3.7 Required: Given, : distance between the charger, r = ? magnitude of equal charges, 9₁ = 92 = 75×10 °c Force between charges, F = 1 N Formula used: F = ± 4760 21%2 342 [Coulomb's Law] Now F = 1 47760 9192 912 05, 2 = 4760 9192 F = 10 x 750 75×10 9 • [uh% = 2xv° _svw/e2] 1 0% 4² = 9x75x75 × 10 ×109 6 2² = 50.625×10 05 M = √50.625×10 - 7.12×10 m or, -3 Criven, Hence, M = 7.12×10 m the required distance is 7.12×103 m 4.7 Required: magnitude of force, F = ? magnitude of equal charges, a₁ = 92 = distance M= 1×103³ m Formula used, F = L URGO 9192 242 Now, F = 9x10 x 35×35 (1x103)2 0% F = 1.1×10 N Hence, the required force is 1. 1×10t N 35 C 9x109 x 35x35 1×106 = 1.1×10 N 7
3.7 Required: Given, : distance between the charger, r = ? magnitude of equal charges, 9₁ = 92 = 75×10 °c Force between charges, F = 1 N Formula used: F = ± 4760 21%2 342 [Coulomb's Law] Now F = 1 47760 9192 912 05, 2 = 4760 9192 F = 10 x 750 75×10 9 • [uh% = 2xv° _svw/e2] 1 0% 4² = 9x75x75 × 10 ×109 6 2² = 50.625×10 05 M = √50.625×10 - 7.12×10 m or, -3 Criven, Hence, M = 7.12×10 m the required distance is 7.12×103 m 4.7 Required: magnitude of force, F = ? magnitude of equal charges, a₁ = 92 = distance M= 1×103³ m Formula used, F = L URGO 9192 242 Now, F = 9x10 x 35×35 (1x103)2 0% F = 1.1×10 N Hence, the required force is 1. 1×10t N 35 C 9x109 x 35x35 1×106 = 1.1×10 N 7
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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Re write again text form not handwriting. Also need substitutions please do it. Both Physics expert
![3.7
Required:
Given,
:
distance between the charger, r = ?
magnitude of equal charges, 2₁ = 92 = 75×10 °c
Force
between charges, F = IN
Formula used:
F = ±
4πGo
2192
+2
[Coulomb's Law]
Now
F =
1
476
9192
912
or, ² =
9192
4760
=
F
10 x 75x100 5x109
• ³
[zum%% = 9x10 _8vw/e2]
1
0%
r² =
9x 75 x 75 x 10
×109
6
or
2² =
50.625×10
6
-3
05
r =
50.625×10
= 7.12×15
m
Criven,
Hence,
or M = 7.12×10³m
the required distance is 7·12×103 m
4.7 Required:
magnitude of force
~ F = ?
magnitude of equal charges, 9₁ = 92 =
distance
35 C
r =
1×10³m
Formula used,
F = 1
9192
42
Now,
F = 9×10 x 35×35
(1x103)2
9×10 x 35×35
7
= 1.1×10 N
1×106
0%
F = 1.1×10 N
Hence, the required force is
1.1×10t N](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fa917bcce-fd4c-4ad0-a3b5-845f6a1c8aca%2F0b7afc26-0f9f-4d4a-883a-e90a3e2abfe7%2F1gojdkl_processed.jpeg&w=3840&q=75)
Transcribed Image Text:3.7
Required:
Given,
:
distance between the charger, r = ?
magnitude of equal charges, 2₁ = 92 = 75×10 °c
Force
between charges, F = IN
Formula used:
F = ±
4πGo
2192
+2
[Coulomb's Law]
Now
F =
1
476
9192
912
or, ² =
9192
4760
=
F
10 x 75x100 5x109
• ³
[zum%% = 9x10 _8vw/e2]
1
0%
r² =
9x 75 x 75 x 10
×109
6
or
2² =
50.625×10
6
-3
05
r =
50.625×10
= 7.12×15
m
Criven,
Hence,
or M = 7.12×10³m
the required distance is 7·12×103 m
4.7 Required:
magnitude of force
~ F = ?
magnitude of equal charges, 9₁ = 92 =
distance
35 C
r =
1×10³m
Formula used,
F = 1
9192
42
Now,
F = 9×10 x 35×35
(1x103)2
9×10 x 35×35
7
= 1.1×10 N
1×106
0%
F = 1.1×10 N
Hence, the required force is
1.1×10t N
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