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
