3. Suppose that the charges in Sample Problems 1.4 item number 3 are situated as shown. Find the resultant electric force on the charge at A. +q b. 1.00 m В 1.00 m C D b- +9
3. Suppose that the charges in Sample Problems 1.4 item number 3 are situated as shown. Find the resultant electric force on the charge at A. +q b. 1.00 m В 1.00 m C D b- +9
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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Suppose that the charges in Sample Problems 1.4 item number 3 are situated as shown, find the resultant electric force on the charge at A.
![Suppose that the charges in Sample Problems 1.4 item number 3 are
situated as shown. Find the resultant electric force on the charge at A.
3.
+q
1.00 m
A
В
1.00 m
D
b.
+q](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F181b4c54-5bd7-4aaa-b706-3a5c053d162a%2F8ba61279-aeac-484a-942c-42ff56ce76a7%2Fp81hr5_processed.png&w=3840&q=75)
Transcribed Image Text:Suppose that the charges in Sample Problems 1.4 item number 3 are
situated as shown. Find the resultant electric force on the charge at A.
3.
+q
1.00 m
A
В
1.00 m
D
b.
+q
![Four point charges (two with q=2.50×10-6C and two with
q=-2.50×10-6 C) are situated at the corners of a square of side
1.00 m as shown. Find the resultant force that the charge at A will
experience due to the charges at the other corners of the square.
+9
3.
+q
Given: 9,=9B=+2.50×10¬6 C
r=1.0 m
Ic=9p=-2.50×10-6C
Solution:
Eq. (1.1) is used to solve for the magnitude of the electric force exerted by each of the
charges at the other corners of the square on the charge at A.
Fs on A = k-
|(+2.50×10¬°C)(+2.50×10¯“C)|
= 19x10° N-m²/C"|
(1.00 m)²
= 0.05625 N20.06 N
Feom A = k
|(+2.50×10“CX-2.50×10°C
=19x10° N m²/C²|
(1.00 m)?
= 0.05625 N20.06 N
Solve first for the distance between D and A. Using the Pythagorean theorem,
'D on A =
("B on A
V(1.00 m)’ +(1.00 m)² = 1.414 m~1.41 m.
Therefore,
Fo on A = k-
=[9x10° N m²/C²| (+2.50×10°C)X-2.50×10°C||
(1.414 m)²
= 0.02813 N20.03 N.
FB on A is repulsive. Since the charges are both positive, the charge
at B will push the charge at A away from it. Therefore, FB on A is directed
to the left. Fcon A is attractive since the charges are opposite. The charge
at C will attract the charge at A toward it. Therefore, Fc on A is directed
down. Fpon A is attractive since the charges are opposite. The charge at D
will attract the charge at A toward it. Therefore, F on A is directed down
at 45° with the +x-axis, with q, placed at the origin of the x-y plane as
shown in the diagram on the right. All other forces are also shown in
this diagram.
Fooma
Foona
Fcon A
The table below presents the respective horizontal and vertical components of these forces
and the resultant electric force.
Force
Horizontal Component
Vertical Component
FoonA = 0.05625 N
–0.05625 N
FconA=0.05625 N
–0.05625 N
Foona = 0.02813 N
+0.02813 N (cos 45°) =+0.0199
-0.02813 N (sin 45°)=-0.0199 N
Resultant electric force F
EF,=-0.03635 N
EF,=-0.07615 N
Solving for the magnitude of the resultant electric force,
F =
V(-0.03635 N)' +(-0.07615 N)²
= 0.0844 N.
Solving for the direction,
|EF,
0 = tan
EF
TA
--0.07615 N
64,5°
= tan
|-0.03635 N
= 64.5°.
Thus, the resultant force is 0.0844 N directed down
at 64.5° with the negative x-axis as shown.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F181b4c54-5bd7-4aaa-b706-3a5c053d162a%2F8ba61279-aeac-484a-942c-42ff56ce76a7%2Fopbvfa_processed.png&w=3840&q=75)
Transcribed Image Text:Four point charges (two with q=2.50×10-6C and two with
q=-2.50×10-6 C) are situated at the corners of a square of side
1.00 m as shown. Find the resultant force that the charge at A will
experience due to the charges at the other corners of the square.
+9
3.
+q
Given: 9,=9B=+2.50×10¬6 C
r=1.0 m
Ic=9p=-2.50×10-6C
Solution:
Eq. (1.1) is used to solve for the magnitude of the electric force exerted by each of the
charges at the other corners of the square on the charge at A.
Fs on A = k-
|(+2.50×10¬°C)(+2.50×10¯“C)|
= 19x10° N-m²/C"|
(1.00 m)²
= 0.05625 N20.06 N
Feom A = k
|(+2.50×10“CX-2.50×10°C
=19x10° N m²/C²|
(1.00 m)?
= 0.05625 N20.06 N
Solve first for the distance between D and A. Using the Pythagorean theorem,
'D on A =
("B on A
V(1.00 m)’ +(1.00 m)² = 1.414 m~1.41 m.
Therefore,
Fo on A = k-
=[9x10° N m²/C²| (+2.50×10°C)X-2.50×10°C||
(1.414 m)²
= 0.02813 N20.03 N.
FB on A is repulsive. Since the charges are both positive, the charge
at B will push the charge at A away from it. Therefore, FB on A is directed
to the left. Fcon A is attractive since the charges are opposite. The charge
at C will attract the charge at A toward it. Therefore, Fc on A is directed
down. Fpon A is attractive since the charges are opposite. The charge at D
will attract the charge at A toward it. Therefore, F on A is directed down
at 45° with the +x-axis, with q, placed at the origin of the x-y plane as
shown in the diagram on the right. All other forces are also shown in
this diagram.
Fooma
Foona
Fcon A
The table below presents the respective horizontal and vertical components of these forces
and the resultant electric force.
Force
Horizontal Component
Vertical Component
FoonA = 0.05625 N
–0.05625 N
FconA=0.05625 N
–0.05625 N
Foona = 0.02813 N
+0.02813 N (cos 45°) =+0.0199
-0.02813 N (sin 45°)=-0.0199 N
Resultant electric force F
EF,=-0.03635 N
EF,=-0.07615 N
Solving for the magnitude of the resultant electric force,
F =
V(-0.03635 N)' +(-0.07615 N)²
= 0.0844 N.
Solving for the direction,
|EF,
0 = tan
EF
TA
--0.07615 N
64,5°
= tan
|-0.03635 N
= 64.5°.
Thus, the resultant force is 0.0844 N directed down
at 64.5° with the negative x-axis as shown.
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