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57 Cards in this Set
- Front
- Back
1/(L-x^2) approximately equal to |
(1/L^2)(1+2x/L) |
|
1/(L+x^2) approximately equal to |
(1/L^s)(1-2x/L) |
|
p is the |
electric dipole moment vector |
|
p = |
qd |
|
positive always points |
away |
|
negative always points |
towards |
|
F = in terms of E field |
qE |
|
angular frequency of oscillation, omega = |
sqrt(k/m) |
|
frequency = in terms of omega |
omega/(2pi) |
|
the field at a point is ... |
the vector sum of the fields of each of the charges in the system |
|
continuous charge distribution (rather than a point charge), requires ... |
an integral |
|
charge (q) = |
density times area or volume (depending on the case) |
|
hemisphere E = |
Ey = (ro k)/(4 e0) |
|
the E field due to hemisphere is ... |
only vertical because horizontal components cancel |
|
the flux is zero when |
E field is parallel to plane |
|
flux = |
E * total area = (kqArea)/r^2 |
|
for sphere, flux = |
q/e0 |
|
flux is independent of |
the size of the surface |
|
e0 = |
1/(4pi k)
|
|
Gauss' law only holds for ... |
spheres, cylinders (infinite line of charge), and planes |
|
For spheres, E is ... |
radially directed and the same at all points on the sphere |
|
volume of a sphere |
(4/3) pi r^3 |
|
always find the volume of ... |
the inner radius |
|
field of line charge ... dq = |
lambda dx |
|
Esphere = |
Q/(4 pi e0 r^2) |
|
Eline = |
lambda / (2 pi e0 r) |
|
E sheet = |
sigma/ (2 e0) |
|
grad f = |
direction of most rapid increase in f |
|
for uniformly charged disk, the total charge of the system = |
pi r^2 sigma |
|
on a cube, E at the surface is |
not constant |
|
E field is zero when |
V(x) has a slope of zero; dV/dx = 0 Ex = -dV/dx |
|
E inside of a conductor = |
0 because charge moves freely on the surface |
|
If E not equal to 0 |
then charge moves and experiences a force (F=qE) |
|
For conductor, V = |
constant |
|
To find force applied by a charged object on a test charge ... |
find E, then use F=qE |
|
Finite charge distribution |
V = int k (dq/r) |
|
For a point charge, E = |
(1/4 pi e0) Q/r^2 |
|
For a point charge, V(r) = |
(1/4 pi e0) Q/r |
|
k = (numerical) |
9 x 10^9 Nm^2/C^2 |
|
Even if charges are unequal, the magnitude of the forces... |
must be equal |
|
Infinite line of charge, charge density = |
lambda |
|
infinite line of charge, Ex = |
int dEx = 0 |
|
infinite line of charge, Ey = |
int d Ey = 2k lamba / r |
|
E field from uniformly charged rod at -a: |
Ex = - int [from 0 to L] kQ/(L(a+x)^2) dx |
|
surface area of a sphere |
4pi r^2 |
|
If E is parallel to dAL |
int E da = 0 |
|
if E is constant at the surface, we can |
pull E out as a constant, E int dA = EA |
|
Infinite line of charge flux = |
(lambda L/e0) = because lambda L = Qenclosed |
|
Gaussian surface over dipoles, E is |
not constant at the surface --> we can't write E int dA |
|
Gaussian surface for sphere (r) that is larger than original sphere (R), E = |
Q/(4 pi e0 r^2) r>R |
|
Gaussian surface for sphere (r) that is smaller than original sphere (R), E = |
ro r/ (3 e0) r |
|
for solid sphere conductor, charges can |
move freely
|
|
solid sphere conductor, Einside = |
0 |
|
solid sphere conductor, qinside = |
0 |
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Gaussian cylindrical surface (r) that is larger than original cylinder (R), E= |
lamda/ (2 pi e0 r) r> R |
|
when finding gaussian surface of a sphere, the volume used for Qenclosed always uses... |
the smaller radius
|
|
gaussian surfaces: if hollow, charge distribution is on the surface, so utilize volume by using ... |
surface area |