In the figure, a nonconducting rod of length L = 8.21 cm has charge q = -4.51 fC uniformly distributed along its length. (a) What is the linear charge density of the rod? What are the (b) magnitude and (c) direction (positive angle relative to the positive direction of the X axis) of the electric field produced at point P, at distance a = 13.5 cm from the rod? What is the electric field magnitude produced at distance a = 76 m by (d) the rod and (e) a particle of charge -q = -4.51 fC that replaces the rod?
The Correct Answer and Explanation is:
Here are the solutions to the problem:
(a) Number: -5.49e-14 Units: C/m
(b) Number: 1.38e-3 Units: N/C or V/m
(c) Number: 180 Units: This answer has no units
(d) Number: 7.01e-9 Units: N/C or V/m
(e) Number: 7.02e-9 Units: N/C or V/m
Explanation
This problem involves calculating the linear charge density and the electric field produced by a uniformly charged rod.
(a) Linear Charge Density
The linear charge density, symbolized by λ (lambda), is the total charge (q) distributed over the total length (L) of the rod. The formula is λ = q / L. First, we convert the length from centimeters to meters: L = 8.21 cm = 0.0821 m. The charge is given as q = -4.51 fC, which is -4.51 x 10⁻¹⁵ C.
Using the formula:
λ = (-4.51 x 10⁻¹⁵ C) / (0.0821 m) = -5.49 x 10⁻¹⁴ C/m.
(b) & (c) Electric Field Magnitude and Direction (a = 13.5 cm)
For a point on the axis of a finite rod of length L and charge q, the magnitude of the electric field at a distance ‘a’ from the nearest end is given by the formula:
|E| = |k * q| / (a * (a + L)), where k is Coulomb’s constant (8.99 x 10⁹ N·m²/C²).
We convert the distance ‘a’ to meters: a = 13.5 cm = 0.135 m.
Plugging in the values:
|E| = (8.99 x 10⁹ * 4.51 x 10⁻¹⁵) / (0.135 * (0.135 + 0.0821))
|E| = (4.055 x 10⁻⁵) / (0.135 * 0.2171) = 1.38 x 10⁻³ N/C.
For the direction, the rod has a negative charge. Electric field lines point towards negative charges. Since point P is on the positive x-axis relative to the rod, the electric field vector at P points towards the rod, which is in the negative x-direction. The angle of the negative x-axis relative to the positive x-axis is 180 degrees.
(d) & (e) Electric Field at a Large Distance (a = 76 m)
This part compares the electric field of the rod to that of a point particle at a large distance.
For the rod (d), we use the same formula as in part (b) with a = 76 m:
|E_rod| = (8.99 x 10⁹ * 4.51 x 10⁻¹⁵) / (76 * (76 + 0.0821)) = 7.01 x 10⁻⁹ N/C.
For the particle (e), we use the formula for the electric field of a point charge: |E| = |k * q| / r², where the distance r is a = 76 m.
|E_particle| = (8.99 x 10⁹ * 4.51 x 10⁻¹⁵) / (76)² = 7.02 x 10⁻⁹ N/C.
The results are very close, which shows that at distances much larger than the object’s size (76 m >> 0.0821 m), a distributed charge can be accurately approximated as a single point charge.
