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Interior of an insulating sphere
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In the case of an insulating sphere with a homogeneous charge distribution, the charges cannot move. The electric field can be calculated by assuming spherical symmetry and defining the Gaussian surface as a sphere with a given radius. In this way, the electric field and potential will depend on the charge enclosed by this surface.
ID:(2077, 0)
Particle in electric field of an sphere, internal
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 ($dS$):
| $\displaystyle\int_S\vec{E}\cdot\hat{n}\,dS=\displaystyle\frac{Q}{\epsilon_0\epsilon}$ |
Since the surface area of the surface of a sphere ($S$) is equal to the pi ($\pi$) and the disc radius ($r$), we have:
| $ S = 4 \pi r ^2$ |
what is shown in the graph
the encapsulated charge on Gauss surface ($q$) with a radius equal to the distance between charges ($r$) and the sphere radius ($R$) with the charge ($Q$) so that:
| $ q =\displaystyle\frac{ r ^3}{ R ^3} Q $ |
For the electric field, sphere, interior ($E_i$), the resulting expression is:
| $ E_i =\displaystyle\frac{1}{4 \pi \epsilon_0 \epsilon }\displaystyle\frac{ Q r }{ R ^3 }$ |
ID:(11840, 0)
Particle in electric potencial of an sphere, internal
Concept
As the potential difference is the reference electrical, insulating sphere, inner ($\varphi_i$) with the electric field, sphere, interior ($E_i$) and the radius ($r$), we get:
| $ \varphi_i = -\displaystyle\int_0^ r du\, E_i $ |
Given that the electric field, sphere, interior ($E_i$) with the pi ($\pi$), the charge ($Q$), the electric field constant ($\epsilon_0$), the dielectric constant ($\epsilon$), the sphere radius ($R$), and the distance between charges ($r$) is equal to:
| $ E_i =\displaystyle\frac{1}{4 \pi \epsilon_0 \epsilon }\displaystyle\frac{ Q r }{ R ^3 }$ |
In spherical coordinates, this is:
$\varphi_i = -\displaystyle\int_0^{r} du \displaystyle\frac{ Q u }{4 \pi \epsilon_0 \epsilon R ^3 }= -\displaystyle\frac{ Q }{ 8 \pi \epsilon_0 \epsilon }\displaystyle\frac{ r ^2 }{ R ^3 }$
Therefore, the reference electrical, insulating sphere, inner ($\varphi_i$) with the distance between charges ($r$) results in:
| $ \varphi_i = -\displaystyle\frac{ Q }{8 \pi \epsilon_0 \epsilon }\displaystyle\frac{ r ^2 }{ R ^3 }$ |
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 }{8 \pi \epsilon_0 \epsilon }\displaystyle\frac{ r_1 ^2 }{ R ^3 }$ |
and the electric potential 2 ($\varphi_2$), according to the equation:
| $ \varphi_2 = -\displaystyle\frac{ Q }{8 \pi \epsilon_0 \epsilon }\displaystyle\frac{ r_2 ^2 }{ R ^3 }$ |
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:(11847, 0)
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Equations
$ E_{i1} =\displaystyle\frac{1}{4 \pi \epsilon_0 \epsilon }\displaystyle\frac{ Q r_1 }{ R ^3 }$
E_i = Q * r /(4 * pi * epsilon_0 * epsilon * R ^3)
$ E_{i2} =\displaystyle\frac{1}{4 \pi \epsilon_0 \epsilon }\displaystyle\frac{ Q r_2 }{ R ^3 }$
E_i = Q * r /(4 * pi * epsilon_0 * epsilon * R ^3)
$ \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 }{8 \pi \epsilon_0 \epsilon }\displaystyle\frac{ r_1 ^2 }{ R ^3 }$
phi_i =- Q * r ^2/(8 * pi * epsilon * epsilon_0 * R ^3)
$ \varphi_2 = -\displaystyle\frac{ Q }{8 \pi \epsilon_0 \epsilon }\displaystyle\frac{ r_2 ^2 }{ R ^3 }$
phi_i =- Q * r ^2/(8 * pi * epsilon * epsilon_0 * R ^3)
