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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Transcribed Image Text:Summary of Useful Formulas, Conversions, and Constants
Formulas:
E =
- clectric field (N/C)
= charge (Coulomb, C)
- distance (meters, m)
F = force (Newtons, N)
= current (Amperes, A)
= time (seconds, s)
V = voltage (volts, V)
R - resistance (Ohms, 2)
E = energy (Joules, J)
f
F = Eq
E
kq 92
F =
r
t
V = RI
Requivalent = R, + R, +R, (series)
1
1
1
+
(parallel)
= frequency (1/s Hertz, Hz)
2 = wavelength (meters, m)
= pressure (Pascals, Pa)
- volume (m')
T = temperature (Kelvin, K)
= number of moles (mol)
R. R R, R,
hc
E = hf =
PV = nRT
If the number of moles
is constant (n, = n2), then
PV P,Vr
%3D
T
Conversions:
1 inch = 2.54 cm
1 m3 = 1000 L
1 atm = 1.013 × 105 Pa
Kelvin = 273 + Celsius
STP = Standard Temperatur & Pressure (0°C and 1 atm)
Constants:
Molar Masses:
N.m2
k =9.0x10
Nitrogen, N:14
mol
e =1.60 x101 C
Air: 29 8
mol
R=8.314-
mol K
h = 6.626 x10 J.s
c = 3.0x10 m
Transcribed Image Text:ne College Sch...
CollegeBo
What is the magnitude of the electric field 1-inch
from a component that has a charge of 0.22 µC?
Report your answer rounded to the nearest .
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