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The connection between electric potential \(V\), and electric field, \(\vec E\).
Gradient field, \(\vec E=-\nabla V\)
Earlier [1,2,3], we found the following equations $$ \newcommand{oiint}{\subset\!\!\supset\kern-1.65em\iint} \begin{align} \oiint_S \vec E \cdot d\vec A &= \frac{\sum_i Q_i}{\varepsilon_0} \label{eq:gauss} \\[1pt] V_P &= \int_P^\infty \vec E\cdot d\vec r \label{eq:electpotential} \\[1pt] V_A – V_B &= \int_A^B \vec E\cdot d\vec l \label{eq:electpotdiff} \end{align} $$
Work over closed line
If the electric field is a static electric field (no moving charges) \(\Longrightarrow\) the forces are conservative forces.
If I take a charge \(q\) from point \(A\), all around the conservative field, and back to point \(A\), then the work based on equation \(\eqref{eq:electpotdiff}\) is $$ V_A-V_A=\int_A^A \vec E\cdot d\vec l = 0 \nonumber $$
We mark the integral with a circle to indicate that it is a closed path. So moving through static electric field, over a closed line, the work is \(0\) $$ \shaded{ \oint \vec E\cdot d\vec l = 0 } $$
From potentials to electric field
If you look at equation \(\eqref{eq:electpotential}\) and \(\eqref{eq:electpotdiff}\), you see that the potential is the integral of the electric field \(\Longrightarrow\) the electric field must be the derivative of the potential.
Let there be a charge \(Q\) a distance \(r\) from point \(P\)
Recall: we derived the electric potential at that location, the scalar
$$ V = \frac{Q}{4\pi\varepsilon_0 r} \nonumber $$
The derivative of this potential $$ \newcommand{dv}[2]{\frac{d #1}{d #2}} \newcommand{pdv}[2]{\frac{\partial #1}{\partial #2}} \dv{V}{r} = \dv{}{r} \frac{Q}{4\pi\varepsilon_0 r} = -\underline{\frac{Q}{4\pi\varepsilon_0 r^2}} \label{eq:pdvV} $$
Recall: the electric field is defined as the vector
$$ \vec E = \underline{\frac{Q\,\hat r}{4\pi\varepsilon_0 r^2}} \nonumber $$
Substituting \(\eqref{eq:pdvV}\) in this, shows that if you know the potential everywhere in space, you can retrieve the electric field $$ \newcommand{dv}[2]{\frac{d #1}{d #2}} \shaded{ \vec E = -\dv{V}{r}\,\hat r } \nonumber $$
That means that if you release a charge at \(0\) speed, it will always start perpendicular to an equipotential surface, because it always starts to move in the direction of a field line.
Going deeper
- Imagine that I am somewhere in space at position \(P\). At that position the potential is \(V_P\). At that position there is an electric field.
- Now I make a tiny step in the \(x\)-direction. If I measure
- no change in potential, over that little step, it means that the component of the electric field in the direction of \(x\) is \(0\),
- a change in potential, it means that the magnitude of the \(x\)-component of the electric field $$ \newcommand{pdv}[2]{\frac{\partial #1}{\partial #2}} |E_x| = \left|\pdv{V}{x}\right| \nonumber $$ The partial derivative \(\frac{\partial}{\partial x}\), means you keep the other variables \(y,z\) constant.
- Equally for the \(y\) and \(z\)-directions $$ \newcommand{pdv}[2]{\frac{\partial #1}{\partial #2}} |E_y| = \left|\pdv{V}{y}\right|,\quad |E_z| = \left|\pdv{V}{z}\right| \nonumber $$
Until now we used the unit \(\tfrac{\rm N}{\rm C}\) for the electric field, but from now on we will use \(\tfrac{\rm V}{\rm m}\) because it gives a bit more insight. You move a bit in meters, and see how much the potential changes in volts.
