HARD Physics Problem (pic incuded)?

Question

http://i333.photobucket.com/albums/m363/djwilliamt12/Scan0002.jpg

It deals with finding experssions of an electric field.

Answer 1

(a)

The charge is uniformly distributed with λ = Q/L

So the charge element in the line element dx is:

dQ = λ dx

(b)

Treat the charge element as point charge.

The Electrical field surrounding a point charge is given by

Ě = (k·Q/r²) · ě_r

with

Ě Electric field vector

k = 4·π·ε₀ Coulomb's constant

r distance charge to E-field evaluation point

ě_r  unit vector pointing from the charge to the evaluation point

Let the location of the E-field evaluation point and the charge given by the point vectors ř₁ and ř₂ respectively.

The the vector

ř = ř₂ - ř₁

is pointing from charge to evaluation point.

Divide by its length and you get  unit vector for that direction:

ě_r = (ř₂ - ř₁) / ||ř₂ - ř₁||

Moreover the length of the vector is identical to distance between the points:

r =  ||ř₂ - ř₁||

Hence:

Ě = k·Q · (ř₂ - ř₁)/||ř₂ - ř₁||³

Let q be the charge element located at x

Q =  dq = λ dx

ř₁ = < x ; 0 >

ř₂ = < 0 ; a >

=>

(ř₂ - ř₁) = < -x ; a >

=>

||ř₂ - ř₁|| = √((-x)² + a²) = √(x² + a²)

So the E-field vector element induced by this charge element is:

dĚ = (k·λ dx) · < -x ; a > / (√(x² + a²))³

Its components are

dE_x = - k·λ·x / (√(x² + a²))³  dx

dE_y = k·λ·a / (√(x² + a²))³  dx

(c)

To get the whole E-field just integrate over the length:

E_y = ∫ dE_y

= ∫0→L [ k·λ·a / (√(x² + a²))³ ] dx

You an express x in terms of the given angle θ. I a right triangle tangent equals the ratio of opposite leg to adjacent leg. Hence:

tan(θ) = x/a

<=>

x = a·tan(θ)

=>

dx = a·(tan²(θ)+1) dθ = a·sec²(θ) dθ

E_y = ∫0→arctan(L/a) [ k·λ·a / (√(a²·tan²(θ) + a²))³ ]  a·sec²(θ) dθ

= ∫0→arctan(L/a) [ k·λ·a·/sec(θ) dθ

= k·λ·a· ∫0→arctan(L/a) cos(θ) dθ

= k·λ·a·sin(arctan(L/a))

This can be simplified

sin(θ) = √(1 - cos²(θ))= (√(1 - cos²(θ))

= (cos(θ) ·√(1/cos²(θ) - 1) = √(sec²(θ) - 1) / sec(θ)

= √(tan²(θ)) / √(tan²(θ) +1)  = tan(θ) / √(tan²(θ) +1)  

=>

sin(arctan(z)) = z / √(z² +1)  

=>

E_y = k·λ·a·(L/a) / √((L/a)² +1) = k·λ·a·L/√((L² +a²)