Superposition Principle: Difference between revisions
No edit summary |
|||
| Line 1: | Line 1: | ||
The Superposition Principle states that the net result of multiple vectors acting on a given place and time is equal to the vector sum of each individual vector. For intro physics, this mostly relates to effect that multiple electric or magnetic fields and forces have on a certain location. | The Superposition Principle states that the net result of multiple vectors acting on a given place and time is equal to the vector sum of each individual vector. For intro physics, this mostly relates to effect that multiple electric or magnetic fields and forces have on a certain location. | ||
Claimed by | Claimed by gweese3(Spring 2018) | ||
Revision as of 08:50, 14 April 2018
The Superposition Principle states that the net result of multiple vectors acting on a given place and time is equal to the vector sum of each individual vector. For intro physics, this mostly relates to effect that multiple electric or magnetic fields and forces have on a certain location. Claimed by gweese3(Spring 2018)
A Mathematical Model
The Superposition Principle is derived from the properities of additivity and homogeneity for linear systems which are defined in terms of a scalar value of a by the following equations:
[math]\displaystyle{ F(x_1 + x_2) = F(x_1) + F(x_2)\quad Additivity }[/math]
[math]\displaystyle{ aF(x) = F(ax)\quad Homogeneity }[/math]
The principle can be applied to any linear system and can be used to find the net result of functions, vectors, vector fields, etc. For the topic of introductory physics, it will mainly apply to vectors and vector fields such as electric forces and fields.
If given a number of vectors passing through a certain point, the resultant vector is given by simply adding all the the vectors at that point. For example, for a number of uniform electric fields passing though a single point, the resulting electric field at that point is given by:
[math]\displaystyle{ \vec{E} = \vec{E}_{1} + \vec{E}_{2} +...+ \vec{E}_{n} = \sum_{i=1}^n\vec{E}_{i} }[/math]
This same concept can be applied to electric forces as well as to magnetic fields and forces. This is more useful when dealing with the effect that multiple point charges have on each other is an area of void of other electric fields. By adding together each of the contributing electric field vectors at a specific point you can figure out the net contribution that all of the present charges have on the overall electric field experienced at said location.
You can calculate the electric field of a single point charge by using the following equation: [math]\displaystyle{ \vec{E} = \frac{1}{4 \pi \epsilon_0}\sum_{i=1}^n\frac{q_i}{r_i^2}\hat{r_i} }[/math]
Where:
[math]\displaystyle{ \frac{1}{4 \pi \epsilon_0} }[/math] is a constant that equals [math]\displaystyle{ 9e9 }[/math].
[math]\displaystyle{ q_i }[/math] is the charge of the particle you are studying.
[math]\displaystyle{ r_i }[/math] is the magnitude of the distance between the particle and the observation location.
[math]\displaystyle{ \hat{r_i} }[/math] is the vector pointing from the particle to the observation location.
Examples
When attempting to solve these problems be sure to show all steps in your solution and include diagrams whenever possible.
Simple - Two Point Charges and One Dimension
If [math]\displaystyle{ Q_1 }[/math] equals 1e-4 C and [math]\displaystyle{ Q_2 }[/math] equals 1e-5C, what is the net electric field at the midpoint of both charges?
Click for Solution
We know that the distance of each charge to the midpoint is [math]\displaystyle{ 1m }[/math]. Since this value is the magnitude of the distance it will serve as our [math]\displaystyle{ r_i }[/math] value for both of the electric field formulas.
Our [math]\displaystyle{ \hat{r_1} }[/math] value for [math]\displaystyle{ E_1 }[/math] will be [math]\displaystyle{ 1 }[/math] and the [math]\displaystyle{ \hat{r_2} }[/math] value for [math]\displaystyle{ E_2 }[/math] will be [math]\displaystyle{ -1 }[/math], if you use the midpoint as your observation point in the x-direction.
Now that we have all the necessary values to use the electric field formula, we can plug them in.
[math]\displaystyle{ \vec{E_1} = \frac{1}{4 \pi \epsilon_0}\frac{q_1}{r_1^2}\hat{r_1} = (9e9)\frac{1e-4C}{1m^2}(1m) = 9e5\frac{N}{C} }[/math] [math]\displaystyle{ \vec{E_2} = \frac{1}{4 \pi \epsilon_0}\frac{q_2}{r_2^2}\hat{r_2} = (9e9)\frac{1e-5C}{1m^2}(-1m) = -9e4\frac{N}{C} }[/math]
By this point, you have the individual contributions of each charge. To get the net contribution of all the charges, you only need to add them.
[math]\displaystyle{ \vec{E_1}+\vec{E_2}=9e5\frac{N}{C}-9e4\frac{N}{C}= 8.1e5\frac{N}{C} }[/math]
This means that the net electric field will point towards the right with a magnitude of [math]\displaystyle{ 8.1e5\frac{N}{C} }[/math].
Medium - Two Point Charges and Two Dimensions
If [math]\displaystyle{ Q_1 }[/math] and [math]\displaystyle{ Q_2 }[/math] are positive point charges with a charge of e, what is the net electric field at point P?
