6. A spherical capacitor is composed of a solid conducting sphere concentric with a larger spherical conducting shell. The inner conductor has positive total charge Q and radius R. The outer conductor has negative total charge -Q, inner radius 3R, and outer radius 4R. The space between the conductors is filled with two spherical dielectric shells. The dielectric in the region R

icon
Related questions
Question
### Problem 6: Spherical Capacitor with Dielectric

A spherical capacitor is composed of a solid conducting sphere concentrically placed within a larger spherical conducting shell. The following characteristics define the setup:

- **Inner Conductor:** 
  - Charge: Positive, total charge \( Q \)
  - Radius: \( R \)

- **Outer Conductor:** 
  - Charge: Negative, total charge \( -Q \)
  - Inner Radius: \( 3R \)
  - Outer Radius: \( 4R \)

The space between the conductors is filled with two spherical dielectric shells:

- **First Dielectric Region:** 
  - Region: \( R < r < 2R \)
  - Dielectric Constant: \( K_1 = 6 \)

- **Second Dielectric Region:** 
  - Region: \( 2R < r < 3R \)
  - Dielectric Constant: \( K_2 = 2 \)

#### Diagram Description

A cross-sectional view of the spherical capacitor system is shown with colored regions:

- Inner conductor at radius \( R \)
- First dielectric shell in green from \( R \) to \( 2R \)
- Second dielectric shell in blue from \( 2R \) to \( 3R \)
- Outer conductor from \( 3R \) to \( 4R \)

#### Tasks

(a) **Electric Field Calculation:**

   Find the electric field in the regions:
   
   - \( R < r < 2R \)
   - \( 2R < r < 3R \)

(b) **Capacitance Calculation:**

   Determine the capacitance of this capacitor system.

(c) **Bound Surface Charge Density:**

   Calculate the bound surface charge density at:
   
   - \( r = R \)
   - \( r = 2R \)
   - \( r = 3R \) 

This problem facilitates understanding of electric fields in dielectric materials and the behavior of capacitors with complex geometries.
Transcribed Image Text:### Problem 6: Spherical Capacitor with Dielectric A spherical capacitor is composed of a solid conducting sphere concentrically placed within a larger spherical conducting shell. The following characteristics define the setup: - **Inner Conductor:** - Charge: Positive, total charge \( Q \) - Radius: \( R \) - **Outer Conductor:** - Charge: Negative, total charge \( -Q \) - Inner Radius: \( 3R \) - Outer Radius: \( 4R \) The space between the conductors is filled with two spherical dielectric shells: - **First Dielectric Region:** - Region: \( R < r < 2R \) - Dielectric Constant: \( K_1 = 6 \) - **Second Dielectric Region:** - Region: \( 2R < r < 3R \) - Dielectric Constant: \( K_2 = 2 \) #### Diagram Description A cross-sectional view of the spherical capacitor system is shown with colored regions: - Inner conductor at radius \( R \) - First dielectric shell in green from \( R \) to \( 2R \) - Second dielectric shell in blue from \( 2R \) to \( 3R \) - Outer conductor from \( 3R \) to \( 4R \) #### Tasks (a) **Electric Field Calculation:** Find the electric field in the regions: - \( R < r < 2R \) - \( 2R < r < 3R \) (b) **Capacitance Calculation:** Determine the capacitance of this capacitor system. (c) **Bound Surface Charge Density:** Calculate the bound surface charge density at: - \( r = R \) - \( r = 2R \) - \( r = 3R \) This problem facilitates understanding of electric fields in dielectric materials and the behavior of capacitors with complex geometries.
Expert Solution
Part a)

From Gauss's law1. For region R<r<2RElectric flux ϕ = E.ds = Qεbut ε = K1ε0so electric flux E.4πr2 = QK1ε0E = QK14πε0r2 we know 14πε0 = 9×109E = Q×9×1096×r2  = 3Q×1092r22. Electric flux ϕ = E.ds = 0εE = 03. For region 3R<r<4RElectric flux ϕ = E.ds = -Qεbut ε = K2ε0so electric flux E.4πr2 = QK2ε0E = QK24πε0r2 we know 14πε0 = 9×109E = Q×9×1092×r2  = 9Q×1092r2

steps

Step by step

Solved in 3 steps

Blurred answer
Similar questions
  • SEE MORE QUESTIONS