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A nonconducting rod of length L = 8.15 cm has charge –q = -4.23 fC uniformly distributed along its length.(a) What is the linear charge density of the rod?What are the(b) magnitude and(c) direction(relative to the positive direction of the s-axis) of the electric field produced at point P, at a distance a = 12.0 cm from the rod? What is the electric field produced at point P, at distance a = 12.0 cm from the rod? What is the electric field magnitude produced at distance a = 50 m by(d) the rod and(e) a particle of charge –q = -4.23 fC that replaces the rod?

Respuesta :

Answer:

a)  λ = 5.19 10⁻⁴ C/m , b)  E = 1,573 10⁻³ N/C , c) the direction of the field is directed to the bar

Explanation:

a) the linear density defined as the ratio between the charge per unit length

       λ = q / l

Let's start by reducing the units to the SI system

     L = 8.15 cm (1m / 100cm) = 8.15 10⁻² m

     a = 12 cm (1m / 100cm) = 12 10⁻² m

    q = -4.23 fC (1 C / 10¹⁵ ft) = -4.23 10⁻¹⁵ C

    λ = -4.23 10⁻¹⁵ C / 8.15 10⁻²

    λ = 5.19 10⁻⁴ C/m

b) Let's look for the electric field for a point at a distance a from the end of the bar

      E = k  dq / r²

To simplify the notation, suppose the bar is the x axis. Since the density is constant, we can write it differentially

     λ = dq/dx

     dq = λ dx

     E = k ∫ λ dx / x²

We integrate and evaluate between the lower limits x = a and higher x = L + a. Here we place the test point at the origin of the system

     E = k λ (-1 / x)

     E = k λ (-1 /(L + a) + 1 /a)

     E = k λ (L /a(L + a)

Let's change the density for its value

     E = k (q / L) (L / a (L + a)

     E = k q  1 /[a(L + a)]

     E = 8.99 10⁹ 4.23 10⁻¹⁵ [1 /12 10⁻²(8.15 10⁻² + ​​12 10⁻²)]

     E = 1,573 10⁻³ N/C  

c) the direction of the field is directed to the bar, because it has a negative charge

d) now we change the distance a = 50 cm = 0.50 m

Bar

      E = 8.99 10⁹ 4.23 10⁻¹⁵ ( 1 /0.5(0.0815 +0.5))

      E = 1,308 10⁻⁴ N/C

Charge point

      q = -4.23 10⁻¹⁵ C

     E = k q / r²

     E = 8.99 10⁹ 4.23 10⁻¹⁵ / 0.5²

     E = 1.521 10⁻⁴ N/C

lhmarianateixeira

In this exercise we have to use the knowledge of density to calculate the electric field, in this way:

a)  λ = 5.19 10⁻⁴ C/m

b)  E = 1,573 10⁻³ N/C

c) the direction of the field is directed to the bar

So calculating some electromagnetism stuff we find that:

a) The linear density is :

[tex]\lambda = q / l\\ L = 8.15 cm (1m / 100cm) = 8.15 10^{-2} m\\ a = 12 cm (1m / 100cm) = 12 10^{-2} m\\ q = -4.23 fC= -4.23 10^{-15} C\\ \lambda = -4.23 10^{-15} C / 8.15 10^{-2}\\ \lambda = 5.19 10^{-4} C/m[/tex]

b) Calculating the electric field of the bar is equal to:

[tex]E = k \lambda (-1 / x)\\ E = k \lambda (-1 /(L + a) + 1 /a)\\E = k \lambda (L /a(L + a)\\ E = k (q / L) (L / a (L + a)\\ E = k q 1 /[a(L + a)]\\ E = 1,573 *10^{-3} N/C[/tex]

c) the direction of the field is directed to the bar, because it has a negative charge.

d) Calculating the distance from the bar is equal to:

      [tex]E = k q / r^2\\ E = 8.99 *10^9* 4.23 10^{-15} / 0.5^2\\ E = 1.521 *10^{-4} N/C[/tex]

See moer about density at brainly.com/question/24650225

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