A thin, horizontal, 11-cm-diameter copper plate is charged to -2.9 nC. Assume that the electrons are uniformly distributed on the surface What is the strength of the electric field 0.1 mm above the center of the top surface of the plate? Answer: 1.7*104N/C What is the direction of the electric field 0.1 mm above the center of the top surface of the plate? Answer: toward the plate What is the strength of the electric field at the plate's center of mass? Answer:170311.8N/C What is the strength of the electric field 0.1 mm below the center of the bottom surface of the plate? Answer:1.7*104N/C What is the direction of the electric field 0.1 mmmm below the center of the bottom surface of the plate? toward the plate

Physics for Scientists and Engineers: Foundations and Connections
1st Edition
ISBN:9781133939146
Author:Katz, Debora M.
Publisher:Katz, Debora M.
Chapter25: Gauss’s Law
Section: Chapter Questions
Problem 47PQ: The infinite sheets in Figure P25.47 are both positively charged. The sheet on the left has a...
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A thin, horizontal, 11-cm-diameter copper plate is charged to -2.9 nC. Assume that the electrons are uniformly distributed on the surface

 

What is the strength of the electric field 0.1 mm above the center of the top surface of the plate?

Answer: 1.7*104N/C 

 

What is the direction of the electric field 0.1 mm above the center of the top surface of the plate?

Answer: toward the plate

 

What is the strength of the electric field at the plate's center of mass?

Answer:170311.8N/C

 

What is the strength of the electric field 0.1 mm below the center of the bottom surface of the plate?

Answer:1.7*104N/C

 

What is the direction of the electric field 0.1 mmmm below the center of the bottom surface of the plate?

toward the plate

with this problem I just want you to show me the steps how to solve 

 

 

 

Expert Solution
Step 1 strength of electric field

Strength of electric field is calculated using charge density. It is defined as the charge contained over a surface area.

Charge density=Chargesurface area

Electric field intensity=charge density2εo

Where, εo=8.85×10-12 F/m

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