Electricity

Question 1

What is the fundamental unit of charge?

Answer 1

e=1.602*10^-19 Coulombs (+ for proton, - for electron)

Question 2

What is the unit for charge?

Answer 2

A coulomb which is a charge carried by current of 1 amp/s. So its Amp*s

Question 3

Coulumb’s Law

Answer 3

F=kQq/r^2

Question 4

Electric Field Equation

Answer 4

E=F/q = (kQq/r^2)/q= kQ/r^2 F=qE -Force divided by charge. -Recall that F in electricity can be calculated using coulombs law so you can sub it in where you see F

Question 5

Torque in an electric field

Answer 5

T=FdsinΘ= qE*d*sinΘ When in a field, a dipole will rotate until it is properly oriented/aligned in the field. Any force that causes rotation is considered torque. Recall the force in an electric field is F=qE

Question 6

Conductivity

Answer 6

σ=1/𝙥

Question 7

Conductance

Answer 7

G=σ(A/l)

Question 8

Resistivity

Answer 8

𝙥=1/σ

Question 9

Resistance

Answer 9

R=𝙥(l/A)

Question 10

Electric Potential Energy

Answer 10

W=F*d=(kQq/r^2 )*r= kQq/r = U Measured in JOULES

Question 11

Electric Potential

Answer 11

V=U/q= (kQq/r)/q= kQ/r measured in JOULES/ COULOMB The electric potential energy divided by charge

Question 12

What is an equation that relates work and voltage?

Answer 12

𝚫V= W/q

Question 13

True or False: If a charge travels along an equipotential line no work is done, if it travels from one equipotential line to another, then it does work

Answer 13

True

Question 14

An atom is electrically neutral until it gains 4 electrons and 1 proton. What would be the final charge on the atom?

Answer 14

The charge carried by a single electron is negative 1.6 x 10-19 C. The charge carried by a single proton is positive 1.6 x 10-19. Therefore, to find the net change in charge of an atom after gaining 4 electrons and 1 proton, the number of total subatomic particles gained must be multiplied by the charge carried by each particle, which is then added together. Charge carried by electrons: -1.6 x 10-19 * 4 electrons = - 6.4 * 10-19 C Charge carried by protons: 1.6 x 10-19 * 1 proton = 1.6 x 10-19 C Adding the charges together: -6.4 x 10-19 C + 1.6 x 10-19 C = -4.8 x 10-19 C

Question 15

How would you calculate a force give the following information: 10 Coulombs, 4 Coulombs, 2 meters

Answer 15

For 10 Coulombs, 4 Coulombs, 2 meters: Use coulumbs law F = k * (10 * 4) / 22 = k * 40 / 4 = k * 10 N

Question 16

True or False: C= 1/R

Answer 16

True, Conductance is equal to 1 divided by resistance. Therefore, this equation correctly describes a way for calculating conductance.

Question 17

An unknown subatomic particle is dropped into the electric field around an unknown charge, at the point indicated by the green arrow.If the unknown particle moves left, what is the charge on it?

Answer 17

Negatively charged particles move opposite the direction of electric field lines. Therefore, the particle is negatively charged.

Question 18

True or false: An electron placed in a dipole will move toward the midpoint of that dipole, and then oscillate forever around that dipole.

Answer 18

This statement is false. A dipole is a linear object with positive and negative charge at either end. An electron would move down the electron field lines of a dipole, toward the positive end of the dipole.

Question 19

A proton travels through an electric field, moving through a potential difference (ΔV) of -20 V. What can be said regarding work in this scenario?

Answer 19

Positive work occurs when a force is exerted on a particle in the same direction in which the particle is moving. Because the potential difference is negative, the proton must have moved from a region of high to low electric potential, which would be along the direction of the electric field lines. Therefore, the electric force exerted on the proton would have been in the direction in which it was moving, so the work done on the proton was positive.

Another way to think about this is that, since the proton moved from a region of high to low potential, it must have gained kinetic energy. This means that the net work done on it must be positive.

Question 20

The voltage of the equipotential line at P2 minus the voltage of the equipotential line at P1 (ΔV) is most likely:

Answer 20

The two points are located on the same equipotential line, meaning they are at regions of equal electric potential. Therefore, P2 – P1 must be equal to zero.

Question 21

Electric Force is measured in….

Answer 21

J/m

The units of force are newtons. Newtons times meters equals joules – therefore, joules divided by meters is an appropriate unit to describe force.

Question 22

Electric potential energy is measured in….

Answer 22

Electric potential energy describes a change in energy of an object as it moves through an electric field, described by joules.

Question 23

Electric potential is measured in…

Answer 23

Electric potential is the amount of energy per unit of charge needed to move a charged object through a field. This can be described by joules per coulomb.

Question 24

Electric field is measured in…

Answer 24

N/C. Electric fields extend outwards from an electric charge. The value for electric field will tell you how many newtons of force the electric field will exert on an object per unit of charge, at some point in that field.

Question 25

What are the units for the following:

Charge

Voltage

Electric Field

Resistance

Capacitance

Current

Answer 25

Charge: coulumbs

Voltage: Volts J/C

Electric Field:

Resistance: Ohms V/A

Capacitance: farrads C/V

Current: amps C/s

Question 26

What is Farraday’s Constant?

Answer 26

96,500 Coulumbs/Mol