PHY2061 Enriched Physics 2 Lecture Notes Electric Field

PHY2061 Enriched Physics 2 Lecture Notes Electric Field

Electric Fields

Disclaimer: These lecture notes are not meant to replace the course textbook. The content may be incomplete. Some topics may be unclear. These notes are only meant to be a study aid and a supplement to your own notes. Please report any inaccuracies to the professor.

Fields

How does one charged particle attract or repel another particle
at a distance? Þ Action at a distance

This is more prevalent than you think. Consider how you touch something and cause it to move by applying a force. In fact, you never actually “touch” the object. The electrons in the atoms of your hand start repelling the electrons in the atoms of the object. The electrons never really touch! (they are point particles anyway…) So touching something is Coulomb repulsion at a distance.

In physics, we say that the force exerted by one object onto another a distance away is conveyed through a field.

Electric Field

The electric field is defined as the force acting on a positive test charge, per unit charge.

Units are thus N/C for the electric field.

It is similar to the gravitational field on the surface of the Earth for a test mass m0 :

The electric field is a vector field. It has a magnitude and a direction. Another example vector field is the wind distribution. Temperature distribution is an example of a scalar field.

Field Lines

Consider a positive test charge immersed in the electric field of a positive charge.

·  Lines indicate the direction of if a test charge is placed there

·  Field lines point away from positive charge, and toward negative charge

·  Density of lines indicates the strength of the field

·  Field lines cannot cross. If they did, there would be more than one possible direction for a particle to travel at a given point.

·  The lines themselves have no meaning. The field is present everywhere.

The magnitude of the electric field a distance r away from a point charge q:

i.e. dropped the q0 from Coulomb’s Law

Electric Field of Several Point Charges

Apply the superposition principle. This principle states that the resulting electric field is the sum of all fields, without any interference of one field upon another. It is generally true for electromagnetism at least for fields that are not enormously strong.

For example, the total electric field at some point A from N charges is:

In other words, the fields from each point particle i a distance riA away from point A all add together, without any field affecting another.

Electric Dipole

Let’s consider the simplest multi-charge system: just two oppositely charged particles—a system referred to as an electric dipole. Let’s calculate the electric field anywhere along the x axis (result will actually apply anywhere in the x-z plane for y=0).

y

Let the top particle have charge q,

And the bottom particle charge -q.

The separation of charges is d.

x

By symmetry, must point in the direction.

Apply the superposition principle to determine it, by adding the electric field from each charge separately to get the net electric field.

Consider a point along the x axis. The distance from either charge is

The magnitude of the electric field from each charge separately is

But the vector force points in a direction away from the
positive charge and toward the negative one. i.e.

So we see explicitly how the x-component of the total field cancels, and we are left with:

since

Now,

So,

By the binomial expansion, for small x.

Let’s define the electric dipole moment

where we have generalized from anywhere along the x axis to anywhere in the x-z plane, a distance r away from the dipole.

Note that this field falls off faster than a single point charge, , because the opposite-sign charges contribute to the canceling of the electric field. This is a general behavior for the dipole field strength in any direction. For example, along the y axis, one gets a dipole field 2 times larger than the expression above (but all other terms the same).

Continuous Charge Distributions

Consider the electric field arising from an infinitesimal charge dq at a point a distance r away:

Projecting along each axis yields the following components of the infinitesimal electric field:

Field from a Continuous Line Charge

Now consider electric charge distributed uniformly along a 1-dimensional line from -L to L along the z-axis.

The charge per unit length is l (units: C/m)

In terms of the total charge q,

So an infinitesimal length of the line, dz, has a charge dq = ldz.

Let’s calculate the electric field along the y-axis (any axis perpendicular to the line will do).

By symmetry, only the component of the field (that is, a field pointing perpendicular to the line) can survive when we integrate the field contributions from each infinitesimal charge dq along the length.

Consider , alternatively that

Then:

where y can be replaced by the distance from the line charge to where the field is measured.

We can consider another limit, the one when (i.e. very far away from a finite length line charge). Then:

But since represents the total charge of the line, then we can re-write this as:

That is, when viewed far away, the field is just that due to a point charge.

Field of a Continuous Ring of Charge

Let’s find the field along the z-axis only. Ring has radius R, charge per unit length l.

By symmetry, only Ez is non-zero (the x-y components cancel)

An infinitesimal length along the ring, ds, has a charge dq = l ds.

It generates an electric field of:

Only the z-component matters, so:

since R and z are held constant when integrating around the ring.

But , the circumference of the circle. We can see this explicitly by switching to polar coordinates:

So:

Now, , the total charge of the ring. So we can re-write the field arising from a ring of charge as:

If , the field of a point charge as expected when the ring becomes very small when observed far away.

Field of a Charged Disk

Let’s find the field along the z-axis again. Disk has radius R,
and a charge per unit area of s, such that an infinitesimal area has a charge of

We can solve for the electric field by first determining the electric field arising from a ring of infinitesimal thickness dr and radius r. (Note that this is a different definition of “r” from the previous example).

In fact, we solved for the field of a charged ring already in the previous example! Just make the following substitutions:

So the electric field of this thin ring points in the z-direction and has a magnitude of:

The total field arising from the entire disk is then:

Let’s change variables:

Then:

If the disk becomes very large, , then we can simplify by the approximation:

This is the electric field from an infinite sheet of charge, and you can see that it is independent of the distance, z, away from the sheet.

Now you should also be able to solve problems with non-uniform charge densities (i.e. . Only the integrals become slightly more complicated, but the techniques are the same.

D. Acosta Page 9 9/1/2005