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7/30/2019 Cheat Sheet Physics 1 Exam
1/2
Power=
V^2/r
If Elec F is only F then
particle accelerating
A=(q/m)E
F(12)=
by q1
on q2
The energy stored is (1/2)CV^2
Parallel conducting plates
The field is E = V/d =
12/0.004 = 3000 V/m =
3000 N/C
force of charge in field =
Eq
7/30/2019 Cheat Sheet Physics 1 Exam
2/2
Electric potential V
V is path independent
V=kq/r
Work place 3 charges
W(q1)=0
W(q2)=q(2)xV
=q(2)xkq(1)/r
W(q3=q(3)((kq(1)/r+kq(2)/r
Wtot=w3+w2Linear charge density electric
field
E(y)= kQ/(y*sqrt((L/2)^2+y^2))
E(x)=kq/(x(o)(x(o)-L))
kQ/x^2 (point part)
Outward flux ? = EAx/2 Flux = enclosed charge/E(o))
A charge is placed uniformly on a square sheet
nonconducting on yz plane:
Surface charge density = Q/A
Mag of E.feild to right and left of sheet
E=s/2E(o)
if the electric field is 0 in some region of space, theelectric potential must also be 0 in that region
FALSE
If the electric potential is 0 in some region of space,
the electric field must also be 0 in that region TRUE
If the electric potential is 1 at a point, the electric
field must also be0 at that point FALSE
Electric field lines always point toward regi ons of
lower potential TRUE
The value of the electric potential can be chosen to
be 0 at any convenient point TRUE
Dielectric breakdown occurs in the air when the
potential is 3*10^6 V true
C=S/d is the initial capacitanceof the parallel-plate capacitor
C1=S/(d+d) is the capacitance
after moving the plates
U=Q/C
U+U=Q/C1 is the voltage after
moving the plates ==>
Qd/(S)+U=Q(d+d)/(S) ==>
U=Q*d/(S)
Q=S*U/d
(b) W=0.5QU initial energy
W1=0.5Q(U+U)
W=W1-W=0.5Q*U
SPEED FRM ORIGIN
The distance from each point to
the origin is a*sqrt(2)
so V = k*q/r
so we get V = k/(a*sqrt(2))*(q +
2q -3q+6q) = 6/sqrt(2)*(kq/a) =
4.24*k*q/a
so the potential energy = V*q =
4.24*k*q^2/a
At infinity this potential energy is
now kinetic
1/2*m*v^2 = 4.24*k*q^2/a
so v = sqrt(2*4.24*k*q^2/(m*a))
= sqrt(8.48*k*q^2/(m*a))
+ -
Electric flux
= EA when Elec Field is perp to
AreaNot perp =EAcos()
For curved surface delta =EdeltaAcos()
When the side is
perp to field there is
no flux
The negative plate is at the
lower potential if we are
using the classical reference
that current (not electrons)
flows from positive to
negative.
The work done by the
electric field is the force
on the electron times
the distance it is moved
through