Now we can write the relation between electric field and potential in Cartesian coordinates (instead of just a function of distance \(r\)) $$ \newcommand{pdv}[2]{\frac{\partial #1}{\partial #2}} \begin{align*} \vec E &= \left(E_x\hat x+ E_y\hat y+ E_z\hat z\right) \\ &= – \left( \pdv{V}{x}\hat x + \pdv{V}{y}\hat y + \pdv{V}{z}\hat z \right) \\ &= -\left\langle \pdv{V}{x}, \pdv{V}{y},\pdv{V}{z} \right\rangle \end{align*} \nonumber $$
Using the symbolic \(\nabla\)-operator, this is written as $$ \shaded{ \vec E = -\rm{grad}\,V = -\nabla\,V } \tag{gradient} \label{eq:gradient} $$
Note that here in physics the sign is opposite as used in the math notation for a gradient field.
Example: given potentials
We have a potential field \(V(x)=10^5 x\) for \(0\lt x\lt 10^{-2}\rm m\). What is the electric field in space?
The electric field \(\vec E\) follows from equation \(\eqref{eq:gradient}\) $$ \newcommand{pdv}[2]{\frac{\partial #1}{\partial #2}} \begin{align*} \vec E &= -\nabla\,V = -\left\langle\pdv{10^5 x}{x},\pdv{10^5 x}{y},\pdv{10^5 x}{z}\right\rangle \\ &= -\left\langle 10^5,0,0\right\rangle = \underline{-10^5\,\hat x} \end{align*} $$
This is constant and opposite to \(\hat x\).
Example: parallel plates
Continue the charged plates example. If the plates are close together \(d\ll \sqrt{A}\) where \(A\) is the area of a plate, we can approximate the field between these plates as those coming from a pair of infinite planes. We saw that the electric field inside is independent of the distance.
Assume this electric field is the \(\vec E=-10^5\hat x\) from the previous example.
The potential difference follows from equation \(\eqref{eq:electpotdiff}\), where going from \(A\) to \(B\) \(\Rightarrow\) \(d\vec l=d\vec x\) $$ \begin{align*} V_A – V_B &= \int_A^B \vec E\cdot d\vec x = \int_A^B \underline{-10^5\,\hat x}\cdot d\vec x \\ &= -10^5 \int_A^B \hat x\cdot\vec x \end{align*} $$
The unit vector \(\hat x\) is in the same direction as \(d\vec x\), so the dot-product becomes a multiplication of scalars \(|\hat x|\,|d\vec x|\,\cos 0=\)\(1.dx.1\) $$ \begin{align*} V_A – V_B &= -10^5 \int_A^B dx = -10^5\Big[x\Big]_A^B \\ &= -10^5 (x_B – x_A) \end{align*} $$
If plates \(A\) and \(B\) are \(10\,\rm{cm}\) apart $$ V_A – V_B = -10^5\,10^{-2} = -1000\,\rm V \nonumber $$
Note
- \(A\) is \(1000\,\rm V\) lower than \(B\).
- From \(A\) to \(B\) the potential increases linearly.
Instead of choosing the \(0\)-potential at \(\infty\), we could choose it at plate \(A\). Then plate \(B\) would be 1000 V, and the potential \(V\) and the electric field \(\vec E\) become $$ \shaded{ \left\{ \begin{align*} V &= 10^5 x \\ \vec E &= -10^5\,\hat x \end{align*} \right. } \nonumber $$
Electrostatic Shielding
As long as there is no charge moving, and we’re dealing with solid conductors \(\Longrightarrow\) we have static electric fields (the charges are not heavily moving) \(\Longrightarrow\) the field inside the conductor is always \(0\). Because conductors have free electrons, and if these free electrons see an electric field inside, then they start to move until they no longer experience a force \(\Longrightarrow\) they kill the electric field inside.
So the charge in a conductor always rearranges itself so that the electric field becomes zero.
Charging a solid conductor
Suppose we bring a plus charge near a solid conductor.