$ q_1 =\displaystyle\frac{ r_1 ^3}{ R ^3} Q $
q = r ^3* Q / R ^3
$ q_2 =\displaystyle\frac{ r_2 ^3}{ R ^3} Q $
q = r ^3* Q / R ^3
ID:(15806, 0)
Charge fraction (1)
Equation
In the case of a the sphere radius ($R$) sphere with homogeneous charge, the Gaussian surface for the distance between charges ($r$) includes the encapsulated charge on Gauss surface ($q$) for the charge ($Q$) :
| $ q_1 =\displaystyle\frac{ r_1 ^3}{ R ^3} Q $ |
| $ q =\displaystyle\frac{ r ^3}{ R ^3} Q $ |
ID:(11461, 1)
Insulating sphere with full volume charge, interior (1)
Equation
The electric field, sphere, interior ($E_i$) is with the pi ($\pi$), the charge ($Q$), the electric field constant ($\epsilon_0$), the dielectric constant ($\epsilon$), the sphere radius ($R$) and the distance between charges ($r$) is equal to:
| $ E_{i1} =\displaystyle\frac{1}{4 \pi \epsilon_0 \epsilon }\displaystyle\frac{ Q r_1 }{ R ^3 }$ |
| $ E_i =\displaystyle\frac{1}{4 \pi \epsilon_0 \epsilon }\displaystyle\frac{ Q r }{ R ^3 }$ |
Considering a spherical Gaussian surface, the electric field is constant. Therefore, the electric eield ($E$) is equal to the charge ($Q$), the electric field constant ($\epsilon_0$), the dielectric constant ($\epsilon$), and the surface of the conductor ($S$) as shown by:
| $ E S = \displaystyle\frac{ Q }{ \epsilon_0 \epsilon }$ |
Since the surface area of the surface of a sphere ($S$) is equal to the pi ($\pi$) and the disc radius ($r$), we have:
| $ S = 4 \pi r ^2$ |
The charge enclosed by the Gaussian surface, with the encapsulated charge on Gauss surface ($q$), the sphere radius ($R$), and the distance between charges ($r$), is given by:
| $ q =\displaystyle\frac{ r ^3}{ R ^3} Q $ |
Therefore, the electric field, sphere, interior ($E_i$) results in:
| $ E_i =\displaystyle\frac{1}{4 \pi \epsilon_0 \epsilon }\displaystyle\frac{ Q r }{ R ^3 }$ |
ID:(11447, 1)
Charge fraction (2)
Equation
In the case of a the sphere radius ($R$) sphere with homogeneous charge, the Gaussian surface for the distance between charges ($r$) includes the encapsulated charge on Gauss surface ($q$) for the charge ($Q$) :
| $ q_2 =\displaystyle\frac{ r_2 ^3}{ R ^3} Q $ |
| $ q =\displaystyle\frac{ r ^3}{ R ^3} Q $ |
ID:(11461, 2)
Insulating sphere with full volume charge, interior (2)
Equation
The electric field, sphere, interior ($E_i$) is with the pi ($\pi$), the charge ($Q$), the electric field constant ($\epsilon_0$), the dielectric constant ($\epsilon$), the sphere radius ($R$) and the distance between charges ($r$) is equal to:
| $ E_{i2} =\displaystyle\frac{1}{4 \pi \epsilon_0 \epsilon }\displaystyle\frac{ Q r_2 }{ R ^3 }$ |
| $ E_i =\displaystyle\frac{1}{4 \pi \epsilon_0 \epsilon }\displaystyle\frac{ Q r }{ R ^3 }$ |
Considering a spherical Gaussian surface, the electric field is constant. Therefore, the electric eield ($E$) is equal to the charge ($Q$), the electric field constant ($\epsilon_0$), the dielectric constant ($\epsilon$), and the surface of the conductor ($S$) as shown by:
| $ E S = \displaystyle\frac{ Q }{ \epsilon_0 \epsilon }$ |
Since the surface area of the surface of a sphere ($S$) is equal to the pi ($\pi$) and the disc radius ($r$), we have:
| $ S = 4 \pi r ^2$ |
The charge enclosed by the Gaussian surface, with the encapsulated charge on Gauss surface ($q$), the sphere radius ($R$), and the distance between charges ($r$), is given by:
| $ q =\displaystyle\frac{ r ^3}{ R ^3} Q $ |
