See the Proof of Various Derivative Formulas section of the Extras chapter to see the proof of this formula. This formula is sometimes called the power rule. All we are doing here is bringing the original exponent down in front and multiplying and then subtracting one from the original exponent. There are actually three different proofs in this section. In this case we have the sum and difference of four terms and so we will differentiate each of the terms using the first property from above and then put them back together with the proper sign.
Notice that in the third term the exponent was a one and so upon subtracting 1 from the original exponent we get a new exponent of zero.
The point of this problem is to make sure that you deal with negative exponents correctly. Here is the derivative. Make sure that you correctly deal with the exponents in these cases, especially the negative exponents. Now in this function the second term is not correctly set up for us to use the power rule. The power rule requires that the term be a variable to a power only and the term must be in the numerator. So, prior to differentiating we first need to rewrite the second term into a form that we can deal with.
Note that we left the 3 in the denominator and only moved the variable up to the numerator. Remember that the only thing that gets an exponent is the term that is immediately to the left of the exponent.
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Here is the derivative for this part. All of the terms in this function have roots in them. In order to use the power rule we need to first convert all the roots to fractional exponents. Again, remember that the Power Rule requires us to have a variable to a number and that it must be in the numerator of the term. In the last two terms we combined the exponents. You should always do this with this kind of term. In a later section we will learn of a technique that would allow us to differentiate this term without combining exponents, however it will take significantly more work to do.
We can now differentiate the function. Make sure that you can deal with fractional exponents. You will see a lot of them in this class. In all of the previous examples the exponents have been nice integers or fractions. They work exactly the same. However, this problem is not terribly difficult it just looks that way initially.
There is a general rule about derivatives in this class that you will need to get into the habit of using. When you see radicals you should always first convert the radical to a fractional exponent and then simplify exponents as much as possible. Following this rule will save you a lot of grief in the future. There are some that we can do. It is still possible to do this derivative however. All that we need to do is convert the radical to fractional exponents as we should anyway and then multiply this through the parenthesis.
So, as we saw in this example there are a few products and quotients that we can differentiate. If we can first do some simplification the functions will sometimes simplify into a form that can be differentiated using the properties and formulas in this section. We know that the rate of change of a function is given by the functions derivative so all we need to do is it rewrite the function to deal with the second term and then take the derivative.
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Note that we rewrote the last term in the derivative back as a fraction. Again, notice that we eliminated the negative exponent in the derivative solely for the sake of the evaluation. Example 1: Consider equal charges of Q whose value in coulombs is not known. The force between two of these charges at distance X is F. What is the net force on the charge at point B? Arrangement of point charges for the example. Solution : This problem can be solved through proportional reasoning. The force of 3 Q on the one charge at B will be 3 F.
The forces of 3 F and 4 F are at right angles, and therefore, the resultant force is 5 F , or. When a small positive test charge is brought near a large positive charge, it experiences a force directed away from the large charge. If the test charge is far from the large charge, the electrostatic force given by Coulomb's law is smaller than when it is near. This data of direction and magnitude of an electrostatic force, due to a fixed charge or set of fixed charges, constitutes an electrostatic field.
The electric field is defined as the force per unit charge exerted on a small positive test charge q 0 placed at that point. Note that both the force and electric field are vector quantities. The test charge is required to be small so that the field of the test charge does not affect the field of the set charges being examined. These lines are called field lines or lines of force. Electric field lines of a positive and b negative point charges. Electric field lines of a two opposite charges, b two like charges, and c two oppositely charged plates. The rules for drawing electric field lines for any static configuration of charges are.
Electric flux is defined as the number of field lines that pass through a given surface.
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In Figure , lines of electric flux emerging from a point charge pass through an imaginary spherical surface with the charge at its center. The accepted convention is that flux lines are positive if leaving a surface and negative if entering a surface. Gauss's law provides a method to calculate any electric field; however, its only practical use is for fields of highly symmetric distributions of fixed charges.
As a result, if no charge exists with a given closed surface, then there are as many flux lines entering the surface as there are leaving it. The imaginary surface necessary to apply Gauss's law is called the gaussian surface. Consider the calculation of the electric field due to a point charge. Figure shows the point charge, the direction of its field, and a gaussian surface. Gauss's law is. The derivation of the expression for the field due to a thin conducting shell of charge follows. When outside the shell of charge, as in Figure a , the left side of Gauss's equation reduces to the following expression for the same reasons given for a point charge:.
Thus, the electric field outside a sphere of charge is the same as if the same amount of charge were concentrated in a point located at the center of the sphere. The gaussian surface inside the sphere encloses no charge, and therefore, there is no electric field inside the uniformly charged spherical shell. The same proof holds within a solid conductor because all the charge of the conductor resides on the surface.
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Because the electric field inside even an irregularly shaped conductor is zero, the charge will not be evenly distributed over an irregular shape. The charge will tend to accumulate on protruding points on the outside of the conductor. In the preceding examples, the charge distributions were spherical, and so the gaussian surface was a sphere. When finding the electric field of either a sheet of charge or a line of charge, a cylinder is the correct gaussian surface to use.
Example 2: Find the electric field for a nonconducting infinite sheet of charge. The electric field is directed outward from the sheet on both sides. See Figure for the electric field and gaussian surface. The electric field is parallel to the wall, which is at right angles to the outward normal of the wall area; thus, the last term on the right is zero. The total charge inside the gaussian surface is the product of the charge per unit area and the area; so.
Note that the magnitude of the electric field does not depend upon the distance from the plate. The electric field is uniform. In the practical case of finite plates of charge, the electric field is uniform relatively close to the charged plate.
The resultant electric field of two parallel plates is double that of one sheet with the same charge:. If the plates have opposite charges, the electric field will exist between the plates and be zero outside the plates. If the charges are of equal sign, the electric field will be zero between the plates and be expressed by the above equation outside the plates. These results can be derived by Gauss's law.
The work done in transferring the charge equals the product of the force on the test charge and the parallel component of displacement, using the same definition of work given in the section on mechanics. See Figure For certain configurations of electric field, it may be necessary to use the integral definition of electrostatic potential:. The electric potential due to a point charge q at a distance r from the point charge is.
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