integration and is typically done by computers because the computational
integration and is typically done by computers because the computational
complexity is dependant upon the size of \delta{y} with respect to L.
complexity is dependant upon the size of \delta{y} with respect to L.
=== Simplifying ===
Using calculus, we can simplify a lot of the math required to compute the
electric field at any given point. Notationally, all we're doing is switching from the
discretely-sized \delta{y} to \textit{dy} and from the sigma notation to
an integral starting from -L/2 (the lower end of the rod) and ending at
L/2 (the upper end of the rod) as follows:
[image 9] [image 10]
By evaluating the integral, we can determine that the x-component of the
electric field at any point is:
[image 11]
Without evaluating the integral for the y-component of the electric field,
we can use symmetry to determine that the y-component of the electric
field at any given point is 0. Let's consider the contributions to the
electric field from the top and bottom halves of the rod at any
observation point.
[image 12]
Since the y-components of E_top and E_bottom are of equal magnitude and
opposite direction, they cancel each other out, and therefore the
y-component of teh electric field at any given point due to the rod is 0.
[image 13]
Finally, because the rod is round and can be rotated, as a convenience,
we'll use d (distance from the rod) as opposed to x (distance along the
x-direction) to refer to the electric field.
Thus we can simplify electric field calculations for a rod into a form
that we can readily use:
[image 14]
===Week 5===
===Week 5===
Revision as of 23:35, 25 November 2018
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The Main Idea
Previously, we've learned about the electric field of a point particle. Often, when analyzing physical systems, it is the case that we're unable to analyze each individual particle that composes an object and need to therefore generalize collections of particles into shapes (in this case, a rod) whereby the mathematics corresponding to electric field calculations can be simplified. This can essentially be done by adding up the contributions to the electric field made by parts of an object, approximating each part of an object as a point charge.
The System in Question
As discussed in the previous section, we're considering a system
abstracted from the particle model we're familiar with, therefore we will
make the generalization that our rod of length L has a total charge of
quantity Q. For this generalization, we will need to assume that the rod
is so thin that we can ignore its thickness.
Since the electric field produced by a charge at any given location is
proportional to the distance from the charge to that location, we will
need to relate the observation location to the source of the charge, which
we will consider the origin of the rod. To do that, we will need to divide
the rod into pieces of length \delta y each containing a charge \delta Q.
In the image below, you can see what this looks like and the relation that
can be found between the observation location and the source, forming the
distance vector \vect{r}.
By the pythagorean theorem, we can find the vector \vect{r} as follows:
And to find the unit vector in the direction of \vect{r}, \hat{r}, we do as
follows:
Finding the Contribution of Each Piece to the Electric Field
Now that we've set up a model for the system, with the rod broken down
into pieces, we can find the contribution of each piece to the electric
field of the system. We will start from the electric field equation you
learned for a point particle but plug in the parameters for the rod system
into the equation.
By mathematically simplifying, we then get the following equation:
Finding the Net Contribution of all Pieces
In the previous section, we found out the contribution to the electric
field at a given location of only one of the pieces constituting the rod.
In order to figure out the net field at any particular location, we need
to add up the electric fields produced by individual pieces along the
length of the rod.
We will switch from vector notation for the electric field to the scalar
notation for the x- and y-components. (From the vector in the equation
above, we can see that the z-component of the electric field at any point
is always 0.) The x-component of the electric
field is the sum of the x-components of every \delta{y} along the rod, and
the y-component of the electric field is the sum of the y-components of
every \delta{y} along the rod. We can show this mathematically:
To make use of this relation, because we don't know \delta{Q}, we need to
relate it to parameters that we already know about the rod system we're
analyzing. We can express \delta{Q} as the charge density of the rod
(which is Q/L) times the \delta{y} we've chosen for the system. Thus,
By plugging the above equation into our equations for the x- and
y-components of the electric field at a point, we can find the electric
field at any point in the system. This technique is called numerical
integration and is typically done by computers because the computational
complexity is dependant upon the size of \delta{y} with respect to L.
