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Example: Electric Field and Potential Inside a Charged Sphere Problem: A sphere of radius R = 0.2 m is uniformly charged with a total charge Q = 5 μC. The sphere is made of a dielectric material with relative permittivity € = 4. Calculate: 1. The electric field intensity E(r) inside and outside the sphere. 2. The electric potential (r) at any point inside the sphere. Solution: Step 1: Given Data Radius of the sphere: R = 0.2m, Total charge: Q-5 μC=5× 10° C. Step 2: Electric Field Inside the Sphere (< Using Gauss's Law:

Example: Electric Field and Potential Inside a Charged Sphere Problem: A sphere of radius R = 0.2 m is uniformly charged with a total charge Q = 5 μC. The sphere is made of a dielectric material with relative permittivity € = 4. Calculate: 1. The electric field intensity E(r) inside and outside the sphere. 2. The electric potential (r) at any point inside the sphere. Solution: Step 1: Given Data Radius of the sphere: R = 0.2m, Total charge: Q-5 μC=5× 10° C. Step 2: Electric Field Inside the Sphere (< Using Gauss's Law:

Delmar's Standard Textbook Of Electricity
Delmar's Standard Textbook Of Electricity
7th Edition
ISBN:9781337900348
Author:Stephen L. Herman
Publisher:Stephen L. Herman
Chapter19: Capacitors
Section: Chapter Questions
Problem 3RQ: A capacitor uses air as a dielectric and has a capacitance of 3 F. A dielectric material is inserted...
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Example: Electric Field and Potential Inside a Charged Sphere Problem: A sphere of radius R = 0.2 m is uniformly charged with a total charge Q = 5 μC. The sphere is made of a dielectric material with relative permittivity € = 4. Calculate: 1. The electric field intensity E(r) inside and outside the sphere. 2. The electric potential (r) at any point inside the sphere. Solution: Step 1: Given Data Radius of the sphere: R = 0.2m, Total charge: Q-5 μC=5× 10° C. Step 2: Electric Field Inside the Sphere (< Using Gauss's Law: <R) Permittivity of free space: Co 8.854 × 10-12 F/m, D.dA=qenclosed A • Relative permittivity: €, = 4, D= €E, and enclosed p. V. Permittivity of the dielectric: = c=coc, 8.854 x 10-12 x 43.54 × 10" F/m. The charge density p is uniform, so: Q 5 × 10-6 P = 0.149 C/m³. R³ (0.2)³ The enclosed charge at radius is:
Transcribed Image Text:Example: Electric Field and Potential Inside a Charged Sphere Problem: A sphere of radius R = 0.2 m is uniformly charged with a total charge Q = 5 μC. The sphere is made of a dielectric material with relative permittivity € = 4. Calculate: 1. The electric field intensity E(r) inside and outside the sphere. 2. The electric potential (r) at any point inside the sphere. Solution: Step 1: Given Data Radius of the sphere: R = 0.2m, Total charge: Q-5 μC=5× 10° C. Step 2: Electric Field Inside the Sphere (< Using Gauss's Law: <R) Permittivity of free space: Co 8.854 × 10-12 F/m, D.dA=qenclosed A • Relative permittivity: €, = 4, D= €E, and enclosed p. V. Permittivity of the dielectric: = c=coc, 8.854 x 10-12 x 43.54 × 10" F/m. The charge density p is uniform, so: Q 5 × 10-6 P = 0.149 C/m³. R³ (0.2)³ The enclosed charge at radius is:
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