Anonconducting 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?

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

like poles repel so keep pushing north to north and repels funny to see this

force tends to be low and long diustance masses are high