Simplifying
Using calculus, we can simplify a lot of the math required to compute the
electric field at any given point. Notationally, all we're doing is switching from the
discretely-sized \delta{y} to \textit{dy} and from the sigma notation to
an integral starting from -L/2 (the lower end of the rod) and ending at
L/2 (the upper end of the rod) as follows:
By evaluating the integral, we can determine that the x-component of the
electric field at any point is:
Without evaluating the integral for the y-component of the electric field,
we can use symmetry to determine that the y-component of the electric
field at any given point is 0. Let's consider the contributions to the
electric field from the top and bottom halves of the rod at any
observation point.
Since the y-components of E_top and E_bottom are of equal magnitude and
opposite direction, they cancel each other out, and therefore the
y-component of teh electric field at any given point due to the rod is 0.
Finally, because the rod is round and can be rotated, as a convenience,
we'll use d (distance from the rod) as opposed to x (distance along the
x-direction) to refer to the electric field.
Thus we can simplify electric field calculations for a rod into a form
that we can readily use:
Further Simplification
By noting the contributions of each variable to the equation for the
electric field, we can make approximations to simplify our math by simply
declaring one variable as insignificant.
For example, if we have a system in which the length of a rod is much
greater than the magnitude of the distance from the rod (denoted L>>d), we
can neglect some of the instances in which d is taken into account as
follows:
Finding the Electric Field from a Rod with Code
Here is some code that you can run which shows the electric field vector
at a given distance from the rod along its length. The rod is shown as a
series of green balls to help emphasize that when using the numerical
integrations mentioned on this page, you are measuring the field produced
by discrete parts of the rod being analyzed.
Notice the edge-effects of the electric field of the rod. For reasons
discussed above, if we used the long rod approximation (L>>d), these
effects would be negligible.
Previously, we've learned about the electric field of a point particle. Often, when analyzing physical systems, it is the case that we're unable to analyze each individual particle that composes an object and need to therefore generalize collections of particles into shapes (in this case, a rod) whereby the mathematics corresponding to electric field calculations can be simplified. This can essentially be done by adding up the contributions to the electric field made by parts of an object, approximating each part of an object as a point charge.
The System in Question
As discussed in the previous section, we're considering a system
abstracted from the particle model we're familiar with, therefore we will
make the generalization that our rod of length L has a total charge of
quantity Q. For this generalization, we will need to assume that the rod
is so thin that we can ignore its thickness.
[image 1]
Since the electric field produced by a charge at any given location is
proportional to the distance from the charge to that location, we will
need to relate the observation location to the source of the charge, which
we will consider the origin of the rod. To do that, we will need to divide
the rod into pieces of length \delta y each containing a charge \delta Q.
In the image below, you can see what this looks like and the relation that
can be found between the observation location and the source, forming the
distance vector \vect{r}.
[image 2]
By the pythagorean theorem, we can find the vector \vect{r} as follows:
[image 3]
And to find the unit vector in the direction of \vect{r}, \hat{r}, we do as
follows:
[image 4]
Finding the Contribution of Each Piece to the Electric Field
Now that we've set up a model for the system, with the rod broken down
into pieces, we can find the contribution of each piece to the electric
field of the system. We will start from the electric field equation you
learned for a point particle but plug in the parameters for the rod system
into the equation.
[image 5]
By mathematically simplifying, we then get the following equation:
[image 6]
Finding the Net Contribution of all Pieces
In the previous section, we found out the contribution to the electric
field at a given location of only one of the pieces constituting the rod.
In order to figure out the net field at any particular location, we need
to add up the electric fields produced by individual pieces along the
length of the rod.
We will switch from vector notation for the electric field to the scalar
notation for the x- and y-components. (From the vector in the equation
above, we can see that the z-component of the electric field at any point
is always 0.) The x-component of the electric
field is the sum of the x-components of every \delta{y} along the rod, and
the y-component of the electric field is the sum of the y-components of
every \delta{y} along the rod. We can show this mathematically:
[image 7]
To make use of this relation, because we don't know \delta{Q}, we need to
relate it to parameters that we already know about the rod system we're
analyzing. We can express \delta{Q} as the charge density of the rod
(which is Q/L) times the \delta{y} we've chosen for the system. Thus,
[image 8]
By plugging the above equation into our equations for the x- and
y-components of the electric field at a point, we can find the electric
field at any point in the system. This technique is called numerical
integration and is typically done by computers because the computational
complexity is dependant upon the size of \delta{y} with respect to L.
