# Jump conditions for electric field across an interface

## Main Question or Discussion Point

Hi, I have some confusion about the jump conditions for an electric field across an interface between two materials with different properties. In general, we have the two jump conditions across an interface:
n.(ɛE)+ - n.(ɛE)- = σ (Normal direction) ; where σ is the surface charge density on the interface
n x E+ - n x E- = 0 (Tangential direction)

Here, + and - subscripts denote the properties outside and inside the interface respectively.

If we define an electric potential V, then electric field E = grad(V) where V satisfies the equation:
where ρ is the volume density of charge

Now, if we use the boundary conditions shown above, we will have two equations for V at the interface, which will overconstrain the system of equations. (We only need one boundary condition at the interface for a scalar quantity). I have seen that people normally use only the jump condition for the normal direction. Does it mean that if the jump condition in the normal direction for the potential V is satisfied, then the tangential boundary condition will be automatically fulfilled? Can anyone provide a logical explanation?

Thanks!

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When the problem simply involves an electric field vector $E$ that is normal to the surface, the other condition is automatically satisfied, because both $E$ vectors that are parallel to the surface are zero. $\\$ For transverse electromagnetic waves incident at normal incidence, only the second condition is necessary. In deriving the Fresnel relations for reflection and transmission, for arbitrary angles of incidence, I believe both boundary conditions are employed. $\\$ And note: Usually $\vec{E}=-\nabla V$ , with a minus sign. This equation only holds in the static case, incidentally.

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When the problem simply involves an electric field vector $E$ that is normal to the surface, the other condition is automatically satisfied, because both $E$ vectors that are parallel to the surface are zero. $\\$ For transverse electromagnetic waves incident at normal incidence, only the second condition is necessary. In deriving the Fresnel relations for reflection and transmission, for arbitrary angles of incidence, I believe both boundary conditions are employed. $\\$ And note: Usually $\vec{E}=-\nabla V$ , with a minus sign. This equation only holds in the static case, incidentally.
For the sake of simplicity, let's just discuss about tthe static case, where E = - grad(V). How would you use both the boundary conditions? Only one boundary condition for V is required as it is a scalar quantity. Using both conditions would lead to overconstraining.

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I believe it can be one of the conditions that the potential function must satisfy. (Some of these boundary potential problems can get quite complex and often involve solutions to Poisson's/ Laplace's equation involving Legendre polynomials). In the case of one of the surfaces being a conductor, I have seen a necessary condition is that the electric field parallel to the surface must be zero as one requirement on the potential function.

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To add to the above, a charged conducting sphere of radius $a$, and charge $Q$ will have a very simple form for the potential that can be written down almost immediately: $\\$ $V_{external}(r,\theta,\phi)=\frac{A}{r}$ for $r>a$ , and $V_{internal}(r)=B$ for $r<a$. $\\$ At $r=a$, $V_{internal}(r, \theta, \phi)=V_{external}(r,\theta,\phi)$, but alternatively, at $r=a$, the boundary conditions you have in the OP could be applied. $\\$ We can readily compute the surface charge density $\sigma=\frac{Q}{4 \pi a^2}$. We also have the two boundary conditions for the electric field of the OP above:$\\$ $-(\frac{\partial{V}_{external}}{\partial{r}})|_{r=a}=\frac{\sigma}{\epsilon_o}=\frac{A}{r^2}|_{r=a}=\frac{A}{a^2}$, $\\$ and $\\$ $-(\frac{\partial{V}}{\partial{\theta}})|_{r=a}=0$.$\\$ With these boundary conditions, we see $V_{external}(r,\theta, \phi)=\frac{Q}{4 \pi \epsilon_o r}=\frac{\sigma a^2}{\epsilon_o r}$ for $r>a$, immediately, and can rule out any of the other forms involving other exponents of $r$, and/or involving any of the Legendre polynomials containing higher powers of $\cos{\theta}$. $\\$ See e.g. the equations in the OP of https://www.physicsforums.com/threads/potential-of-sphere-given-potential-of-surface.887477/#post-5582019 $\\$ The electric field boundary conditions make the problem very simple, and the need for more complex solutions can be ruled out immediately. $\\$ Perhaps there is no need here for the electric field boundary conditions, but they did not overconstrain the problem. The tangential boundary condition is really just a statement of the spherical symmetry.

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clem
Each boundary condition follows from the equation ${\bf E}=-\nabla\phi$.
$E_{tan}$ is continuous because ${\bf E}=-\nabla\phi$ gives curl E=0,
and then Stokes' theorem gives $E_{tan}$ continuous.
The fact that the E in ${\bf E}=-\nabla\phi$ is finite means that $\phi$ is continuous.
Since each boundary condition follows directly from the same equation they are not independent, and either one may be used by itself.

• Each boundary condition follows from the equation ${\bf E}=-\nabla\phi$.
$E_{tan}$ is continuous because ${\bf E}=-\nabla\phi$ gives curl E=0,
and then Stokes' theorem gives $E_{tan}$ continuous.
The fact that the E in ${\bf E}=-\nabla\phi$ is finite means that $\phi$ is continuous.