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Wire
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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.
ID:(2073, 0)
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 $ |
ID:(11844, 0)
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Equations
$ E_{w1} =\displaystyle\frac{1}{2 \pi \epsilon_0 \epsilon }\displaystyle\frac{ \lambda }{ r_1 }$
E_w = lambda /(2 * pi * epsilon_0 * epsilon * r )
$ E_{w2} =\displaystyle\frac{1}{2 \pi \epsilon_0 \epsilon }\displaystyle\frac{ \lambda }{ r_2 }$
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_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 )
ID:(15802, 0)
Infinite wire (1)
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_{w1} =\displaystyle\frac{1}{2 \pi \epsilon_0 \epsilon }\displaystyle\frac{ \lambda }{ r_1 }$ |
| $ 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, 1)
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)
Infinite wire (2)
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_{w2} =\displaystyle\frac{1}{2 \pi \epsilon_0 \epsilon }\displaystyle\frac{ \lambda }{ r_2 }$ |
| $ 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, 2)
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)