Cheat Sheet Physics 1 Exam

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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

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    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