Electric Field Lines Picture
An alternative way to see why the flux through a closed spherical surface is independent of the radius of the surface is to look at the electric field lines. Note that every field line from q that pierces the surface at radius \ also pierces the surface at \ ).
Therefore, the net number of electric field lines passing through the two surfaces from the inside to outside direction is equal. This net number of electric field lines, which is obtained by subtracting the number of lines in the direction from outside to inside from the number of lines in the direction from inside to outside gives a visual measure of the electric flux through the surfaces.
You can see that if no charges are included within a closed surface, then the electric flux through it must be zero. A typical field line enters the surface at \ and leaves at \. Every line that enters the surface must also leave that surface. Hence the net flow of the field lines into or out of the surface is zero ). The same thing happens if charges of equal and opposite sign are included inside the closed surface, so that the total charge included is zero ). A surface that includes the same amount of charge has the same number of field lines crossing it, regardless of the shape or size of the surface, as long as the surface encloses the same amount of charge ).
Determining Tension On A Single Strand
Statement Of Gausss Law
Gausss law generalizes this result to the case of any number of charges and any location of the charges in the space inside the closed surface. According to Gausss law, the flux of the electric field \ through any closed surface, also called a Gaussian surface, is equal to the net charge enclosed \\) divided by the permittivity of free space \\):
This equation holds for charges of either sign, because we define the area vector of a closed surface to point outward. If the enclosed charge is negative ), then the flux through either \ or \ is negative.
The Gaussian surface does not need to correspond to a real, physical object; indeed, it rarely will. It is a mathematical construct that may be of any shape, provided that it is closed. However, since our goal is to integrate the flux over it, we tend to choose shapes that are highly symmetrical.
If the charges are discrete point charges, then we just add them. If the charge is described by a continuous distribution, then we need to integrate appropriately to find the total charge that resides inside the enclosed volume. For example, the flux through the Gaussian surface \ of Figure \ is
Note that \ is simply the sum of the point charges. If the charge distribution were continuous, we would need to integrate appropriately to compute the total charge within the Gaussian surface.
Example \: Electric Flux through Gaussian Surfaces
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How To Calculate Tension In Physics
This article was co-authored by Bess Ruff, MA. Bess Ruff is a Geography PhD student at Florida State University. She received her MA in Environmental Science and Management from the University of California, Santa Barbara in 2016. She has conducted survey work for marine spatial planning projects in the Caribbean and provided research support as a graduate fellow for the Sustainable Fisheries Group.There are 7 references cited in this article, which can be found at the bottom of the page.wikiHow marks an article as reader-approved once it receives enough positive feedback. This article has 15 testimonials from our readers, earning it our reader-approved status. This article has been viewed 1,633,307 times.
In physics, tension is the force exerted by a rope, string, cable, or similar object on one or more objects. Anything pulled, hung, supported, or swung from a rope, string, cable, etc. is subject to the force of tension.XResearch source Like all forces, tension can accelerate objects or cause them to deform. Being able to calculate tension is an important skill not just for physics students but also for engineers and architects, who, to build safe buildings, must know whether the tension on a given rope or cable can withstand the strain caused by the weight of the object before yielding and breaking. See Step 1 to learn how to calculate tension in several physical systems.
Conduction Of Electricity And Heat
The free-electron collisions transfer energy to the atoms of the conductor. The electric field does work in moving the electrons through a distance, but that work does not increase the kinetic energy of the electrons. The work is transferred to the conductors atoms, possibly increasing temperature. Thus a continuous power input is required to keep a current flowing. An exception, of course, is found in superconductors, for reasons we shall explore in a later chapter. Superconductors can have a steady current without a continual supply of energya great energy savings. In contrast, the supply of energy can be useful, such as in a lightbulb filament. The supply of energy is necessary to increase the temperature of the tungsten filament, so that the filament glows.
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Magnitude Of Electric Field Created By A Charge
Let us now first look at finding magnitude of electric field created by a charge.
To find electric field due to a single charge we make use of Coulombs Law.
If a point charge $q$ is at a distance $r$ from the charge $q$ then it will experience a force
Electric field at this point is given by relation
This is electric field at a distance $r$ from a point charge $q$ and $\hat$ is the unit vector along the direction of electric field.
Above relation defining electric field at a distance $r$ tels about both magnitude and direction of the field.
