A particle of charge 3 x 10^-9 is located at point (0,0) in a certain coordinate system. Assuming all lengths are in meters, calculate the potential at the following points: A (3,3) and B (0,2). How much work is needed to take a particle of charge 2 x 10^-5 C from point A to point B?
A particle of charge 3 x 10^-9 is located at point (0,0) in a certain coordinate system. Assuming all lengths are in meters, calculate the potential at the following points: A (3,3) and B (0,2). How much work is needed to take a particle of charge 2 x 10^-5 C from point A to point B?
College Physics
11th Edition
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
Chapter1: Units, Trigonometry. And Vectors
Section: Chapter Questions
Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...
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A particle of charge 3 x 10^-9 is located at point (0,0) in a certain coordinate system. Assuming all lengths are in meters, calculate the potential at the following points: A (3,3) and B (0,2). How much work is needed to take a particle of charge 2 x 10^-5 C from point A to point B?
![Sample Problems 2.3
1. A point charge of – 6.00×10 ° C is 3.00 m from point A and 5.00 m from point B. (a) Find the
potential at point A and point B. (b) How much work is done by the electric field in moving a
2.00 nC particle from point A to point B?
Given: q=-6.0×10-°C
distance r, of point A from charge =3.00 m
distance rg of point B from charge=5.00 m
Solution:
(9×10° N·m²/C³X-6.00×10~° C)
kq
VA
"A
=-18.0 V
а.
%3D
3.00 m
(9×10° N-m²/C³X-6.00×10° C)
kq
VB
'B
=-10.8 V
5.00 m
b.
Using Eq. (2.3),
W
= (V – Vg)
W = q(V – Vg) =(2.00×10°C)[(-18.0 V)-(-10,8 V)]=1.44×10* J
5.00x10-7 C, q, = -3.00x10-7 C, q, = -2.00x10-7 C,
Four charges, 91 =
and q4 = 6.00×10-7 C, are situated at the corners of a square of side 4.00 m.
Find the potential at the center of the square.
2.
4 m
92
Given: 9, = 5.00×10-7 C
92 = -3.00×10-7 C
94
--- 4 m
93 = -2.00×10 ’ C
94 = 6.00×10-7 C
Solution:
All four charges are equidistant from the center of the square. Let d be the distance from
the center of the square to a corner. This distance is one half of the diagonal. The diagonal of the
square is computed using the Pythagorean theorem.
d =V(1.0 m)' +(4.0 m)² = 2.83 m
Using Eq. (2.4),
kq
V
d
(9×10° N m²/C²X5.00×10 'C)
=1590 V
2.83 m
(9×10’ N m²/C³)X-3.00×10 'C)
kq,
V2
d
=-954 V
2.83 m
kq,_ (9×10° N m² 'C³)(-2.00×10¯"C)
V,
-636 V
%3D
2.83 m
(9×10°N·m²/c³X6.00×10-°C)
V =
1910 V
2.83 m
Therefore, the potential V
at the center of the square is
center
= 1590 V – 954 V – 636 V + 1910 V = +1910 V.
V center
Practice Exercises 2.3
1. A particle of charge 3×10-9 C is located at point (0, 0) in a certain
coordinate system. Assuming all lengths are in meters, calculate the
potential at the following points: A (3, 3) and B (0, 2). How much work is
needed to take a particle of charge 2x10-5 C from point A to point B?
2. Equal charges of 3.00×10-9 C are situated at the three corners of a
of side 5.20 m. Find the potential at the unoccupied corner.
В
square
q= 3x10-9 C](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F324227e3-27b0-47cd-923a-655e2a6a59c8%2Febe5d76a-0e03-4867-a9b8-ad279f7e0b36%2Fnvazeay_processed.png&w=3840&q=75)
Transcribed Image Text:Sample Problems 2.3
1. A point charge of – 6.00×10 ° C is 3.00 m from point A and 5.00 m from point B. (a) Find the
potential at point A and point B. (b) How much work is done by the electric field in moving a
2.00 nC particle from point A to point B?
Given: q=-6.0×10-°C
distance r, of point A from charge =3.00 m
distance rg of point B from charge=5.00 m
Solution:
(9×10° N·m²/C³X-6.00×10~° C)
kq
VA
"A
=-18.0 V
а.
%3D
3.00 m
(9×10° N-m²/C³X-6.00×10° C)
kq
VB
'B
=-10.8 V
5.00 m
b.
Using Eq. (2.3),
W
= (V – Vg)
W = q(V – Vg) =(2.00×10°C)[(-18.0 V)-(-10,8 V)]=1.44×10* J
5.00x10-7 C, q, = -3.00x10-7 C, q, = -2.00x10-7 C,
Four charges, 91 =
and q4 = 6.00×10-7 C, are situated at the corners of a square of side 4.00 m.
Find the potential at the center of the square.
2.
4 m
92
Given: 9, = 5.00×10-7 C
92 = -3.00×10-7 C
94
--- 4 m
93 = -2.00×10 ’ C
94 = 6.00×10-7 C
Solution:
All four charges are equidistant from the center of the square. Let d be the distance from
the center of the square to a corner. This distance is one half of the diagonal. The diagonal of the
square is computed using the Pythagorean theorem.
d =V(1.0 m)' +(4.0 m)² = 2.83 m
Using Eq. (2.4),
kq
V
d
(9×10° N m²/C²X5.00×10 'C)
=1590 V
2.83 m
(9×10’ N m²/C³)X-3.00×10 'C)
kq,
V2
d
=-954 V
2.83 m
kq,_ (9×10° N m² 'C³)(-2.00×10¯"C)
V,
-636 V
%3D
2.83 m
(9×10°N·m²/c³X6.00×10-°C)
V =
1910 V
2.83 m
Therefore, the potential V
at the center of the square is
center
= 1590 V – 954 V – 636 V + 1910 V = +1910 V.
V center
Practice Exercises 2.3
1. A particle of charge 3×10-9 C is located at point (0, 0) in a certain
coordinate system. Assuming all lengths are in meters, calculate the
potential at the following points: A (3, 3) and B (0, 2). How much work is
needed to take a particle of charge 2x10-5 C from point A to point B?
2. Equal charges of 3.00×10-9 C are situated at the three corners of a
of side 5.20 m. Find the potential at the unoccupied corner.
В
square
q= 3x10-9 C
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