Problem 63 - University Physics with Modern Physics

Question

22.63. Positive charge \(Q\) is distributed uniformly over each of two spherical volumes with radius \(R\) . One sphere of charge is centered at the origin and the other at \(x=2 R\) (Fig. 2244 ). Find the magnitude and direction of the net electric field due to these two distributions of charge at the following points on the \(x\) -axis: (a) \(x=0 ;\) (b) \(x=R / 2 ;(c) x=R ;\) (d) \(x=3 R\) .

Step-by-Step Solution

Answer

(a) E = \( \frac{kQ}{4R^2} \) left, (b) E = \( -\frac{11kQ}{72R^2} \) left, (c) E = 0, (d) E = \( \frac{10kQ}{9R^2} \) right.

Step 1: Understanding the Problem

We need to find the net electric field due to two uniformly charged spheres at different locations along the x-axis. Both spheres have charge \( Q \) and radius \( R \). The spheres are centered at \( x=0 \) and \( x=2R \). We will calculate the electric field at various points along the x-axis: \( x=0 \), \( x=R/2 \), \( x=R \), and \( x=3R \). Remember, inside a uniformly charged sphere, the electric field is given by \( E = \frac{kQr}{R^3} \), and outside, it is \( E = \frac{kQ}{r^2} \), where \( k \) is Coulomb's constant and \( r \) is the distance from the center.

Step 2a: Calculating Electric Field at x=0

At \( x=0 \), we are inside the first sphere. Here, the electric field is \( E_1 = 0 \) since we are at the center. For the second sphere at \( x=2R \), we are outside, so \( E_2 = \frac{kQ}{(2R)^2} \), directed towards negative x-axis. Therefore, the net electric field is the field from the second sphere, \( E_{net} = \frac{kQ}{4R^2} \), directed left.

Step 2b: Calculating Electric Field at x=R/2

For the first sphere at \( x=R/2 \), we are inside, so \( E_1 = \frac{kQ(R/2)}{R^3} = \frac{kQ}{8R^2} \), directed right. For the second sphere, \( E_2 = \frac{kQ}{(3R/2)^2} = \frac{4kQ}{9R^2} \), directed left. The net field is \( E_{net} = \frac{kQ}{8R^2} - \frac{4kQ}{9R^2} = -\frac{11kQ}{72R^2} \) (negative, so directed left).

Step 2c: Calculating Electric Field at x=R

At \( x=R \), for the first sphere, \( E_1 = \frac{kQ}{R^2} \), directed right (outside region). For the second sphere at \( R \), \( E_2 = \frac{kQ}{R^2} \), directed left. Therefore, \( E_{net} = \frac{kQ}{R^2} - \frac{kQ}{R^2} = 0 \).

Step 2d: Calculating Electric Field at x=3R

At \( x=3R \), both spheres are treated as point charges. For the first sphere, \( E_1 = \frac{kQ}{(3R)^2} = \frac{kQ}{9R^2} \), directed right. For the second sphere, \( E_2 = \frac{kQ}{R^2} \), directed right, since we are outside the second sphere. The net electric field \( E_{net} = \frac{kQ}{9R^2} + \frac{kQ}{R^2} = \frac{10kQ}{9R^2} \), directed right.

Key Concepts

Uniform Charge Distribution

When discussing electric fields, one critical concept to understand is the **uniform charge distribution** within a sphere. This means that the charge is evenly spread throughout the volume of the sphere, not just on its surface. This is important because it affects how the electric field behaves both inside and outside the charged object.

• **Inside a uniformly charged sphere**: The electric field increases linearly with distance from the center. At the very center, the electric field is zero.
• **Outside the sphere**: The sphere behaves as if all its charge were concentrated at its center due to symmetry, and the field decreases with the square of the distance from the center.

Understanding this allows us to use formulas to determine the electric field at various points around charged spheres, as seen in the exercise. It helps to visualize the charge acting as if it were at a single central point when outside the sphere.

Coulomb's Law

Coulomb's law is fundamental in calculating electric forces and fields. It states that the electric force (F) between two point charges is proportional to the product of the magnitudes of charges and inversely proportional to the square of the distance between them. Mathematically, it's expressed as: \[F = k \frac{|q_1 q_2|}{r^2}\]where

• *k* is Coulomb's constant
• *q₁* and *q₂* are the point charges
• *r* is the distance between charges

In our exercise, Coulomb’s law assists in finding the electric field from each spherical charge distribution, treating the spheres as point charges when determining the field at points outside the spheres. This simplification helps in mathematical calculations and provides insights into how charges interact at a distance.

Spherical Coordinates

In physics, we often use **spherical coordinates** instead of Cartesian coordinates to solve problems involving spheres or circular symmetry. Spherical coordinates (*\(r, \theta, \phi\)*) are suitable for situations where the problem has symmetry about a point or an axis, like determining electric fields of spherical charge distributions.

• *r* is the radial distance from the point to the origin.
• *\(\theta\)*, the polar angle, is the angle from the positive z-axis.
• *\(\phi\)*, the azimuthal angle, is measured from the positive x-axis in the x-y plane.

By using spherical coordinates in problems like the one given, it simplifies the analysis since we focus on distances and symmetry, ultimately aiding in understanding how electric fields radiate from uniformly charged spheres.

Problem-Solving in Physics

**Problem-solving in physics** often involves a systematic approach to break down a complex problem into manageable parts. To solve physics problems effectively, students should:

1. **Identify what is given** and what needs to be found. In our exercise, the spheres' charge and their positions are given, and we need the electric field at various points. 2. **Apply relevant principles** like symmetry and related laws – Coulomb's law, for instance, is used for calculating the electric field here. 3. **Use appropriate coordinates** systems like spherical coordinates to exploit any symmetry, simplifying calculations. 4. **Check units and dimensions**, ensuring that they are consistent throughout the calculations. 5. **Interpret the results** in a meaningful way to understand the physical implications of your calculations. Approaching physics problems step by step and using these strategies encourages deeper understanding and instills confidence in tackling challenging exercises.