Chapter 5: Electrostatics and Magnetism Flashcards

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1
Q

unit of charge (e)

A

e = 1.60 x 10^-19 C (coulomb)

protons are positively chraged and electrosn are negatively charged

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2
Q

insulator

A

will not easily distribute charge

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2
Q

conductor

A

will distribute approximately evenly upon the surface of the conductor

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3
Q

coulomb’s law

A

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

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4
Q

coulomb’s constant (k)

A

8.99 x 10^9 N . m^2 / C^2

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5
Q

eletric field

A

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

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6
Q

positive charges electric field

A

lines point outward

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7
Q

negative charges electric field

A

lines point inwards

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7
Q

electric potential energy

A

U = kQq / r

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7
Q

electric potential energy

A

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.

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8
Q

charges and electriv potential energy

A

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.

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8
Q

eletric potential

A

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.

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9
Q

1 Volt

A

1 J / C

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10
Q

potential difference

A

ΔV = Vb -Va = Wab / q

Wab = Work needed to move a test charge q through an electric field from A to point B.

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11
Q

For a positive test charge.

A

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.

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12
Q

For a negative test charge.

A

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)

13
Q

dipole moment (p)

A

p = qd

product ofcharge and separation distance

14
Q

cos 90 =

A

0

14
Q

magnetic field

A

tesla (T)
1T = 104 gauss

14
Q

torque on a dipole in an electric field

A

τ = p E sinθ

14
Q

magnetic field from a loop of wire

A

B =

15
Q

magnetic force on a movieng point charge

A

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

15
Q

magnetic force on current-carrying wire

A

Fb = I L B sinθ

I = current
L = lenght of the wire in the field

current is considered the flow of positive charge

16
Q

positive charges will move spontaneously from ___ potential to ___ potential

A

high / low

17
Q

negative charges will move spontaneously from ___ potential to ___ potential

A

low / high

18
Q

diamagnetic

A

no unpaired electrons

19
Q

paramagnetic

A

some umpaired, weakly magnetic

20
Q

ferromagnetic

A

some umpaired, strong magnetic