Question

Un alambre de plástico, aislante y recto mide 10 cm de longitud y tiene una densidad de carga de +150 nC/m, distribuidos de manera uniforme por toda su longitud. Se encuentra sobre una mesa horizontal. A) Encuentre la magnitud y la dirección del campo eléctrico que produce este alambre en un punto que está 8 cm directamente arriba de su punto medio. B) Si el alambre ahora se dobla para formar un círculo que se coloca aplanado sobre la mesa, calcule la magnitud y la dirección del campo eléctrico que produce en un punto que se encuentra 6 cm directamente arriba de su centro.

Answer 1

English only

Step-by-step explanation:

Answer 2

When solving problems related to Electric Fields, care must be taken about symmetries. In our particular case when we take a look to at the drawings of the attached file, we realize:

1.-By symmetry each dx associated at a, has an opposite dx with point b as reference. The respective dE ( the charge is uniform ) is the same, as the charge of the wire is positive the force and the Field on a test charge (+) located at h will be upward, therefore the components dEx will cancel each other and the Electric Field becomes E = Ey = ∫ 2×dE× cosθ

The solutions:

A) Ey = 4623 N/C

B) Ey = 19.34 N/C

E = Ey = ∫ 2×dE× cosθ

Here cosθ = h/ d ⇒ cosθ = h/√h² + x² dE = K× dQ / d²

d² = h² + x²

k = 8.9 ×10⁹ Nm²C⁻² ; dQ = λ×dx λ = 150×10⁻⁹ C h = 0.08 m

Then by substitution

Ey = 2 ∫[K× λ×dx/ (h² + x²) ] × h / √h² + x²

reordering that equation:

Ey = 2×K×λ×h ∫ dx / [√ ( h² + x² ) ]³ (2)

To solve the integral we make use of a change of variables

x = h × tanα then x² = h² ×tan²α and dx = h× sec²α dα

plugging that values in equation (2)

Ey = 2×K×λ×h ∫ h× sec²α× dα / [√ ( h² + h²tan²α)]³

Ey = 2×K×λ×h² ∫ sec²α× dα / [ h × √ (1 + tan²α)]³ 1 + tan²α = sec²α

Ey = 2×K×λ×h²× ∫ (sec²α / h³× sec³α )×dα

Ey = 2×K×λ/h × ∫ ( 1 / secα dα

Ey = 2×K×λ/h × sinα now we αneed to come back to our original variables:

as x = h × tanα tanα = x/h then x is the opposite leg in a right triangle and h the adjacent one then the hypothenuse is √ (h² + x²) then sin α = x/ √ (h² + x²)

Ey = 2×K×λ/h × x/ √ (h² + x²) |₀⁰°⁰⁵

Ey = 2×8.9×10⁹× 150×10⁻⁹× 5×10⁻²/8× 10⁻²× √ 10⁻² ( 8 + 5 ) N/C

Ey = 4623 N/C

To answer the second question again we will make use of symmetries if you look at drawing ( Figure 2 ) you see that again the components in direction of x-axis cancel each other and the components in y-axis direction will add. Then

Ey = ∫ dE× cosθ

following the same procedure we will find:

Ey = ∫ [K×λ × dl/d²] × h/ d

The importan point here is that the radius of the circle is

2×π×r = 0.01 ( the length of the wire) ⇒ r = 0.16×10⁻² m

And we need to take into account that the integration is over the circle and the length of the circle is 0.01 m or ××2×π×r. All other factors are constant. Then by substitution

Ey = [K×λ ×h× / ( √ r² + h²)³ ] × 10⁻² N/C

Ey = 8.9 × 10⁹ × 150× 10⁻⁹ × 6× 10⁻² × 10⁻² / √ 10⁻² ( 0.16 + 6)

Ey = 0.8 × 10² / 6

Ey = 19.34 N/C