Problem 34 - University Physics with Modern Physics

Question

Two uniformly charged insulating rods are bent in a semicircular shape with radius \(r=10.0 \mathrm{~cm} .\) If they are positioned so they form a circle but do not touch and have opposite charges of \(+1.00 \mu \mathrm{C}\) and \(-1.00 \mu \mathrm{C}\) find the magnitude and direction of the electric field at the center of the composite circular charge configuration.

Step-by-Step Solution

Answer

Answer: The magnitude of the electric field at the center of the composite circular charge configuration is 1.44×10^5 N/C, and its direction is to the left.

Step 1: 1. Identify the givens

Given that two semicircular rods with uniform charge are bent to form a circle of radius \(r=10.0\,\mathrm{cm}\). One rod has a positive charge \(+1.00\,\mu\mathrm{C}\) while the other has a negative charge \(-1.00\,\mu\mathrm{C}\). We need to find the electric field at the center of the composite circular configuration.

Step 2: 2. Calculate the electric field due to each rod

In the case of a semicircular rod, the electric field at the center (O) of the semicircle can only have horizontal components. We will find the horizontal component of the electric field due to each rod at the center. Let's analyze the positive rod first. Due to symmetry, each small charge segment of the positive rod contributes to the horizontal component of the electric field at the center. The positive rod generates an electric field pointing towards its plane, i.e., towards the left. Let's assume \(E_+=E_{\mathrm{pos}}\). For the negative rod, the electric field is repulsive, and hence the electric field produced by the negative rod will point away from its plane, i.e., towards the left. Let's assume \(E_-=E_{\mathrm{neg}}\).

Step 3: 3. Formula for electric field due to a charged semicircular rod

The electric field generated by a semicircular rod with a charge of \(q\) and radius \(r\) at its center is given by the formula: \(E=\frac{1}{4\pi\epsilon_{0}}\frac{2q}{r^2}\), where \(\epsilon_{0}=8.85\times10^{-12}\,\mathrm{C^2/N\cdot m^2}\) is the permittivity of free space.

Step 4: 4. Calculate electric fields due to positive and negative rods

Using the formula, we'll calculate the electric field due to the positive rod, \(E_+\), and the negative rod, \(E_-\) as follows: \(E_{\mathrm{pos}}=\frac{1}{4\pi\epsilon_{0}}\frac{2(+1.00\,\mu\mathrm{C})}{(10.0\,\mathrm{cm})^2}\) and \(E_{\mathrm{neg}}=\frac{1}{4\pi\epsilon_{0}}\frac{2(-1.00\,\mu\mathrm{C})}{(10.0\,\mathrm{cm})^2}\)

Step 5: 5. Use the principle of superposition to find the net electric field at the center

The net electric field at the center will be the algebraic sum of the horizontal components of the electric fields due to the positive and negative rods. Since both \(E_{\mathrm{pos}}\) and \(E_{\mathrm{neg}}\) are in the same direction (to the left), their magnitudes add up: \(E_{\text{net}}=E_{\mathrm{pos}}-E_{\mathrm{neg}}= 2 E_{\mathrm{pos}}=2 E_{\mathrm{neg}}\) Now, using the calculated values, we will find the net electric field at the center: \(E_{\text{net}}=2 E_{\mathrm{pos}}=2\left(\frac{1}{4\pi\epsilon_{0}}\frac{2(+1.00\,\mu\mathrm{C})}{(10.0\,\mathrm{cm})^2}\right)\)

Step 6: 6. Compute the final value for the net electric field and direction

After calculating the net electric field, we find its magnitude and direction: \(E_{\text{net}}=2 E_{\mathrm{pos}}=\frac{1}{4\pi\epsilon_{0}}\frac{4(+1.00\,\mu\mathrm{C})}{(10.0\,\mathrm{cm})^2}=1.44\times10^{5}\,\mathrm{N/C}\) The net electric field is directed to the left, away from the center of the composite circular charge configuration. In conclusion, the magnitude of the electric field at the center of the composite circular charge configuration is \(1.44\times10^{5}\,\mathrm{N/C}\), and its direction is to the left.

Key Concepts

Electric Field Due to Uniform Charge

Understanding the electric field due to a uniform charge distribution is critical for grasping various electrostatic phenomena. A uniform charge means that the charge density is consistent throughout the material. When dealing with rod-like objects, which can be approximated as one-dimensional charge distributions, the electric field calculation involves integrating the contributions of each infinitesimal charge element along the length of the rod.

In our example with semicircular rods, we can imagine slicing the rod into tiny pieces, each with a small charge \(dq\). Each segment's electric field contribution at the center, due to its symmetry, will have a horizontal component. The resulting total electric field is then calculated by adding up (integrating) these contributions from each segment, a process simplified by the rod's uniform charge distribution.

Superposition Principle in Electrostatics

The superposition principle is a fundamental concept in electrostatics that allows us to calculate the net electric field created by multiple charges. It states that the total electric field produced by a collection of charges is the vector sum of the fields produced by each charge independently. In other words, the individual electric fields simply add up.

This principle is essential when dealing with situations like the two semicircular rods in our exercise. Each rod creates its own electric field, and the net electric field at the center is the sum of the electric fields due to each rod. If the fields point in the same direction, their magnitudes add. If in opposite directions, they subtract.

Symmetry in Electric Field Calculations

Symmetry plays an important role in simplifying electric field calculations. When a charge distribution is symmetric, it's possible to make predictions about the electric field's behavior without detailed calculations. For semicircular rods, such symmetry allows us to infer that only the horizontal components of the electric field at the center contribute to the net field.

In our exercise, the symmetry of the setup—two identically shaped rods with opposite charges—means that the vertical components of the electric fields cancel each other out. This leaves us with only the horizontal components to consider, greatly simplifying our calculation.

Permittivity of Free Space

The permittivity of free space \(\epsilon_0\), also known as the electric constant, is a physical constant that appears in the equations of electromagnetism. It represents the ability of the vacuum of space to permit electric field lines. This value is crucial in calculating the force between electric charges with Coulomb's law, as well as in determining the electric field created by a charge distribution.

In our problem, the value of \(\epsilon_0\) enters the formula for the electric field due to a charged rod. Its value, \(8.85 \times 10^{-12} \mathrm{C^2/N\cdot m^2}\), ensures the correct units and magnitude for the electric field in our calculations. Remember that \(\epsilon_0\) is a constant, providing consistency in electrostatics computations across different scenarios.