Q: Given $$q = -1\ \mathrm{nC}$$, $$Q = 5\ \mathrm{nC}$$, $$s = 6\ \mathrm{cm}$$, and the rod in Case B is $$10\ \mathrm{cm}$$ long, what is the magnitude of the electric force on $$q in $$Case A? Use Coulomb's Law: $$F = k_e \frac{|qQ|}{s^2$}$$, where $$k_e = 8.99 \times 10^9$\ \mathrm{N\cdot m^2$/C^2$$}$.

$$F = 1.25 \times 10$$^{-6}$$\ \mathrm{N}$$

Question 1

In the figure above, Case A shows a point charge $$+q a $$distance $$s$$ above a point charge $$+Q$$, and Case B shows $$+q a $$distance $$s$$ above the center of a uniformly charged rod with total charge $$+Q. In $$which case is the magnitude of the electric force on $$+q$$ greater, and why?

Accepted Answer

Case A, because all of $$+Q is $$concentrated at a single point, resulting in a stronger electric field at distance $$s.$$

Question 2

Given $$q = -1\ \mathrm{nC}$$, $$Q = 5\ \mathrm{nC}$$, $$s = 6\ \mathrm{cm}$$, and the rod in Case B is $$10\ \mathrm{cm}$$ long, what is the magnitude of the electric force on $$q in $$Case A? Use Coulomb's Law: $$F = k_e \frac{|qQ|}{s^2$}$$, where $$k_e = 8.99 	imes 10^9$\ \mathrm{N\cdot m^2$/C^2$$}$.

Accepted Answer

$$F = 1.25 	imes 10$$^{-6}$$\ \mathrm{N}$$

Question 3

For Case B, with $$q = -1\ \mathrm{nC}$$, $$Q = 5\ \mathrm{nC}$$, $$s = 6\ \mathrm{cm}$$, and the rod length $$L = 10\ \mathrm{cm}$$, which of the following best describes how to calculate the electric force on $$q$$ due to the rod?

Accepted Answer

Integrate the electric field contributions from each infinitesimal charge element $$dq$$ along the rod: $$F = k_e |q| \int_{-L/2}^{L/2} \frac{\lambda dx}{(x^2$ + s^2$)} \frac{s}{\sqrt{x^2$ + s^2$}}$$ where $$\lambda = Q/L.$$

Question 4

Suppose $$q = -1\ \mathrm{nC}$$ and $$Q = 5\ \mathrm{nC}. In $$Case B, is the magnitude of the force on $$q$$ greater, less than, or equal to the force in Case A? Why?

Accepted Answer

Less than, because the rod's charge is distributed, so each part is farther from $$q$$ than a point charge would be.

Comparing Electric Forces: Point Charge vs. Uniformly Charged Rod

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