How Does the Electric Field Vary at Point Z in a Dipole Setup?

[Dell]

given a dipole as in the diagram (http://picasaweb.google.com/devanlevin/DropBox?authkey=Gv1sRgCL_4l4PpvP_YsQE#5311991019489958690 ) with charges of +q and -q placed a distance of "d" from one another.
a) what is the electric field at point z, on a line 90 degrees to the middle of the dipole?
b)what is the electric field at point z, on a line 90 degrees to the middle of the dipole when z>>d? (use the taylor series -> f(x)=f(0)+(1/1!)f'(0)+(1/2!)f''(0)... x<<1)

for a)

the distace from each charge to z is r=[(0.5d)²+z²]^0.5
let alpha be the angle between the field of each charge and the horizontal z line,
sin(alpha)= (0.5d)/r= d/2[(0.5d)²+z²]^0.5

for my x-axis i can see that my field will be 0 since E1_x=-E2_x
E_y=[K(q/r²)+K(q/r²)]sin(alpha)

E_y=(2kq/[(0.5d)²+z²])*(d/2=[(0.5d)²+z²]^0.5)

E=kqd/[(0.5d)²+z²]^1.5

now for b) it is clear that if z>>d then d has no real meaning in the denominator so E=kqd/z³ but how do i prove this using the taylor series

 

[AvroArrow]

?To answer b) using the Taylor series, we can begin by considering the formula for E at point z: E = kqd/[(0.5d)2+z2]1.5We can use the Taylor series to approximate this expression when z >> d. We can write the expression as: E = kqd/[(0.5d)2+(z-d + d)2]1.5 We can then expand the numerator and denominator using the Taylor series. We have: E = kqd/[(0.5d)2+(z-d)2 + 2(z-d)+d2]1.5 Since z>>d, the terms involving d are negligible and we can simplify the expression to: E = kqd/[z2 + 2z]1.5 Finally, we can use the Taylor series to approximate the denominator when z>>d. We have: E = kqd/z3 Therefore, the electric field at point z, on a line 90 degrees to the middle of the dipole when z>>d is given by E= kqd/z3.

 

[nlightner]

First, let's define the electric field at point z as E(z). From the diagram, we can see that the electric field at point z is the sum of the electric fields from the two charges, E1 and E2, which are placed at a distance of d/2 from each other.

a) In order to find the electric field at point z, we need to consider the distance from each charge to z, which is given by r=[(0.5d)2+z2]0.5. We also need to consider the angle between the field of each charge and the horizontal z line, which is denoted by alpha. Using trigonometry, we can find that sin(alpha) = (0.5d)/r = d/2[(0.5d)2+z2]0.5. Therefore, the electric field at point z can be written as E(z) = E1 + E2 = [K(q/r2)+K(q/r2)]sin(alpha) = (2kq/[(0.5d)2+z2])*(d/2=[(0.5d)2+z2]0.5). Simplifying this expression, we get E(z) = kqd/[(0.5d)2+z2]1.5.

b) Now, let's use the Taylor series to find the electric field at point z when z>>d. The Taylor series expansion for a function f(x) is given by: f(x)=f(0)+(1/1!)f'(0)+(1/2!)f''(0)... x<<1. In our case, the function is E(z) = kqd/[(0.5d)2+z2]1.5. We can rewrite this as E(z) = kqd/[0.25d2(1+z2/d2)1.5]. Since z>>d, we can assume that z2/d2 is very small. Using the Taylor series expansion, we can write (1+z2/d2)1.5 as 1 + (1.5)(z2/d2) + (1.5)(1.5-1)(z2/d2)2 + ... x<<1. Since we are only considering the first two terms in this expansion, we can write (1+z2/d2)1.5 as 1 + (1.