Electric Charge and Electric Field: Step-by-Step Physics Guidance

Q1. Three Charges on the x-axis: Forces and Fields

Background

Topic: Coulomb's Law, Electric Field Vectors, Superposition Principle

This question explores the calculation of electric forces and fields between discrete point charges, using vector addition and the principle of superposition. It also introduces the concept of finding equilibrium points for forces along a line.

Key Terms and Formulas

• Coulomb's Law: $F = k \frac{|q_1 q_2|}{r^2}$

• Electric Field from a Point Charge: $\vec{E} = k \frac{q}{r^2} \hat{r}$

• Superposition Principle: The net force or field is the vector sum of individual contributions.

• Unit vectors: Used to indicate direction in vector calculations.

• k (Coulomb's constant): $k = 8.99 \times 10^9 \ \mathrm{N \cdot m^2/C^2}$

Step-by-Step Guidance

1. Draw a diagram: Place $q_1 = +5$ nC at the origin (0,0), $q_2 = -3$ nC at $x_2 = 4$ cm, and $q_3 = -2$ nC at $x_3 = 3$ cm. Indicate the direction of forces $\vec{F}_{13}$ (from 1 on 3) and $\vec{F}_{23}$ (from 2 on 3).

2. Calculate the force $\vec{F}_{13}$: Use Coulomb's law to find the magnitude and direction of the force exerted by $q_1$ on $q_3$. Remember to consider the sign of the charges to determine if the force is attractive or repulsive.

3. Calculate the force $\vec{F}_{23}$: Similarly, use Coulomb's law for the force from $q_2$ on $q_3$. Again, pay attention to the sign and direction.

4. Sum the forces: Add $\vec{F}_{13}$ and $\vec{F}_{23}$ vectorially to find the net force $\vec{F}_3$ on $q_3$.

5. For equilibrium location: Set up the equation where the magnitudes of the forces from $q_1$ and $q_2$ on $q_3$ are equal and opposite, and solve for the position $x_3$ where this occurs. Consider the regions along the x-axis where this is possible.

Try solving on your own before revealing the answer!

Q2. Electric Field from a Uniformly Charged Rod

Background

Topic: Electric Field from Continuous Charge Distributions, Integration in Physics

This question involves calculating the electric field at a point on the x-axis due to a rod with uniform charge distribution. It requires setting up and evaluating an integral, using calculus to sum the contributions from each infinitesimal segment of the rod.

Key Terms and Formulas

• Line Charge Density: $\lambda = \frac{Q}{2a}$

• Infinitesimal Charge: $dQ = \lambda dx$

• Electric Field from $dQ$: $d\vec{E} = k \frac{dQ}{r^2} \hat{r}$

• Integration: $\vec{E}(x_0) = \int d\vec{E}$ over the length of the rod

• Limits for the rod: $x$ runs from $-a$ to $+a$

Step-by-Step Guidance

1. Express the line charge density: $\lambda = \frac{Q}{2a}$, where $Q$ is the total charge and $2a$ is the length of the rod.

2. Write the infinitesimal charge element: $dQ = \lambda dx$ for a small segment $dx$ at position $x$ along the rod.

3. Set up the expression for the infinitesimal electric field: At point $x_0$ on the x-axis, $d\vec{E} = k \frac{dQ}{(x_0 - x)^2} \hat{i}$, since all field contributions point along the x-axis.

4. Set up the integral for the net electric field: $\vec{E}(x_0) = \int_{-a}^{+a} k \lambda \frac{dx}{(x_0 - x)^2} \hat{i}$

5. Make a substitution to simplify the integral: Let $u = x_0 - x$, so $du = -dx$. Change the limits accordingly and rewrite the integral in terms of $u$.

Try solving on your own before revealing the answer!

Final Answers

Q1: The net force on $q_3$ is $-640 \hat{i}$ ($\mu$N). The equilibrium position is at $x_3 = 13.3$ cm to the right of both $q_1$ and $q_2$.

Q2: The net electric field at $x_0$ is $\vec{E}(x_0) = \frac{kQ}{x_0^2 - a^2} \hat{i}$. For $x_0 \gg a$, the field approaches that of a point charge: $\vec{E}(x_0) \approx \frac{kQ}{x_0^2} \hat{i}$.

These results follow from careful application of Coulomb's law and integration for continuous charge distributions.