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' =

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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
.
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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