Magnitude of electric field would be
In above equation you could notice the missing $\hat$ part. This $\hat$ tells us about the direction of the electric field. Since electric force is a central force and we have defined electric field using Coulombs law we can conclude that electric field acts along the line joining the charge $q$ and field point at which it is being measured.
This vector This $\hat$ is known as unit vector of our displacement vector $\vec$ and is given by relation
Where $r=|\vec|$ which is the distance between our source point and the field point.
Example 2 Calculating The Force Exerted On A Point Charge By An Electric Field
What force does the electric field found in the previous example exert on a point charge of 0.250 C?
Since we know the electric field strength and the charge in the field, the force on that charge can be calculated using the definition of electric field \mathbf=\frac}\\;rearranged to F =;qE.
The magnitude of the force on a charge q = 0.250 C;exerted by a field of strength E;= 7.20 × 105 N/C is thus,
Because q is negative, the force is directed opposite to the direction of the field.
The force is attractive, as expected for unlike charges. The charges in this example are typical of common static electricity, and the modest attractive force obtained is similar to forces experienced in static cling and similar situations.
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How Is A Compass Used To Find Directions
A compass has a magnetic needle that can rotate freely. When a compass is kept at a place, the magnetic needle aligns in a north-south direction. The red arrow of the compass needle is termed as the North Pole and the other end as the South Pole.
- A simple compass is a magnetic needle mounted on a pivot, or short pin.
- The needle, which can spin freely, always points north. The pivot is attached to a compass card.
- A special kind of compass called a gyrocompass does point to true north.
- The gyrocompass uses a device called a gyroscope, which always points in the same direction.
- Today, large ships carry both magnetic compasses and gyrocompasses.
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Example 3 Calculating Drift Velocity In A Common Wire
Calculate the drift velocity of electrons in a 12-gauge copper wire carrying a 20.0-A current, given that there is one free electron per copper atom. The density of copper is 8.80 × 103;kg/m3.
We can calculate the drift velocity using the equation;I = nqAvd.;The current I = 20.0 A is given, and;q;=;1.60×1019;C;is the charge of an electron. We can calculate the area of a cross-section of the wire using the formula A = r2 , where r is one-half the given diameter, 2.053 mm. We are given the density of copper,;8.80;×;103;kg/m3;and the periodic table shows that the atomic mass of copper is 63.54 g/mol. We can use these two quantities along with Avogadros number,;6.02 × 1023 atoms/mol,;to determine n, the number of free electrons per cubic meter.
First, calculate the density of free electrons in copper. There is one free electron per copper atom. Therefore, is the same as the number of copper atoms per m3. We can now find n as follows:
\beginn& =& \frac^}}\times \frac\text\times }^}\text}}\times \frac}\text\text}\times \frac}}\times \frac\times }^\text}}^}\\ & =& \text\text\text\times }^}^}^\end\\.
The cross-sectional area of the wire is
\beginA& =& \pi ^\\ & =& \pi }^\text}\right)}^\\ & =& \text\times }^}}^\text\end\\
Rearranging I = nqAvd to isolate drift velocity gives
\begin_}=\frac}}\\ =\frac}\text\times }^}}^\right)\left\left}\\ =\text\text\text\times }^}\text\end\\
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What Does An Electric Field Look Like
An electric field can be visualized on paper by drawing lines of force, which give an indication of both the size and the strength of the field. Lines of force are also called field lines. Field lines start on positive charges and end on negative charges, and the direction of the field line at a point tells you what direction the force experienced by a charge will be if the charge is placed at that point. If the charge is positive, it will experience a force in the same direction as the field; if it is negative the force will be opposite to the field.
The fields from isolated, individual charges look like this:
When there is more than one charge in a region, the electric field lines will not be straight lines; they will curve in response to the different charges. In every case, though, the field is highest where the field lines are close together, and decreases as the lines get further apart.
Calculating Electric Flux With Charge
The electric flux through a planar area A of an electric field E is the field multiplied by the component of the area perpendicular to the field. To get this perpendicular component, you use the cosine of the angle between the field and the plane of interest in the formula for flux, represented by = EA cos, where is the angle between the line perpendicular to the area and the direction of the electric field.
This equation, known as Gauss’s Law, also tells you that, for surfaces like these ones, which you call Gaussian surfaces, any net charge would reside on its surface of the plane because it would be necessary to create the electric field.
