Summary of Gauss Law Problems, Hollow Charged Spherical Conductor With Cavity, Electric Field, Physics

Summary of Main Ideas and Concepts

This video explains how to apply Gauss's Law to analyze electric fields in hollow spherical conductors, both with and without charges inside cavities. The key points covered include:

Electric Field inside a Hollow Spherical Conductor with no charge in the cavity is zero.
How to calculate Surface Charge Density on the conductor.
How to find the Electric Field outside the conductor, treating it as a Point Charge.
Effect of placing a charge inside the cavity on the distribution of charges on the conductor and the resulting electric fields at various distances.

Detailed Explanation and Methodologies

Part A: Electric Field at the Center of a Hollow Spherical Conductor (No Charge Inside)

Given: Hollow Spherical Conductor radius \( R = 3 \, m \).
Electric Field inside a conductor = 0.
Even if there is a cavity inside the conductor, if no charge is placed inside the cavity, the Electric Field at the center is zero.
Use Gauss's Law:
- \(\Phi_E = E \times A = \frac{Q_{\text{enclosed}}}{\epsilon_0}\)
- Construct a Gaussian Surface inside the cavity.
- Since \(Q_{\text{enclosed}} = 0\), and \(A\) is finite, \(E = 0\).

Part B: Surface Charge Density on the Hollow Sphere

Surface Charge Density \(\sigma = \frac{Q}{A}\).
Surface area of sphere \(A = 4 \pi R^2\).
Given charge \(Q = 20 \, \mu C = 20 \times 10^{-6} \, C\), radius \(R = 3 \, m\).
Calculation:
\[ \sigma = \frac{20 \times 10^{-6}}{4 \pi (3)^2} = 1.768 \times 10^{-7} \, C/m^2 \]

Part C: Electric Field at a Distance Outside the Sphere (5 m from center)

For \(r > R\), the hollow charged sphere behaves like a Point Charge.
Use Gauss's Law again:
\[ E \times 4 \pi r^2 = \frac{Q}{\epsilon_0} \implies E = \frac{Q}{4 \pi \epsilon_0 r^2} \]
Using Coulomb's Constant \(k = \frac{1}{4 \pi \epsilon_0} = 9 \times 10^9\),
\(Q = 20 \times 10^{-6} C\), \(r = 5 m\),
Calculate:
\[ E = k \frac{Q}{r^2} = 9 \times 10^9 \times \frac{20 \times 10^{-6}}{25} = 7200 \, N/C \]

Case Study: Hollow Sphere with Charge Inside the Cavity

Inner charge \(q = -8 \, \mu C\) (negative charge at center).
Total charge on spherical conductor \(Q = 15 \, \mu C\) (positive).

Charge distribution:

Induced charge on inner surface of conductor = \(+8 \, \mu C\) (opposite to inner charge).
Remaining charge on outer surface = \(15 - 8 = 7 \, \mu C\).

Electric Field calculations at different radii:

Define three radii:
- \(R_1 = 1\, m\) (inside cavity, near inner charge)
- \(R_2 = 3\, m\) (inside conductor material)
- \(R_3 = 5\, m\) (outside conductor)
At \(R_1\):
- Gaussian Surface encloses only inner charge \(-8 \, \mu C\).
- Electric Field:
  \[ E_1 = k \frac{8 \times 10^{-6}}{R_1^2} \]
At \(R_2\):
- Gaussian Surface encloses inner charge \(+8 \, \mu C\) induced on conductor interior plus \(-8 \, \mu C\) inside cavity.
- Total enclosed charge = 0.
- Electric Field \(E_2 = 0\).
At \(R_3\):
- Gaussian Surface encloses total net charge on conductor \(7 \, \mu C\).
- Electric Field:
  \[ E_3 = k \frac{7 \times 10^{-6}}{R_3^2} \]

Key Lessons and Concepts

Gauss's Law is a powerful tool for analyzing electric fields in conductors and cavities, allowing simplification of complex charge distributions into manageable calculations.

Notable Quotes

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