Homework Help is Here – Start Your Trial Now!
Engineering
Electrical Engineering
Two 1.9 cm-diameter disks face each other, 1.3 mm apart. They are charged to   ±± 16 nC.   1. What is the electric field strength between the disks?Express your answer to two significant figures and include the appropriate units. 2. A proton is shot from the negative disk toward the positive disk. What launch speed must the proton have to just barely reach the positive disk? Express your answer to two significant figures and include the appropriate units.

Two 1.9 cm-diameter disks face each other, 1.3 mm apart. They are charged to   ±± 16 nC.   1. What is the electric field strength between the disks?Express your answer to two significant figures and include the appropriate units. 2. A proton is shot from the negative disk toward the positive disk. What launch speed must the proton have to just barely reach the positive disk? Express your answer to two significant figures and include the appropriate units.

Introductory Circuit Analysis (13th Edition)
Introductory Circuit Analysis (13th Edition)
13th Edition
ISBN:9780133923605
Author:Robert L. Boylestad
Publisher:Robert L. Boylestad
Chapter1: Introduction
Section: Chapter Questions
Problem 1P: Visit your local library (at school or home) and describe the extent to which it provides literature...
See similar textbooks
icon
Related questions
Question

Two 1.9 cm-diameter disks face each other, 1.3 mm apart. They are charged to   ±± 16 nC.

 

1. What is the electric field strength between the disks?Express your answer to two significant figures and include the appropriate units.

2. A proton is shot from the negative disk toward the positive disk. What launch speed must the proton have to just barely reach the positive disk?
Express your answer to two significant figures and include the appropriate units.
Expert Solution
Step 1

Given

q=16 nc D=1.9cm

or D=0.019m

d=1.3mm

or 

d=0.0013m

Since these parallel plates each other they produces uniform field inside in ot

The electric Field formula is given by

E=qεoA

Where A=πD24

Therefore E=4qεoD2π

 

Therefore E=4×16×10-98.854×10-12×0.0192×π

Therefore E=6.37357×106 N/C

trending now

Trending now

This is a popular solution!

steps

Step by step

Solved in 3 steps

SEE SOLUTIONCheck out a sample Q&A here
Blurred answer
Knowledge Booster
Magnetic moment
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, electrical-engineering and related others by exploring similar questions and additional content below.
Similar questions
Recommended textbooks for you
Introductory Circuit Analysis (13th Edition)
Introductory Circuit Analysis (13th Edition)
Electrical Engineering
ISBN:
9780133923605
Author:
Robert L. Boylestad
Publisher:
PEARSON
Delmar's Standard Textbook Of Electricity
Delmar's Standard Textbook Of Electricity
Electrical Engineering
ISBN:
9781337900348
Author:
Stephen L. Herman
Publisher:
Cengage Learning
Programmable Logic Controllers
Programmable Logic Controllers
Electrical Engineering
ISBN:
9780073373843
Author:
Frank D. Petruzella
Publisher:
McGraw-Hill Education
Fundamentals of Electric Circuits
Fundamentals of Electric Circuits
Electrical Engineering
ISBN:
9780078028229
Author:
Charles K Alexander, Matthew Sadiku
Publisher:
McGraw-Hill Education
Electric Circuits. (11th Edition)
Electric Circuits. (11th Edition)
Electrical Engineering
ISBN:
9780134746968
Author:
James W. Nilsson, Susan Riedel
Publisher:
PEARSON
Engineering Electromagnetics
Engineering Electromagnetics
Electrical Engineering
ISBN:
9780078028151
Author:
Hayt, William H. (william Hart), Jr, BUCK, John A.
Publisher:
Mcgraw-hill Education,
SEE MORE TEXTBOOKS