Chapter 23: Problem 40
Figure 23-42 shows, in cross section, two solid spheres with uniformly distributed charge throughout their volumes. Each has radius \(R\). Point \(P\) lies on a line connecting the centers of the spheres, at radial distance \(R / 4.00\) from the center of sphere 1 . If the net electric field at point \(P\) is zero, what is the ratio \(q_{2} / q_{1}\) of the total charges?
Short Answer
Step by step solution
Understanding the problem
Write expression for electric field due to sphere 1
Write expression for electric field due to sphere 2
Set net electric field to zero
Simplify and solve for \( q_2 / q_1 \)
Conclusion
Unlock Step-by-Step Solutions & Ace Your Exams!
-
Full Textbook Solutions
Get detailed explanations and key concepts
-
Unlimited Al creation
Al flashcards, explanations, exams and more...
-
Ads-free access
To over 500 millions flashcards
-
Money-back guarantee
We refund you if you fail your exam.
Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Coulomb's Law
- \( F \) is the magnitude of the force between the charges,
- \( k \) is Coulomb's constant, approximately \(9 \times 10^9\, \text{Nm}^2/\text{C}^2\),
- \( q_1 \) and \( q_2 \) are the charges, and
- \( r \) is the distance between the charges.
Uniform Charge Distribution
For a sphere with uniformly distributed charge, the charge density (\( \rho \)) is constant. The electric field inside and outside the sphere is calculated differently:
- Inside the Sphere: The electric field at a distance \( r \) from the center is calculated using the volume charge density. Given by: \( E = \frac{kq}{R^3}r \), when \( r \leq R \).
- Outside the Sphere: The sphere can be treated as a point charge with total charge \( q \), and the electric field is calculated using: \( E = \frac{kq}{r^2} \).
Ratio of Charges
The process involves setting the electric fields produced by both spheres equal because their combination results in a net zero field at point \( P \). For uniform spheres, you would use the inside sphere formula for sphere 1 and the outside sphere formula for sphere 2. After equating the two expressions for electric fields:\[ \frac{k q_1}{R^3} \frac{R}{4} = \frac{k q_2}{\left(\frac{5R}{4}\right)^2} \]By simplifying and rearranging, the desired equation gives you:\[ \frac{q_2}{q_1} = \frac{16}{25} \]This calculated ratio tells us how much charge each sphere needs for the electric fields to cancel each other out at point \( P \). Understanding this concept is critical in electric field applications where balance or neutralization of forces is necessary.