Question:
Find the magnitude and direction of the electric field at points that are at the following distances from the center of the aluminum rod:
(a) 0.5 cm, (b) 1.5 cm, (c) 2.5 cm, (d) 3.5 cm, (e) 7.0 cm
Answer:
(a) 0
(b) 4.795 N/C
(c) 2.877 N/C
(d) 2.055 N/C
(e) 1.027 N/C
Step-by-step explanation:
The given parameters are;
The inner radius of the copper = 2 cm
The outer radius of the copper - 3 cm
The radius of the aluminum rod, r = 1 cm
The radius of the aluminum rod = 4 pC/m
a) The charge at the 0.5 cm from the center of the aluminum rod, E = 0
There is no charge inside the rod given that the aluminum road is a conductor of electricity, having all the charge on its surface
b) At 1.5 cm, we have;

Where;

r = 1.5 cm = 0.015 m
∴ E = (2 × 8.99 × 10⁹ × 4 × 10⁻¹²)/0.015 ≈ 4.795
The magnitude of the electric field, E = 4.795 N/C
The direction of the charge depends on the charge sign on the rod
c) At 2.5 cm, we have;
r = 2.5 cm = 0.025 m
E = (2 × 8.99 × 10⁹ × 4 × 10⁻¹²)/0.025 ≈ 2.877
The magnitude of the electric field, E ≈ 2.877 N/C
d) At r = 3.5 cm, we have;
r = 3.5 cm = 0.035 m
E = (2 × 8.99 × 10⁹ × 4 × 10⁻¹²)/0.035 ≈ 2.055
The magnitude of the electric field, E ≈ 2.055 N/C
e) At
r = 7.0 cm = 0.070 m
E = (2 × 8.99 × 10⁹ × 4 × 10⁻¹²)/0.07 ≈ 1.027
The magnitude of the electric field, E ≈ 1.027 N/C