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Coulomb's Law, which describes the electrostatic force between two charges. how Coulomb established the inverse-square law of force and the proportionality to charge magnitude. Additionally, it discusses the concept of the coulomb as the unit of charge and provides an example of calculating the force between two charges. The document also covers Coulomb's Law in vector form, the principle of superposition, and the electric field.
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Michael Fowler, UVa, 8/29/
Using the two small hanging spheres, we can even find just how the attraction varies with distance, by measuring the angle the string makes with the vertical and doing a simple calculation for varying distances. This is tricky, though—the charge slowly leaks away, especially in summer, moisture in the air dampens the surfaces slightly, and they conduct.
In an experiment essentially equivalent to this, Coulomb in the 1780’s established that the electrostatic force decreased with distance as the inverse square, exactly like gravity (but of course it’s much stronger!). He also found the force to be proportional to the magnitude of the charge. He accomplished that by using a charged sphere, then removing it and putting it in contact with an identical but uncharged sphere, so the charge would be equally shared. Now putting the sphere back, he found the force had been halved.
In his honor, the unit of charge is called the coulomb. The charge on the electron is 1.6x10 -19^ coulombs. The coulomb is a practical unit for dealing with batteries and electric currents, it’s the amount of charge flowing down a wire per second when the current is one ampere (we’ll discuss these units in detail later). Unfortunately, though, in electrostatics we never deal with charge on this scale, and the microcoulomb is more typical.
To picture the strength of electrostatic repulsion, imagine taking an ounce of water, and imagine you could pull all the electrons off the atoms, and put them is a separate glass. (Of course, this can’t be done—as you’ll see!) Now put our glass of electrons and your glass of nuclei one thousand kilometers apart, say here and Orlando. What’s the strength of the attractive force between them?
Now Avogadro’s number of molecules, 6X10 23 , weigh a gram mole, that’s 18 grams for water, so an ounce, 28 grams, is about 10 24 atoms, or 10 25 electrons (a water molecule has ten electrons total). One electron has charge 1.6x10-19^ coulombs, so we have Q =1.6x10 6 coulombs. Here to Orlando r = 10 6 meters, so
F = kQQ / r^2 =9x10 9 x(1.6x10^6 ) 2 /10 12 , about 2x 10 10 N … 2,000,000 tons.
From this, we can definitely conclude that for charged spheres repelling each other, the imbalance in electron numbers from neutral is very, very small: this is why typical electrostatic charges are micro coulombs, but total electron charge in a sphere is of order mega coulombs—the imbalance is of order 10 -12^ or so.
Electric force vectors add : if charge Q 2 is repelled by charge Q 1 and charge Q 3 , the total repulsive force on it is the vector sum of the separate repulsive forces:
This may look obvious, but must be experimentally verified—the Law of Superposition is in fact not true for nuclear forces!
Just why a charge can affect the motion of another charge some distance away is rather mysterious, as indeed is the gravitational attraction between two masses. Einstein was the first to realize that a mass distorts space time in such a way that other masses, instead of moving at constant velocity, accelerate. The earth’s gravitational field slightly distorts space and time here so all masses free to fall accelerate downwards at the same rate (ignoring other forces, such as air resistance, of course).
An electric charge does not distort spacetime, but does have a surrounding energy density in space, called its electric field. Another charge placed in this field experiences the inverse-square force.
1 2 12 (^12 ) 12
32 3 2 32 2 32
1 2 12 2 12
One way to visualize the electric field is to draw lines of force. These are lines drawn so that at any point on the line, the electric force on a positive test charge is in the direction of the line.
Exercise: try sketching the lines of force for the two equal charges in the diagram above.
We already know the x -axis is a line of force, since the field everywhere on it is along the axis, but in different directions for positive and negative x. We also know that anywhere on the y -axis, the force is along the y -axis, pointing away from the nearer charge. We know that close to one of the charges, its force will dominate, so the field lines initially come out close to radially from the charge. Finally, far away the two charges look like one, so again the field lines will be radial at large distances.
Here’s a sketch:
To get a better idea of electric fields in this and other cases, go to this website. Sometimes attempts are made to indicate the strength of the field by how close together the lines are drawn. This is easy to see for the simplest case of a single charge, although even there one would think from a diagram on paper that the field was decreasing as 1/ r , because that’s how the lines thin out—but really the field is in three dimensions, so actually they would thin out faster, as 1/ r^2 , in a (more realistic) 3D model. The other problem with this approach is that where the field is weak, no lines appear at all, so it’s difficult to figure
out what’s going on. We’ve added some dotted local lines of force near the midpoint, the place where the field strength goes to zero.
Given what we’ve just done, this is very easy: if a ring has total charge Q , uniformly spread around the ring, it can be replaced by a large number of small pairs of charges arranged as in the above example, and the axis of the ring is the x -axis in the above picture for all these pairs of charges, the electric fields of all the pairs add, they’re all along the axis for a point on the axis, so the answer has to be, for a ring of radius d ,