Therefore, the electric field, sphere, interior ($E_i$) results in:
| $ E_i =\displaystyle\frac{1}{4 \pi \epsilon_0 \epsilon }\displaystyle\frac{ Q r }{ R ^3 }$ |
ID:(11447, 2)
Calculation of electric potential with spherical geometry, internal (1)
Equation
The reference electrical, insulating sphere, inner ($\varphi_i$) is with the pi ($\pi$), the charge ($Q$), the electric field constant ($\epsilon_0$), the dielectric constant ($\epsilon$), the distance between charges ($r$) and the sphere radius ($R$) is equal to:
| $ \varphi_1 = -\displaystyle\frac{ Q }{8 \pi \epsilon_0 \epsilon }\displaystyle\frac{ r_1 ^2 }{ R ^3 }$ |
| $ \varphi_i = -\displaystyle\frac{ Q }{8 \pi \epsilon_0 \epsilon }\displaystyle\frac{ r ^2 }{ R ^3 }$ |
As the potential difference is the reference electrical, insulating sphere, inner ($\varphi_i$) with the electric field, sphere, interior ($E_i$) and the radius ($r$), we get:
| $ \varphi_i = -\displaystyle\int_0^ r du\, E_i $ |
Given that the electric field, sphere, interior ($E_i$) with the pi ($\pi$), the charge ($Q$), the electric field constant ($\epsilon_0$), the dielectric constant ($\epsilon$), the sphere radius ($R$), and the distance between charges ($r$) is equal to:
| $ E_i =\displaystyle\frac{1}{4 \pi \epsilon_0 \epsilon }\displaystyle\frac{ Q r }{ R ^3 }$ |
In spherical coordinates, this is:
$\varphi_i = -\displaystyle\int_0^{r} du \displaystyle\frac{ Q u }{4 \pi \epsilon_0 \epsilon R ^3 }= -\displaystyle\frac{ Q }{ 8 \pi \epsilon_0 \epsilon }\displaystyle\frac{ r ^2 }{ R ^3 }$
Therefore, the reference electrical, insulating sphere, inner ($\varphi_i$) with the distance between charges ($r$) results in:
| $ \varphi_i = -\displaystyle\frac{ Q }{8 \pi \epsilon_0 \epsilon }\displaystyle\frac{ r ^2 }{ R ^3 }$ |
ID:(11583, 1)
Calculation of electric potential with spherical geometry, internal (2)
Equation
The reference electrical, insulating sphere, inner ($\varphi_i$) is with the pi ($\pi$), the charge ($Q$), the electric field constant ($\epsilon_0$), the dielectric constant ($\epsilon$), the distance between charges ($r$) and the sphere radius ($R$) is equal to:
| $ \varphi_2 = -\displaystyle\frac{ Q }{8 \pi \epsilon_0 \epsilon }\displaystyle\frac{ r_2 ^2 }{ R ^3 }$ |
| $ \varphi_i = -\displaystyle\frac{ Q }{8 \pi \epsilon_0 \epsilon }\displaystyle\frac{ r ^2 }{ R ^3 }$ |
As the potential difference is the reference electrical, insulating sphere, inner ($\varphi_i$) with the electric field, sphere, interior ($E_i$) and the radius ($r$), we get:
| $ \varphi_i = -\displaystyle\int_0^ r du\, E_i $ |
Given that the electric field, sphere, interior ($E_i$) with the pi ($\pi$), the charge ($Q$), the electric field constant ($\epsilon_0$), the dielectric constant ($\epsilon$), the sphere radius ($R$), and the distance between charges ($r$) is equal to:
| $ E_i =\displaystyle\frac{1}{4 \pi \epsilon_0 \epsilon }\displaystyle\frac{ Q r }{ R ^3 }$ |
In spherical coordinates, this is:
$\varphi_i = -\displaystyle\int_0^{r} du \displaystyle\frac{ Q u }{4 \pi \epsilon_0 \epsilon R ^3 }= -\displaystyle\frac{ Q }{ 8 \pi \epsilon_0 \epsilon }\displaystyle\frac{ r ^2 }{ R ^3 }$
Therefore, the reference electrical, insulating sphere, inner ($\varphi_i$) with the distance between charges ($r$) results in:
| $ \varphi_i = -\displaystyle\frac{ Q }{8 \pi \epsilon_0 \epsilon }\displaystyle\frac{ r ^2 }{ R ^3 }$ |
ID:(11583, 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)