The electric field

 

Fields

imageA field is a way of explaining action at a distance.  Massive particles attract each other.  How do they do this, if they are not in contact with each other?  We say that massive particles produce gravitational fields.  A field is a condition in space that can be probed.  Massive particle are also the probes that detect a gravitational field.  A gravitational field exerts a force on massive particles.  The magnitude of the gravitational field produced by a massive object at a point P is the gravitational force per unit mass it exerts on another massive object located at that point.  The direction of the gravitational field is the direction of that force

The magnitude of the gravitational force near the surface of Earth is F = mg, the gravitational field has magnitude F/m = g.  Its direction is downward.


Charged particles attract or repel each other, even when not in contact with each other.  We say that charged particles produce electric fields.  Charged particles are also the probes that detect an electric field.  An electric field exerts a force on charged particles.  The magnitude of the electric field E produced by a charged particle at a point P is the electric force per unit positive charge it exerts on another charged particle located at that point.  The direction of the electric field is the direction of that force on a positive charge.  The actual force on a particle with charge q is given by F = qE.  It points in the opposite direction of the electric field E for a negative charge.

In the presence of many other charges, a charge q is acted on by a net force F, which is the vector sum of the forces due to all the other charges.  The electric field due to all the other charges at the position of the charge q is E = F/q, i.e. it is the vector sum of the electric fields produce by all the other charges.  To measure the electric field E at a point P due to a collection of charges, we can bring a small positive charge q to the point P and measure the force on this test charge.  The test charge must be small, because it interacts with the other charges, and we want this interaction to be small.  We divide the force on the test charge by the magnitude of the test charge to obtain the field.

imageConsider a point charge Q located at the origin.
The force on a test charge q at position r is F = (keQq/r2) (r/r).
The electric field produced by Q is  E = F/q = (keQ/r2) (r/r).

If Q is positive, then the electric field points radially away from the charge.
The electric field decreases with distance as 1/(distance)2.

imageIf Q is negative, then the electric field points radially towards the charge.
The electric field decreases with distance as 1/(distance)2.

 

We obtain the electric field due to a collection of charges using the principle of superposition.
E
= E(Q1) + E(Q2) + E(Q3) + ... .


Field lines

Field lines were introduced by Michael Faraday to help visualize the direction and magnitude of he electric field.  The direction of the field at any point is given by the direction of a line tangent to the field line, while the magnitude of the field is given qualitatively by the density of field lines.  The field lines converge at the position of a point charge.  Near a point charge their density becomes very large.  The magnitude of the field and the density of the field lines scale as the inverse of the distance squared.  

Field lines start on positive charges and end on negative charges.

Rules for drawing field lines:

imageExamples:

The field lines of an electric dipole, i.e. a positive and a negative charge of equal magnitude, separated by a distance d.
The electric field decreases with distance as 1/(distance)3, much faster than the field of a point charge.

imageThe field lines of two positive charges of equal magnitude separated by a distance d.
The electric field decreases with distance as 1/(distance)2.

Polar molecules do not have a net charge, but the centers of the positive and negative charge do not coincide.  Such molecules produce a dipole field and interact via the electrostatic force with their neighbors.

Example:

imageEvery water molecule (H2O) consists of one oxygen atom and two hydrogen atoms.  A water molecule has no net charge because the number of positively charged protons equals the number of negatively charged electrons in each atom.  In the water molecule, each hydrogen atom is bound to the oxygen atom by a covalent bond.  The hydrogen atom and the oxygen atom share an electron, which has a slightly higher probability to be closer to the oxygen atom than to the hydrogen atom.  This results in a polar molecule, in which the oxygen "side" of the molecule has a slight negative charge, while the hydrogen "sides" have a slight positive charge.  Because water is a polar molecule it dissolves many inorganic and organic materials.  Organisms can build macromolecules to attract or repel water as needed simply by varying the charge on side chains.

imageProblem:

The figure on the right shows the electric field lines for a system of two point charges.
(a) What are the relative magnitudes of the charges?
(b) What are the signs of the charges?

Solution:
(a) There are 32 lines coming from the charge on the left, while there are 8 converging on that on the right.  Thus, the magnitude of the charge on the left is 4 times larger than the magnitude of the charge on the right.
(b) The charge on the left is positive; the charge on the right is negative.


Forces and fields due to continuous charge distributions

Even though charge is quantized, we often can treat it as being continuously distributed inside some volume, since one quantum of charge is a tiny amount of charge.  On a macroscopic scale we define the volume charge density ρ = limΔV-->0(ΔQ/ΔV) as the charge per unit volume, the surface charge density σ = limΔA-->0(ΔQ/ΔA) as the charge per unit area, and the line charge density λ = limΔL-->0(ΔQ/ΔL) as the charge per unit length.

We then have for the electric field of a distribution of charges in a volume V

E(r) =  [1/(4πε0)]∑iqi(r - ri)/|r - ri|3] --> [1/(4πε0)][∫v' dV' ρ(r')(r - r')/|r - r'|3.

Here (r - ri)/|r - ri| is the unit vector pointing from ri to r, and (r - r')/|r - r'| is a unit vector pointing from the volume element dV' at r' to r.

If the charge is distributed over a surface, then ρdV --> σdA, where σ is the surface charge density and dA is an element of surface area.  For a line charge distribution we have ρdV --> λdl, where λ is the line charge density and dl is an element length.

Problem:

Consider a line charge with line charge density λ = Q/2a that extends along the x-axis from x = -a to x = +a.  Find the electric field on the y-axis.

Solution:
We are asked to find the electric field due to a line charge distribution.
image
The field on the y-axis due to an infinitesimal element of charge λdx is given by
dEx = (keλdx/(x2 + y2)) cosθ,   dEy = (keλdx/(x2 + y2)) sinθ,
where cosθ = x/(x2 + y2)½, sinθ = y/(x2 + y2)½.

On the y-axis the field due to the line charge therefore is given by
Ex = keλ∫-aaxdx/(x2 + y2)3/2 = 0 from symmetry, and
Ey = keλy∫-aadx/(x2 + y2)3/2.
Using ∫-aadx/(x2 + y2)3/2 = 2a/(y2(a2 + y2)½)
we have  Ey = keλ 2a/(y(a2 + y2)½) = keQ/(y(a2 + y2)½)

The electric field on the y-axis points in the positive y-direction for y > 0 and in the negative y-direction for y < 0.
As y becomes very large, we can neglect a2 compared to y2 under the square root and then Ey = keQ/y2 ∝ 1/y2.  From very far away, the line charge looks like a point charge.
If, on the other hand, the line is very long, and we y << a, then we can neglect y2 compared to a2 under the square root and then Ey = keQ/(ay) ∝ 1/y.  Near a very long line charge the electric field falls off as 1/distance, not as 1/distance2.


Simulation:  Electric Field Hockey  (A homework problem refers to this simulation.)


Link to other web material:  The Physics Classroom: Static Electricity