Exam 1 Flashcards
q=Ne
q=
charge
q=Ne
N=
number of electrons
q=Ne
e=
charge on a single electron
charge is a _____ force
conservative
if an object is polarized, it means…
charges shift in a neutral object
conductors vs insulators
conductors:
metals
allow e- to transfer
insulators:
plastic
hold e- more tightly
µ =
10^-6
F=k (q1q2/r^2)
F=
force
F=k (q1q2/r^2)
k=
Coulomb’s constant
9x10^9
F=k (q1q2/r^2)
q=
charge on that specific object
F=k (q1q2/r^2)
r=
the distance between q1 and q2
given q=18µC and d=15cm, find magnitude and direction on force 1.
q1 = +q
q2 = -2.0q
q3 = +3.0q
-3.00µC in the center of a compass. 2 additional charges are on the center (r=.100m). -4.0µC due north and +5.0µC due east.
find the direction and magnitude of the charge in the center.
an electron moves at 8.5x10^5m/s. what is the radius?
+q placed in an electric field will accelerate _____ the field (gravity)
with
-q placed in an electric field will accelerate _____ the field (gravity)
against
density of electric field lines relates to the ______ of E
magnitude
2 point charges are placed on an x-axis. -5.5µC at the origin and 8.5µC ar x=10cm.
what is the net electric field at x=-4cm?
2 point charges are placed on an x-axis. -5.5µC at the origin and 8.5µC ar x=10cm.
what is the net electric field at x=4cm?
2 point charges are placed on an x-axis. -5.5µC at the origin and 8.5µC ar x=10cm.
where is the net field =0?
E+ = E-
0.41m left
an electron enters a capacitor with v=5.75x10^6m/s. it has a downward distance of d=0.638 and a horizontal length of 2.25cm.
what is the magnitude of the electric field?
an electron is placed between plates of a parallel plate capacitor. when released, it remains balanced.
what is the magnitude and direction of the electric field
F=ma
FE-Fg = 0
FE=Fg
qE=mg
E=mg/q
E= 5.58x10^-11 N/C downward
an electron is placed between plates of a parallel plate capacitor. when released, it remains balanced.
if gravity is turned off, what is the acceleration if a proton is released?
F=ma
0-FE=ma
-qE=ma
a=-qE/m
a=5.34x10^-3 m/s^2 downward
ideal conductors have
_____ charge
fields _____ and at _____
and _____ shielding
excess
inside and at surface
electron shielding
point charge:
E=kd/r^2
voltage = ________
electrical potential
do you use signs in k(q1q2/r^2)
NO!
moving against gravity = ______ height
gaining
moving against electric field = ______ voltage
gain
moving with the electric field = ______ voltage
lose
W = -q∆V
W=
work
W = -q∆V
q=
charge
W = -q∆V
V=
change in potential/voltage
E=- ∆V/∆d
E=
electric field
E=- ∆V/∆d
V=
potential difference
E=- ∆V/∆d
d=
distance between plates
____ decreases when you move with gravity
height
______ decreases when you move with the electric field
potential/voltage
a ______ mass accelerates against the gravitational field from a high to low gravitational height
positive
a positive mass accelerates _____ the gravitational field form a high to low gravitational height
against
a ______ charge accelerates with the electric field from a high to low electrical height
positive
a positive charge accelerates _____ the electric field from a high to low electrical height
with
a positive charge is moving with the field
q∆V = -We = ∆EPE
q∆V = -We = ∆EPE
+ - - (+) -
a negative charge is moving with the field
q∆V = -We = ∆EPE
q∆V = -We = ∆EPE
- - - (-) +