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Point charge
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A point charge is an idealized model in physics where a charge is concentrated at a single point with no dimensions. It generates an electric field that radiates uniformly outward, decreasing in strength with the square of the distance.
ID:(2074, 0)
Particle in electric field of point charge
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 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, 0)
Particle in electric potencial of point charge
Concept
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, 0)
Model
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Equations
$ E_{p1} =\displaystyle\frac{1}{4 \pi \epsilon_0 \epsilon }\displaystyle\frac{Q}{ r_1 ^2}$
E_p = Q /(4 * pi * epsilon_0 * epsilon * r ^2)
$ E_{p2} =\displaystyle\frac{1}{4 \pi \epsilon_0 \epsilon }\displaystyle\frac{Q}{ r_2 ^2}$
E_p = Q /(4 * pi * epsilon_0 * epsilon * r ^2)
$ \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{ Q }{4 \pi \epsilon_0 \epsilon }\displaystyle\frac{1}{ r_1 } $
phi_p = - Q /(4 * pi * epsilon * epsilon_0 * r )
$ \varphi_2 = -\displaystyle\frac{ Q }{4 \pi \epsilon_0 \epsilon }\displaystyle\frac{1}{ r_2 } $
phi_p = - Q /(4 * pi * epsilon * epsilon_0 * r )
ID:(15803, 0)
Point charge (1)
Equation
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_{p1} =\displaystyle\frac{1}{4 \pi \epsilon_0 \epsilon }\displaystyle\frac{Q}{ r_1 ^2}$ |
| $ E_p =\displaystyle\frac{1}{4 \pi \epsilon_0 \epsilon }\displaystyle\frac{Q}{ r ^2}$ |
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:(11442, 1)
Point charge (2)
Equation
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_{p2} =\displaystyle\frac{1}{4 \pi \epsilon_0 \epsilon }\displaystyle\frac{Q}{ r_2 ^2}$ |
| $ E_p =\displaystyle\frac{1}{4 \pi \epsilon_0 \epsilon }\displaystyle\frac{Q}{ r ^2}$ |
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:(11442, 2)
Electrical potentials, point charge (1)
Equation
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_1 = -\displaystyle\frac{ Q }{4 \pi \epsilon_0 \epsilon }\displaystyle\frac{1}{ r_1 } $ |
| $ \varphi_p = -\displaystyle\frac{ Q }{4 \pi \epsilon_0 \epsilon }\displaystyle\frac{1}{ r } $ |
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 } $ |
ID:(11576, 1)
Electrical potentials, point charge (2)
Equation
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_2 = -\displaystyle\frac{ Q }{4 \pi \epsilon_0 \epsilon }\displaystyle\frac{1}{ r_2 } $ |
| $ \varphi_p = -\displaystyle\frac{ Q }{4 \pi \epsilon_0 \epsilon }\displaystyle\frac{1}{ r } $ |
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 } $ |
ID:(11576, 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)