Re write again text form not handwriting. Also need substitutions please do it. Both Physics expert F1-7 Required :- Electric field intensity, E= ?Criven, Force, F = 5NChange, 9 = 6×10 6 cFormula used,E = F/8Now,E = F5.6×10605E = 8.33×105 N/CHence, the required electric field ins8.33× 105 N/C 2.7Required, magnitude of charge, 9=?Given,Electric field, E = 3.4 N/CFormula used,distance,Bulting the value.E =r =4760 423.4 = 9x10° x $(25×102)29x109x9625 x 104or 3.4=3)♡ = 3.4×625×15425×102т=9x10³ Nrt/c²471609x10918 = 2.36 × 10" CHence, the magnitude of charge is2.36×10с

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Re write again text form not handwriting. Also need substitutions please do it. Both Physics expert

University Physics Volume 2

18th Edition

ISBN:9781938168161

Author:OpenStax

Publisher:OpenStax

Chapter6: Gauss's Law

Section: Chapter Questions

Problem 68P: Two parallel conducting plates, each of cross-sectional area 400 cm2, are 2.0 cm apart and...

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Question

Re write again text form not handwriting. Also need substitutions please do it. Both Physics expert

F
1-7 Required :- Electric field intensity, E= ?
Criven, Force, F = 5N
Change, 9 = 6×10 6 c
Formula used,
E = F/8
Now,
E = F
5.
6×106
05
E = 8.33×105 N/C
Hence, the required electric field ins
8.33× 105 N/C

2.7
Required, magnitude of charge, 9=?
Given,
Electric field, E = 3.4 N/C
Formula used,
distance,
Bulting the value.
E =
r =
4760 42
3.4 = 9x10° x $
(25×102)2
9x109x9
625 x 104
or 3.4
=
3)
♡ = 3.4×625×154
25×102
т
=
9x10³ Nrt/c²
47160
9x109
18 = 2.36 × 10" C
Hence, the magnitude of charge is
2.36×10
с

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