Integrals

# Integrals - Introduction This document contains lots of...

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Unformatted text preview: Introduction This document contains lots of stuff. It contains 5 different charge configurations that can be readily solved by integration. Here, readily means that the problem fits very naturally into a coordinate system, and results in an integral that is simple to solve. All of these problems are readily solved for either the electric field OR the electric potential. On your midterm exam, you would be asked to only do one of these calculations ( i.e. one charge distribution and either a calculation of field or potential). You would not be asked to solve for both the field and the potential. Please note also, that the solutions to these problems are written with MUCH MUCH MORE detail than is necessary on your midterm exam. This is done on purpose so that you can learn these problems by understanding them instead of by memorizing them. Prof H PSFor any problem, you can check these relations: E x =- V x ), V =- integraltext vector E vector dl by either integrating the field result or taking the derivative of the potential. 1 1. Finite Line of Charge. Shown in the figure below is a line of charge. The line has a length L and a charge Q. Determine the electric field at the position x=D. L Q D x Figure 1: A line charge of length L and charge Q. Solving for the Electric Field at the Origin When solving for the electric field, we need to solve the integral: vector E = integraldisplay k dQ s 2 away (1) To solve the integral, we must deal with three pieces: dQ , s , and away . First we deal with dQ. Step 1: dQ To deal with the small piece of charge, we need to decide upon the dimensionality of the object. The identification of the dimensionality leads us to one of three choices for the amount of charge dQ: Dims dQ Density 1D dl = charge length 2D dA = charge area 3D dV = charge volume 2 Since our object is a line, it is a TWO DIMENSIONAL object and well represent the small charge dQ as: dQ = dl (2) = Q L (3) This object naturally fits into the CARTESIAN coordinate system and we note that: Different pieces of charge have different x spanning (0 < x < L ). All pieces of charge have the same y ( y = 0). All pieces of charge have the same z ( z = 0). Thus, the length of a small piece will be dx yielding: dQ = Q L dx (4) Step 2: s Next we should deal with the distance from the charge piece to the location at which we do the calculation. From the figure, it is apparent that: s = D- x (5) Step 3: away Dealing with away , means identifying the direction of symmetry in the problem. Looking at the figure, we recognize that the final electric field must be pointed in the x direction. This, all contributions are in the CORRECT direction and there is no away correction....
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## This note was uploaded on 06/05/2011 for the course PHY 132 taught by Professor Rijssenbeek during the Spring '04 term at SUNY Stony Brook.

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Integrals - Introduction This document contains lots of...

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