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The surface electric field i ($\vec{E}_i$) and the versor normal to surface i ($\hat{n}_i$), multiplied by the surface element i ($dS_i$) for each element $i$, which is then summed over the entire section, is equal to the total charge ($Q_t$) divided by the electric field constant ($\epsilon_0$) and the dielectric constant ($\epsilon$):
| $ \displaystyle\sum_i \vec{E}_i \cdot \hat{n}_i dS_i = \displaystyle\frac{ Q }{ \epsilon_0 \epsilon }$ |
Using the surface element ($dS$) for the dot product of the electric field ($\vec{E}$) and the versor normal to the section ($\hat{n}$), we obtain the continuous version of Gauss's law:
| $\displaystyle\int_S\vec{E}\cdot\hat{n}\,dS=\displaystyle\frac{Q}{\epsilon_0\epsilon}$ |
(ID 3213)
The force ($\vec{F}$) on the test charge ($q$) at the position ($\vec{r}$) will depend on the number of charges ($N$), indexed by $i$ and represented by the charge of the ion i ($Q_i$) located at the position of a charge i ($\vec{u}_i$). With the parameters the dielectric constant ($\epsilon$) and the electric field constant ($\epsilon_0$), this can be written as:
| $ \vec{F} =\displaystyle\frac{1}{4 \pi \epsilon_0 \epsilon }\sum_i^N\displaystyle\frac{ q Q_i }{| \vec{r} - \vec{u}_i |^3}( \vec{r} - \vec{u}_i )$ |
With the definition of the electric field ($\vec{E}$) given by
| $ \vec{E} =\lim_{q\rightarrow 0}\displaystyle\frac{ \vec{F} }{ q }$ |
it follows that the electric field of a charge distribution is
| $ \vec{E} =\displaystyle\frac{1}{4 \pi \epsilon_0 \epsilon }\sum_ i ^ N \displaystyle\frac{ Q_i }{| \vec{r} - \vec{u}_i |^3}( \vec{r} - \vec{u}_i )$ |
(ID 3726)
None
(ID 3842)
The potential difference ($\Delta\varphi$) is equal to the sum of the electric field ($\vec{E}$) along an integrated path over the path element traveled ($d\vec{s}$):
| $ \Delta\varphi = -\displaystyle\int_C \vec{E}\cdot d\vec{s} $ |
As the potential difference ($\Delta\varphi$) is calculated by considering the electric potential ($\varphi$) minus the base electrical potential ($\varphi_0$):
| $ \Delta\varphi = \varphi - \varphi_0 $ |
therefore
| $ \varphi =\varphi_0 - \displaystyle\int_C \vec{E}\cdot d\vec{s}$ |
(ID 3844)
None
(ID 3863)
None
(ID 3872)
Examples
The electric potential ($\varphi$) can be calculated from the base electrical potential ($\varphi_0$) and the electric field ($\vec{E}$) integrated along a path over the path element traveled ($d\vec{s}$):
| $ \varphi =\varphi_0 - \displaystyle\int_C \vec{E}\cdot d\vec{s}$ |
(ID 3844)
The magnitude of the force with constant mass ($F$) generated between two charges, represented by the test charge ($q$) and the charge ($Q$), which are at a distance of the distance ($r$), is calculated using the electric field constant ($\epsilon_0$) and the dielectric constant ($\epsilon$) as follows:
| $ F =\displaystyle\frac{1}{4 \pi \epsilon_0 \epsilon }\displaystyle\frac{ q Q }{ r ^2}$ |
(ID 3212)
The force ($\vec{F}$) for the test charge ($q$) is defined as the electric field ($\vec{E}$), which is expressed as:
| $ \vec{E} =\lim_{q\rightarrow 0}\displaystyle\frac{ \vec{F} }{ q }$ |
(ID 3724)
The di-polar moment depends on the
| $ P = r Q $ |
where
(ID 3863)
A dipole generates a potential that at a distance
| $ \varphi = \displaystyle\frac{1}{4 \pi \epsilon_0 \epsilon }\displaystyle\frac{ P \cos \theta }{ r ^2}$ |
where
(ID 3862)
The electric field ($\vec{E}$) at the position ($\vec{r}$) will depend on the number of charges ($N$), accounted for with the index $i$ represented by the charge of the ion i ($Q_i$) located at the position of a charge i ($\vec{u}_i$). With the parameters the dielectric constant ($\epsilon$) and the electric field constant ($\epsilon_0$), this can be written as:
| $ \vec{E} =\displaystyle\frac{1}{4 \pi \epsilon_0 \epsilon }\sum_ i ^ N \displaystyle\frac{ Q_i }{| \vec{r} - \vec{u}_i |^3}( \vec{r} - \vec{u}_i )$ |
(ID 3726)
Si
| $ F =\displaystyle\frac{1}{4 \pi \epsilon_0 \epsilon }\displaystyle\frac{ q Q }{ r ^2}$ |
que introducida en la definici n de campo el ctrico
| $ \vec{E} =\lim_{q\rightarrow 0}\displaystyle\frac{ \vec{F} }{ q }$ |
se obtiene con charge $C$, electric field $V/m$ and force $N$
| $ E_p =\displaystyle\frac{1}{4 \pi \epsilon_0 \epsilon }\displaystyle\frac{ Q }{ r ^2}$ |
(ID 3725)
En el caso de una geometr a esf rica, el camino en la integral del camino es con base electrical potential $V$, electric field $V/m$, electric potential $V$ and infinitesimal distance $m$
| $ \varphi =\varphi_0 - \displaystyle\int_C \vec{E}\cdot d\vec{s}$ |
el campo es inversamente proporcional al radio al cuadrado
$E_f\propto\displaystyle\frac{1}{r^2}$
por lo que el camino mas simple es el radial. Sin embargo el potencial de referencia no puede ser en el origen ya que en dicho punto el integral es infinita. Por ello el potencial de referencia debe ser referida al radio infinito (
| $ \varphi_f = -\displaystyle\int_r^{\infty} du\,E_f(u)$ |
(ID 3887)
Consider a hollow charge, i.e., a hollow sphere with charges on its surface. In this case, we can define an internal surface within the sphere. Since the amount of charge the total charge ($Q_t$) contained within the volume is zero, the electric field the electric eield ($E$) will also be zero:
| $ E =0$ |
(ID 3842)
Once the electric eield ($E$) is known, the force with constant mass ($F$), which acts on the charge ($q$), can be calculated using:
| $ F = q E $ |
(ID 3872)
Using the surface element ($dS$) for the dot product of the electric field ($\vec{E}$) and the versor normal to the section ($\hat{n}$), and the total charge ($Q_t$) divided by the electric field constant ($\epsilon_0$) and the dielectric constant ($\epsilon$), we arrive at the expression for Gauss's law:
| $\displaystyle\int_S\vec{E}\cdot\hat{n}\,dS=\displaystyle\frac{Q}{\epsilon_0\epsilon}$ |
(ID 3213)
The number of electrons ($n_e$) can be determined from the charge of all electrons ($Q_e$) divided by the electron Charge ($e$), resulting in:
| $ n_e =\displaystyle\frac{ Q_e }{ e }$ |
(ID 3211)
ID:(332, 0)