Question

A plastic rod of length d = 1.5 m lies along the x-axis with its midpoint at the origin. The rod carries a uniform linear charge density λ = 3.5 nC/m. The point P is located on the postive y-axis at a distance y0 = 15 cm from the origin. The z-axis points out of the screen. 25% Part (a) By symmetry the electric field at point P has no component in the _____________. 25% Part (b) Choose the correct expression for the y-component of the electric field at P due to a thin slice of the rod of thickness dx located at point x. 25% Part (c) Integrate your correct choice in part (b) over the length of the rod and choose the correct expression for the y-component of the electric field at point P. 25% Part (d) Calculate the magnitude of the electric field at point P, in newtons per coulomb.

Answer 1

a) The electric field at point P has no component in the x and z directions.

b) dEy = kλdxy₀ / (√( x² + y₀²))^3/2

c)Ey = = 2kλd / y₀( d² + 4y₀²))^1/2

d) Ey = 411.84 N/C

Step-by-step explanation:

a)

from the uploaded image;

the electric field will only be in y-direction.

Therefore the electric field at P have no component in the x and z directions.

b)

dEy = dEsin θ

dE = kdq / r²

from figure

sinθ = y₀ / √( x² + y₀²)

r = √( x² + y₀²)

dq = λdx

dEy = kdq sinθ / r²

dEy = kλdxy₀ / (√( x² + y₀²))^3/2

c)

the net electric field at p is ,

Ey = ∫^d/2_-d/2 kλdxy₀ / (√( x² + y₀²))^3/2

= kλy₀ ∫^d/2_-d/2 dx / (√( x² + y₀²))^3/2

= 2kλd / y₀( d² + 4y₀²))^1/2

d)

let y₀ = 15 × 10⁻²m, λ = 3.5 nC/m , d = 1.5m

Ey = 2kλd / y₀( d² + 4y₀²))^1/2

Ey = 2(9×10⁹ N.m²/C²)(3.5×10⁻⁹m)(1.5m) / 0.15×(1.5)² + 4(0.15)²))^1/2

Ey = 411.84 N/C