wire ($\varphi_w$) is with the pi ($\pi$), the electric field constant ($\epsilon_0$), the dielectric constant ($\epsilon$), the linear charge density ($\lambda$), the axle distance ($r$) and the reference radius ($r_0$) is equal to:

$ \varphi_w = -\displaystyle\frac{ \lambda }{ 2 \pi \epsilon_0 \epsilon }\ln\left(\displaystyle\frac{ r }{ r_0 }\right)$

(ID 15813)

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)