For a very short moment, there will be an electric field inside the conductor. This field will move the free electrons close to the plus charge (induction), leaving net positive charge behind. Once these charges cancel out the electric field, there is no longer an electric field to move them, so the electrons stay still. [MIT]
So, after this very brief moment, the electric field inside is zero. This implies
- This inside of the solid contains no charge. (Assume a small Gaussian surface inside the conductor, then equation \(\eqref{eq:gauss}\) tell us that this inside surface contains no charge.)
- The entire solid is an equipotential. (Based on equation \(\eqref{eq:electpotdiff}\).)
The charge has to be on the surface. It however is not uniformly distributed with this odd shape (as we see later). With the solid being an equipotential, the field lines are perpendicular to the surface of the solid.
External electric field lines are perpendicular to the surface. Because if a component of the \(\vec E\)-field was tangent to the surface, charges would flow along the surface until there was no longer any tangential component.
Charging a hollow conductor
The same shape, but hollow. Once more, we charge it.
Again, for a very short moment, there will be an electric field inside the that moves the free electrons so they cancel out the “external” electric field. After that, there is no longer an electric field to move them.
So, after this very brief moment, the electric field inside is zero. This implies
- This inside of the shell and cavity contains no charge. (Assume a small Gaussian surface \(S\) just under the surface of the conductor, then equation \(\eqref{eq:gauss}\) tell us that it encloses no charge.)
- The whole shell including this cavity is an equipotential. (Based on equation \(\eqref{eq:electpotdiff}\).)
There is never any electric field anywhere inside the shell or cavity. The only electric field is on the outside of the shell. So, again, the charge has to be on the outside.
Pretty amazing: one side becomes negative, the other side becomes positive; the whole things becomes an equipotential; no charge inside.
Faraday cage
Earlier we saw that a sphere has no electric field inside. Here, we demonstrated that it doesn’t have to be a sphere (and also can be solid). Any conductor shape will give you an electric field of \(0\) inside.
An external electric field (such as from a van der Graaff generator), can cause polarization on the outside of the conductor, but inside, even the inside wall will have no charge (no electric field). After a very brief moment, the free moving electrons rearranged themselves, in such a way that the electric field is zero everywhere inside the conductor.
The inside is electrically shielded from the outside world. You would not notice a strong electric field outside. This is called electrostatic shielding. The object will be called a Faraday cage. Named after the English scientist Michael Faraday, who invented them in 1836
credit: wikipedia.org
A charge inside a conductor
A hollow metallic sphere is initially uncharged. Now imagine that a positive point charge \(q\) is placed somewhere (not in the middle) in the sphere without touching the walls.
Now there is positive charge inside, so there has be an electric field inside. For a very short moment the free electrons in the conducting wall will move so that the electric field in the wall becomes zero.
There must be charge on the inside surface, because if we take a Gaussian surface \(S\) halfway the shell, the electric field is zero at that surface, so the enclosed charge is zero. That means that negative charge must have accumulated on the inside wall of the conductor, to make the net charge inside surface \(S\) is zero.
Originally the conductor had no charge. On the outside surface we now see charge \(q\), because the minus charge on the inside surface came from electrons that moved from the outside surface.
The plus charge inside, creates an electric field inside, that creates a negative charge on the inside, and a plus charge on the outside.
Because of the spherical symmetry, the charge on the outside will be uniformly distributed. That is the only way nature can obey the laws of physics. (The conductor must become an equipotential; there can be no electric field inside the conductor; the electric field lines have to perpendicular to the surface; the \(\newcommand{oiint}{\subset\!\!\supset\kern-1.65em\iint} \oiint_S \vec E \cdot d\vec l\) must be zero everywhere.)
The uniform outside charge is independent of the position of the charge \(q\) inside. The outside world will not notice if you move the charge \(q\) around. It has no way of knowing what happens on the inside. We call that electrostatic shielding, the effect of the Faraday cage.