Chapter 5: Electrostatics and Magnetism Flashcards
unit of charge (e)
e = 1.60 x 10^-19 C (coulomb)
protons are positively chraged and electrosn are negatively charged
insulator
will not easily distribute charge
conductor
will distribute approximately evenly upon the surface of the conductor
coulomb’s law
Fe = k q1 q2 / r^2
Fe = magitude of the electrostatic force
Cq1 and q2 = magnitud eof two charges
k = coulomb’s constant
r = distance between the harges
coulomb’s constant (k)
8.99 x 10^9 N . m^2 / C^2
eletric field
E = Fe / q = kQ / r^2
E = eletric field magnitude
Fe = magnitude of the force felt by the test charge q
k = coulomb’s constant
Q = souce charge magnitude
r = distance between charges
positive charges electric field
lines point outward
negative charges electric field
lines point inwards
electric potential energy
U = kQq / r
electric potential energy
Electric potential energy is the work necessary to move a test charge from Infinity to a point in space in an electric field surrounding a source charge.
charges and electriv potential energy
The electric potential energy of a system will increase when two like chargers move towards each other or when two opposite charges move apart. The electric potential energy for a system would decrease when two like charges move apart or when two pause opposite charges move towards each other.
eletric potential
V = kQ / r
V = electric potential measured in volts (V)
For a positive source charge, V is positive but for a negative sorce charge V is negative.
Electric potential is the ratio of the work done to move a test charge from Infinity to a point in an electric field surrounding a source charge divided by the magnitude of the test charge.
1 Volt
1 J / C
potential difference
ΔV = Vb -Va = Wab / q
Wab = Work needed to move a test charge q through an electric field from A to point B.
For a positive test charge.
Moving from a position of higher electric potential to a position of lower electric potential. The voltage is negative in this case because q is positive. Thus, the Wab must be negative, which represents a decrease in electric potential energy.
For a negative test charge.
Move from a position of lower electric potential to a position of high electric potential. The voltage is positive in this case because q is negative. And was also be negative, which again represents a decrease in electric potential energy. The take away is that positive charges will spontaneously move from the direction that decreases their electric potential. ( negative voltage). In negative charges will spontaneously move in a direction that increases their electric potential (positive voltage)
dipole moment (p)
p = qd
product ofcharge and separation distance
cos 90 =
0
magnetic field
tesla (T)
1T = 104 gauss
torque on a dipole in an electric field
τ = p E sinθ
magnetic field from a loop of wire
B =
magnetic force on a movieng point charge
Fb = q v B sinθ
v = magnitud eof its velocity
B = magnitude of the magnetic field
θ = angle between the velocity anf thr magnetic field
magnetic force on current-carrying wire
Fb = I L B sinθ
I = current
L = lenght of the wire in the field
current is considered the flow of positive charge
positive charges will move spontaneously from ___ potential to ___ potential
high / low
negative charges will move spontaneously from ___ potential to ___ potential
low / high
diamagnetic
no unpaired electrons
paramagnetic
some umpaired, weakly magnetic
ferromagnetic
some umpaired, strong magnetic