Problem 1: Consider a parallel-plate capacitor made up of two conducting plates with dimensions 39 mm × 32 mm. Part (c) What is the magnitude of the electric field, in newtons per coulomb, inside this capacitor? E=5.34 * 104 E = 5.340 x 104 ✓ Correct! Part (d) If the separation between the plates doubles, what will the electric field be if the charge is kept constant? E' =
Problem 1: Consider a parallel-plate capacitor made up of two conducting plates with dimensions 39 mm × 32 mm. Part (c) What is the magnitude of the electric field, in newtons per coulomb, inside this capacitor? E=5.34 * 104 E = 5.340 x 104 ✓ Correct! Part (d) If the separation between the plates doubles, what will the electric field be if the charge is kept constant? E' =
Physics for Scientists and Engineers, Technology Update (No access codes included)
9th Edition
ISBN:9781305116399
Author:Raymond A. Serway, John W. Jewett
Publisher:Raymond A. Serway, John W. Jewett
Chapter24: Gauss’s Law
Section: Chapter Questions
Problem 24.46P: A thin, square, conducting plate 50.0 cm on a side lies in the xy plane. A total charge of 4.00 108...
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![Problem 1: Consider a parallel-plate capacitor made up of two conducting plates with dimensions 39 mm × 32 mm.
Part (c) What is the magnitude of the electric field, in newtons per coulomb, inside this capacitor?
E=5.34 * 104
E = 5.340 x 104 ✓ Correct!
E' =
Part (d) If the separation between the plates doubles, what will the electric field be if the charge is kept constant?
tan() I ( ) 7 8 9 HOME
acos()
EMAL
4 5 6
sinh()
1 2
3
cotanh()
0
NO BACKSPACE
sin()
cos)
cotan()
asin()
atan() acotan()
cosh()
tanh()
Degrees O Radians
+
-
END
DEL CLEAR
.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F3cff6d5e-6d7b-4ef6-9838-a544f8e2903b%2F7b118339-ee9a-45c1-a34b-f4b9c83a37c7%2Fy2hwrks_processed.png&w=3840&q=75)
Transcribed Image Text:Problem 1: Consider a parallel-plate capacitor made up of two conducting plates with dimensions 39 mm × 32 mm.
Part (c) What is the magnitude of the electric field, in newtons per coulomb, inside this capacitor?
E=5.34 * 104
E = 5.340 x 104 ✓ Correct!
E' =
Part (d) If the separation between the plates doubles, what will the electric field be if the charge is kept constant?
tan() I ( ) 7 8 9 HOME
acos()
EMAL
4 5 6
sinh()
1 2
3
cotanh()
0
NO BACKSPACE
sin()
cos)
cotan()
asin()
atan() acotan()
cosh()
tanh()
Degrees O Radians
+
-
END
DEL CLEAR
.
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