Extra Problem A3 for 7D [just one this week]. Please upload your solution to the course website by 11:59 Tuesday 4/25/2023 This problem illustrates one the interesting uses of the electric potential, V : it is often easier to compute the electric field from a distributed charged distribution by a two-step process (1) compute the electric potential V and then (2) take the partial derivate to get the electric field (equation 23.19) than to directly compute the Electric field from the vector integral of the distributed charge (the procedure you used in Chap 21). The reason is that the integration involved in computing the potential is a scaler integration, whereas the direct calculation of the Electric field is a vector integration which usually requires as many as 3 different integrations for x, y, and z components. Of course, if there is enough symmetry, then Gauss's Law is your best bet to obtain the Electric field vector anywhere in space. P . ( x y ) -> * (origin ) A rod of length "a" carries uniform charge +Q, and is oriented along the x axis as shown in the figure. Not enough symmetry to use Gauss's Law, unfortunately. Take the potential to be 0 at infinity. a) Compute the electric potential V(x, y) at point P, arbitrarily located at some point x, y. Hint: The definite integral that you need to perform is located on page A-4 of your text, or google it. One detail, if you integrate from the origin to the left end of the stick, then the symbol dx is negative. Usually we take this into account by writing it as -dx, so the symbol dx becomes a magnitude, and the integration goes from 0 to -a. b) Find the x component of the electric field by taking the partial derivate with respect to x. A partial derivative is just like a full derivate except that all the other variables are treated as constants. So if V(x,y) = 2x-3y, the partial with respect to x is 2. c) Since the previous parts require a complicated calculation (though still a bit simpler than the vector integration), it is important to figure out techniques to check your answer. One technique is to compute the Potential or Electric field under specific conditions where you know the answer. For example: What does your answer for V become when the value of y is much larger than "a" or x. From the perspective of a location at large distance from the stick, the stick looks more and more like a point, and you know the equation for the potential of a point charge. Does your answer to part a) reduce to the expected equation if y is much greater than x or a? le, does the equation V(y) look like the equation from a point charge? The key is to rewrite the expression for V(x,y) in terms of ratios where the ratio is a small number. For example, if you have y+atx, and y is much larger than a or x, then you can factor out the y to produce an expression with asmall number. So y+a+x = y(1+(a+x)/y) where the term (a+x)/y is a small number. So this for your actual expression for V(x,y) and use the series expansion formula found on page A-4 to expand some of the terms to 1st order. You should also remember that In(n*m)=In(n)+In(m), so if you have In(n*(1+r)), where r is a small number, then In(n*(1+r)) = In(n)+In(1+r) ~In(n)+r. So this check requires that you get comfortable implementing math that you learned previously. d) Another check: If y=0, and x is positive and large compared to "a", then what is the Electric field? You should realize that the Electric field should only have an x component (so is Ey=0?) and the formula for Ex should look like a point source, or kQ/x2. Apply the same procedure outlined in part c) to your answer in b). e) Suppose x=0, Find Ex as a function of y. Notice that you cannot first evaluate V(0,y) and then take the derivative dV/dx since this produces Ex=0. So why not