Because this depends on the geometry of the area of the surface used in calculating flux, it varies depending on the shape. For a circular area, the flux area A would be _r_2 with r as the radius of the circle, or for the curved surface of a cylinder, the flux area would be Ch in which C is the circumference of the circular cylinder face and h is the cylinder’s height.
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Electric Charge And Gravity: Similarities
Coulomb’s law bears striking similarity to Newton’s law for gravitational force FG = G m1m2 / r2 for gravitational force FG, masses m1and m2, and gravitational constant G = 6.674 × 10 11 m3/ kg s2. They both measure different forces, vary with greater mass or charge and depend upon the radius between both objects to the second power. Despite the similarities, it’s important to remember gravitational forces are always attractive while electric forces can be attractive or repulsive.
You should also note that the electric force is generally much stronger than gravity based on the differences in the exponential power of the constants of the laws. The similarities between these two laws are a greater indication of symmetry and patterns among common laws of the universe.
What Is Time Period
In sinusoidal wave motion, as shown above, the particles move about the mean equilibrium or mean position with the passage of time. The particles rise till they reach the highest point that is the crest and then continue to fall till they reach the lowest point that is the trough. The cycle repeats itself in a uniform pattern. The time period of oscillation of a wave is defined as the time taken by any element of the string to complete one such oscillation. For a sine wave represented by the equation: y = -a sin The time period formula is given as:
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How To Find The Lagrangian Of This System
I am trying to find the Lagrangian $L$ of a system I am studying. The equations of motion is:
$$\left\ r \ddot + 2\dot \dot+k \cdot r \dot \dot = 0\\ \ddot – r \dot^2 – k \cdot r^2 \dot^2 = 0\end\right.$$
I have tried a general Ansatz $L=L_1+L_2=\Sigma_ C_ r^m \dot^n \phi^p \dot^q+L_2)$ and plugged into the Euler-Lagrange equation but find the calculation extremely tedious. Is there some systematic way to find it?
I’d really appreciate any hints. Thank you!
If we assume$$L=A \dot^2 + B \dot^2 +C \dot \dot$$
Then$$\left\ \mathscr_r L = 2A \ddot -B_r \dot^2+C\ddot +A_r \dot^2\\ \mathscr_\phi L = 2B \ddot +2B_r \dot\dot+C_r\dot^2 + C \ddot\end\right.$$where $\mathscr_q L \equiv \frac \left-\frac}}$
It seems fine except for $A_r=0$ is conflicting with the others.
Implications Of Gauss’ Law
Gauss’ Law is a powerful method of calculating electric fields. If you have a solid conducting sphere that has a net charge Q on it, you know all the excess charge lies on the outside of the sphere. Gauss’ law tells us that the electric field inside the sphere is zero, and the electric field outside the sphere is the same as the field from a point charge with a net charge of Q. That’s a pretty neat result.
The result for the sphere applies whether it’s solid or hollow. Let’s look at the hollow sphere, and make it more interesting by adding a point charge at the center.
What does the electric field look like around this charge inside the hollow sphere? How is the negative charge distributed on the hollow sphere? To find the answers, keep these things in mind:
- The electric field must be zero inside the solid part of the sphere
- Outside the solid part of the sphere, you can find the net electric field by adding, as vectors, the electric field from the point charge alone and from the sphere alone
We know that the electric field from the point charge is given by kq / r2. Because the charge is positive, the field points away from the charge.
If we took the point charge out of the sphere, the field from the negative charge on the sphere would be zero inside the sphere, and given by kQ / r2 outside the sphere.
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Selected Solutions To Problems & Exercises
2.; 0.263 N;; If the charges are distributed over some area, there will be a concentration of charge along the side closest to the oppositely charged object. This effect will increase the net force.
4.;The separation decreased by a factor of 5.
8.;\beginF&=&k\frac__\mid}^}=ma\Rightarrow=\frac^}}\\\text&=&\frac^\text\cdot\text^/}^\right)^\text\right)}^}^\text\right)^\text\right)}^}\\ & =& 3.45\times ^\text^\end\\
9.; 3.2;; If the distance increases by 3.2, then the force will decrease by a factor of 10 ; if the distance decreases by 3.2, then the force will increase by a factor of 10. Either way, the force changes by a factor of 10.
11.; 1.04 × 109 C;; This charge is approximately 1 nC, which is consistent with the magnitude of charge typical for static electricity
16.; 0.859 m beyond negative charge on line connecting two charges; ;0.109 m from lesser charge on line connecting two charges
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