Wire
Storyboard
The geometry referred to as a wire can be understood as an infinitely long cylinder where the distance to the axis is much greater than the radius of the cylinder. Essentially, this corresponds to a case where the radius approaches zero, effectively becoming an infinitely thin line of charge.
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Particle in electric field of a wire
Concept
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, 0)
Particle in electric potencial of a wire
Concept
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 $ |
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Model
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Parameters
Variables
Calculations
Calculations
Calculations
Equations
$ E_w =\displaystyle\frac{1}{2 \pi \epsilon_0 \epsilon }\displaystyle\frac{ \lambda }{ r }$
E_w = lambda /(2 * pi * epsilon_0 * epsilon * r )
$ \lambda = \displaystyle\frac{ Q }{ L }$
lambda = Q / L
$ \displaystyle\frac{1}{2} m v_1 ^2 + Q \varphi_1 = \displaystyle\frac{1}{2} m v_2 ^2 + Q \varphi_2 $
m * v_1 ^2/2 + Q * phi_1 = m * v_2 ^2/2 + Q * phi_2
$ \varphi_w = -\displaystyle\frac{ \lambda }{ 2 \pi \epsilon_0 \epsilon }\ln\left(\displaystyle\frac{ r }{ r_0 }\right)$
phi_w = - lambda * log( r / r_0 )/(2 * pi * epsilon * epsilon_0 )
$ \varphi_1 = -\displaystyle\frac{ \lambda }{ 2 \pi \epsilon_0 \epsilon }\ln\left(\displaystyle\frac{ r_1 }{ r_0 }\right)$
phi_w = - lambda * log( r / r_0 )/(2 * pi * epsilon * epsilon_0 )
$ \varphi_2 = -\displaystyle\frac{ \lambda }{ 2 \pi \epsilon_0 \epsilon }\ln\left(\displaystyle\frac{ r_2 }{ r_0 }\right)$
phi_w = - lambda * log( r / r_0 )/(2 * pi * epsilon * epsilon_0 )
$ \varphi_w = -\displaystyle\int_{r_0}^r du E_w$
phi_w = -@INT( E_w , u , r_0 , r )
ID:(15802, 0)
Infinite wire
Equation
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 }$ |
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 $ |
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:(11444, 0)
Linear load density
Equation
The linear charge density ($\lambda$) is calculated as the charge ($Q$) divided by the conductor length ($L$):
$ \lambda = \displaystyle\frac{ Q }{ L }$ |
ID:(11459, 0)
Electric potential of a wire
Equation
The electric potential, infinite wire ($\varphi_w$) is with the electric field of an infinite wire ($E_w$), the reference radius ($r_0$) and the radius ($r$) is equal to:
$ \varphi_w = -\displaystyle\int_{r_0}^r du E_w$ |
The electric potential ($\varphi$), in the context of cylindrical geometry, is equal to the base electrical potential ($\varphi_0$) plus the integral along the path of the electric field ($\vec{E}$), taking the dot product with the infinitesimal distance ($ds$):
$ \varphi =\varphi_0 - \displaystyle\int_C \vec{E}\cdot d\vec{s}$ |
the electric field of an infinite wire ($E_w$) is inversely proportional to the radius ($r$):
$E_w\propto\displaystyle\frac{1}{r}$
Therefore, the simplest path is radial. However, the base electrical potential ($\varphi_0$) cannot be at the origin or at infinity, as the integral diverges at both points. Therefore, the base electrical potential ($\varphi_0$) must be referenced to a radius where the potential is zero, which in this case is the reference radius ($r_0$):
$ \varphi_w = -\displaystyle\int_{r_0}^r du E_w$ |
ID:(15814, 0)
Calculation of electric potential of a wire
Equation
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)$ |
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)$ |
ID:(15813, 0)
Calculation of electric potential of a wire (1)
Equation
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_1 = -\displaystyle\frac{ \lambda }{ 2 \pi \epsilon_0 \epsilon }\ln\left(\displaystyle\frac{ r_1 }{ r_0 }\right)$ |
$ \varphi_w = -\displaystyle\frac{ \lambda }{ 2 \pi \epsilon_0 \epsilon }\ln\left(\displaystyle\frac{ r }{ r_0 }\right)$ |
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)$ |
ID:(15813, 1)
Calculation of electric potential of a wire (2)
Equation
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_2 = -\displaystyle\frac{ \lambda }{ 2 \pi \epsilon_0 \epsilon }\ln\left(\displaystyle\frac{ r_2 }{ r_0 }\right)$ |
$ \varphi_w = -\displaystyle\frac{ \lambda }{ 2 \pi \epsilon_0 \epsilon }\ln\left(\displaystyle\frac{ r }{ r_0 }\right)$ |
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)$ |
ID:(15813, 2)
Energy of a particle
Equation
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, 0)