Comparing Electric Forces: Point Charge vs. Uniformly Charged Rod

Q4. Comparing the Force on a Charge Due to a Point Charge and a Uniformly Charged Rod

Background

Topic: Electric Force due to Different Charge Distributions

This question asks you to compare the electric force experienced by a charge q placed a distance s above (A) a point charge Q and (B) the center of a uniformly charged rod with total charge Q. You are also asked to calculate the force in each case for specific values of q and Q.

Key Terms and Formulas

• Coulomb's Law (Point Charge): $F = k_e \frac{|qQ|}{s^2}$ where $k_e$ is Coulomb's constant ($8.99 \times 10^9$ N·m2/C2), $q$ and $Q$ are the charges, and $s$ is the separation distance.

• Electric Force due to a Uniformly Charged Rod: For a rod of length $L$ and total charge $Q$, the force on a charge $q$ located a distance $s$ above the center is found by integrating the contributions from each infinitesimal charge element $dq$ along the rod.

• Linear Charge Density: $\lambda = \frac{Q}{L}$

Step-by-Step Guidance

1. Case A (Point Charge): Write the expression for the force on $q$ due to $Q$ using Coulomb's Law:

   $F_A = k_e \frac{|qQ|}{s^2}$

2. Case B (Uniformly Charged Rod): Set up the integral for the force on $q$ due to the rod. Consider a small element $dx$ at position $x$ along the rod, with charge $dq = \lambda dx$.

   The distance from $q$ to $dq$ is $r = \sqrt{x^2 + s^2}$.

3. Write the expression for the infinitesimal force $dF$ due to $dq$:

   $dF = k_e \frac{|q\,dq|}{r^2} \cdot \cos\theta$

   where $\cos\theta = \frac{s}{\sqrt{x^2 + s^2}}$ projects the force along the vertical direction.

4. Set up the limits of integration for $x$ from $-L/2$ to $L/2$ (since $q$ is above the center of the rod):

   $F_B = \int_{-L/2}^{L/2} k_e \frac{|q\lambda dx|}{x^2 + s^2} \cdot \frac{s}{\sqrt{x^2 + s^2}}$

   Simplify the integrand as much as possible before integrating.

5. Plug in the given values: $q = -1$ nC, $Q = 5$ nC, $L = 10$ cm, $s = 6$ cm. Calculate $\lambda$ and substitute all known quantities into your expressions for $F_A$ and $F_B$.

Try solving on your own before revealing the answer!

Final Answer:

Case A: $F_A = k_e \frac{|qQ|}{s^2}$

Case B: $F_B = 2k_e \frac{|q\lambda| s}{L} \left[ \frac{1}{\sqrt{(L/2)^2 + s^2}} \right]$ (after evaluating the integral)

Numerically, the force in Case A (point charge) is greater than in Case B (rod), because the point charge concentrates all the charge at one location, while the rod spreads it out, reducing the force at the same distance.