Question

where λ is the charge density per unit length on the rod and εο is called the permittivity of free space (it is a universal constant with the value 8.854 x 10-12 F/m (farads per metre)) The integral for the electric field can be evaluated exactly using a method called trigonometric substitution with the result AL We won't learn the method of trigonometric substitution in this course; however, you will approximate the value of the integral using methods we introduced in class (a) Find the linear approximation to the integrand in Equation (1 centered at x = 0 You should have found that your linear approximation from part (a) contained only a constant ternm Consequently, the linear approximation for this particular problem is not very accurate. To obtain a more accurate result, we must use a higher order approximation The next higher order approximation is called the quadratic approximation. The quadratic approxi- mation to a function f(z) centered at x-a is given by: Qa(x) = f(a) + f, (a)(_ a) + Ta)(x-a)2 f"(a) linear approximation quadratic term (b) Determine the quadratic approximation to the integrand in Equation ), centered at 0 For parts (c) and (d), consider a rod of length 2 m (meters) with a charge density of 1.4 × 10-6 C/m (coulombs per meter You will use your results from (a) and (b) to approximate the electric field pro- duced by this rod at the point P(0,2 m). The exact value of the electric field at this point is obtained by substituting these parameters into Equation (2), which gives E(2) 5627.2 V/m (volts per meter) Note that 1F 1 C/V (coulombs per volt) (c) Using the given parameters, approximate the value of the electric field at P(0, 2 m) by taking the integral of your linear approximation from part (a). What is the percentage error in this approximation to the electric field? (d) Using the given parameters, approximate the value of the electric field at P(0, 2 m) by taking the integral of your quadratic approximation from part (b). What is the percentage error in this approx imation to the electric field? (e) Using the given parameters, make a plot of the linear approximation to the integrand and the quadratic approximation to the integrand along with the exact expression for the integrand on a single set of axes Comment on how the shapes of the approximations compare to the shape of the actual integrand. Does the linear approximation or the quadratic approximation give a better approximation to the exact value of the electric field? Justify your answer. (See the additional instructions on LEARN for help with making the plot.) 1 Approximating an integral In this question we wil consider the electric field of a charged rod of length L at a point P located a distance b from the center of the rod along its perpendicular bisector, a illustrated in Figure ·P(0,b) L. Figure 1: A charged rod of length L It can be shown that the magnitude of the electric field at P(O, b) is given by the following integral Ab E(b) = 3/2

Answer 1

(a,b) Around x=0, we can be sure that x/b<<1, so we can use the generalised Binomial theorem to expand such expressions (which is nothing but the Taylor's expansion).

2、-3/2 332 154

Linear term here corresponds to only 1 and with the quadratic term it becomes

3 22 2 62

(c) Using linear app. the electric field is given by

·L/2 K XL

where

K= 4TE0

Thus,

9 × 10" × 1.4 × 10-0 × 2 = 6.3 kV/m EL

Error

6300 5627.2 100 11.96% 5627.2

(d)

Using quadratic app. the electric field is given by

EQb2 J-L/2

Putting the values,

EQ = 5512.5 V/m

Error

5512.5 - 5627.2 5627.2 × 100 =-2.04 %

(e)

For simplicity, I'll factor out

$\frac{K\lambda}{b^2}$

from the integrands and plot only

$\left ( 1+\frac{x^2}{b^2} \right )^{-3/2},1-\frac{3}{2}\frac{x^2}{b^2},1$

with b=2.

1.5 Linear Quadratic Exact 0.5 -1 0.80.6-0.40.2 02 04 0.6 0.8