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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_i =\displaystyle\frac{1}{4 \pi \epsilon_0 \epsilon }\displaystyle\frac{ Q r }{ 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_i = -\displaystyle\int_0^ r du\, E_i $
phi_i = -@INT( E_i , u ,0, r )
$ \varphi_i = -\displaystyle\frac{ Q }{8 \pi \epsilon_0 \epsilon }\displaystyle\frac{ r ^2 }{ R ^3 }$
phi_i =- Q * r ^2/(8 * pi * epsilon * epsilon_0 * R ^3)
$ \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 =\displaystyle\frac{ r ^3}{ R ^3} Q $
q = r ^3* Q / R ^3
ID:(15806, 0)
Charge fraction
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 =\displaystyle\frac{ r ^3}{ R ^3} Q $ |
ID:(11461, 0)
Insulating sphere with full volume charge, interior
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_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, 0)
Electric potential with spherical geometry, internal
Equation
The reference electrical, insulating sphere, inner ($\varphi_i$) is with the electric field, sphere, interior ($E_i$) and the radius ($r$) is equal to:
| $ \varphi_i = -\displaystyle\int_0^ r du\, E_i $ |
In the case of a spherical geometry of an insulator, the path of the integral is the base electrical potential ($\varphi_0$) with the electric potential ($\varphi$), the infinitesimal distance ($ds$), and the electric field ($\vec{E}$), resulting in:
| $ \varphi =\varphi_0 - \displaystyle\int_C \vec{E}\cdot d\vec{s}$ |
The field is proportional to the radius
$E_i\propto r$
so the simplest path is radial. The reference potential can be at the origin since the integral is finite at that point. Therefore, the reference potential should be referred to zero radius ($r\rightarrow 0$) and can be chosen as zero ($\varphi_0=0$). With this, along with the reference electrical, insulating sphere, inner ($\varphi_i$), the electric field, sphere, interior ($E_i$), and the radius ($r$), the result is:
| $ \varphi_i = -\displaystyle\int_0^ r du\, E_i $ |
ID:(11579, 0)
Calculation of electric potential with spherical geometry, internal
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_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, 0)
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)