Click for Solution
To begin this problem, the first step is to find [math]\displaystyle{ r_1 }[/math] and [math]\displaystyle{ r_2 }[/math], the vectors from the charges to point P as well as their magnitudes:
[math]\displaystyle{ \vec{r_1} = 3\hat{i}+2\hat{j}-(0\hat{i}+0\hat{j})\Rightarrow\vec{r_1} = 3\hat{i}+2\hat{j} }[/math]
[math]\displaystyle{ ||\vec{r_1}|| = \sqrt{3^2 + 2^2} =\sqrt{13} }[/math]
[math]\displaystyle{ \vec{r_2} = 3\hat{i}+2\hat{j}-(4\hat{i}+0\hat{j})\Rightarrow\vec{r_2} = -1\hat{i}+2\hat{j} }[/math]
[math]\displaystyle{ ||\vec{r_2}|| = \sqrt{-1^2 + 2^2} =\sqrt{5} }[/math]
Using these in the equation for an electric field from a point charge, you get:
[math]\displaystyle{ \vec{E_1} = \frac{1}{4 \pi \epsilon_0}\frac{Q_1}{||r_1||^2}\hat{r_1}=\frac{1}{4 \pi \epsilon_0}\frac{e}{13}\lt \frac{3}{\sqrt{13}}\hat{i}+\frac{2}{\sqrt{13}}\hat{j}\gt = \lt 9.21E-11\hat{i}+6.14E-11\hat{j}\gt }[/math]N/C
[math]\displaystyle{ \vec{E_2} = \frac{1}{4 \pi \epsilon_0}\frac{Q_2}{||r_2||^2}\hat{r_2}=\frac{1}{4 \pi \epsilon_0}\frac{e}{5}\lt \frac{-1}{\sqrt{5}}\hat{i}+\frac{2}{\sqrt{5}}\hat{j}\gt = \lt -1.29E-10\hat{i}+2.58E-10\hat{j}\gt }[/math]N/C
Then, simply add the two electric fields together:
[math]\displaystyle{ \vec{E}=\vec{E_1}+\vec{E_2} = \lt -3.69E-11\hat{i}+3.19E-10\hat{j}\gt }[/math]
Difficult - Five Point Charges and Three Dimensions
If all point charges have a charge of e, what the the net electric field present at point L?
Click for Solution
This problem is similar to the previous examples but this one includes the z axis and more points. Since there are 5 points and you have already had some practice, you will only see the procedure for the first one. We will still work through the whole problem. Again, first each vector and magnitude:
[math]\displaystyle{ \vec{d_5} = 0\hat{i}+0\hat{j}+0\hat{k}-(2\hat{i}-1\hat{j}-1\hat{k})-\Rightarrow\vec{d_5} = -2\hat{i}+1\hat{j}+1\hat{k} }[/math]
[math]\displaystyle{ ||\vec{d_5}|| = \sqrt{(-2)^2 + 1^2 + 1^2} = 2 }[/math]
Do the same for each of the other point charges and plug them into the electric field formula:
[math]\displaystyle{ \vec{E_5} = \frac{1}{4 \pi \epsilon_0}\frac{Q_5}{||d_5||^2}\hat{d_5}=\frac{1}{4 \pi \epsilon_0}\frac{e}{4}\lt \frac{-2}{2}\hat{i}+\frac{1}{2}\hat{j}+\frac{1}{2}\hat{k}\gt = \lt -3.6E-10\hat{i}+1.8E-10\hat{j}+1.8E-10\hat{k}\gt }[/math]N/C
Follow the same steps for the other electric fields and add them all together to get your final answer:
[math]\displaystyle{ \vec{E_1} = \lt 1.44E-9\hat{i}+0\hat{j}+0\hat{k}\gt }[/math]N/C
[math]\displaystyle{ \vec{E_2} = \lt 0\hat{i}-1.44E-9\hat{j}+0\hat{k}\gt }[/math]N/C
[math]\displaystyle{ \vec{E_3} = \lt -1.29E-10\hat{i}+0\hat{j}+2.58E-10\hat{k}\gt }[/math]N/C
[math]\displaystyle{ \vec{E_4} = \lt 0\hat{i}+0\hat{j}-1.44E-9\hat{k}\gt }[/math]N/C
[math]\displaystyle{ \vec{E} = \vec{E_1}+\vec{E_2}+\vec{E_3}+\vec{E_4}+\vec{E_5}=\lt 9.51E-10\hat{i}-1.26E-9\hat{j}+1.00E-9\hat{k}\gt }[/math]N/C
Connectedness
The superposition principle is critical as you progress to more complicated topics in Physics. This principle simplifies many topics that you will encounter further ahead, by assuming that their net contribution will be equal to the addition of all the individual ones. You will use superposition with electric fields and forces, as well as magnetic fields and forces. In addition, if in a problem you are only given the final net contribution of any type of field, you can inversely solve it by constructing equations that relate the individual formulas. You can then plug in values, and subtract in order to find the magnitudes of each individual contributor. This will also enable you to better analyze different types of circuits.
However, the superposition principle also aids outside of the world of abstract electromagnetism. It is also useful for static problems, which you might encounter if you are a Civil or Mechanical Engineer, or systems and circuitry, which you might encounter if you are a Biomedical engineer.
History
The superposition principle was supposedly first stated by Daniel Bernoulli, a famous scientist known for his work in fluid mechanics, statistics, and the Bernoulli Principle. In 1753, he stated that, "The general motion of a vibrating system is given by a superposition of its proper vibrations." This idea was at first rejected by some other popular scientists until it became widely accepted due to the work of Joseph Fourier, a famous scientist known for his work on the Fourier Series, which is used in heat transfer and vibrations, as well as his discovery of the Greenhouse Effect.
See also
Next Topics
Now that you have a solid understanding of the Superposition Principle, you can move on to these topics:
Further reading
Books, Articles or other print media on this topic
Schrodingers' Cat from a Superposition Point of View
External Links
Visual Explanation of Superposition in Electric Fields
Electric Fields and Superposition Principle Crash Course Video