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PHY2054

Terms in this set (79)

A small metal ball is given a negative charge, then brought near to end A of the rod (positive charge on end, but mixed with neg. charge in center). What happens to end A of the rod when the ball approaches it closely this first time?
It is attracted.
After a great many contacts with the charged ball, how is the charge on the rod arranged (positive charge on end, but mixed with neg. charge in center);(when the charged ball is far away)?
There is negative charge spread evenly on both ends.
How does end A of the rod react when the charged ball approaches it after a great many previous contacts with end A? Assume that the phrase "a great many" means that the total charge on the rod dominates any charge movement induced by the near presence of the charged ball.
It is repelled. on both ends
When a honeybee flies through the air, it develops a charge of +14pC.How many electrons did it lose in the process of acquiring this charge?
8.75*10^7electrons
note: N=q/e-> e=1.6*10^-19 C
pC=10^-12C
Two identical metal spheres A and B are in contact. Both are initially neutral. 1.0×10^12 electrons are added to sphere A, then the two spheres are separated.
a. Afterward, what is the charge of sphere A?
b. Afterward, what is the charge of sphere B?
-8*10^-8C or 80 nC
note: qa=qb
Q=ne
1*10^12 electrons will split evenly into A and B
Q will be negative because we added electrons to the neutral system
If two identical conducting spheres are in contact, any excess charge will be evenly distributed between the two. Three identical metal spheres are labeled A, B, and C. Initially, A has charge q, B has charge −q/2, and C is uncharged.What is the final charge on each sphere if C is touched to B, removed, and then touched to A?
3q/8
note:
Q=(q1+q2)/2
C->B then C->A
Coulomb's law for the magnitude of the force F between two particles with charges Q and Q′ separated by a distance d
is |F|=K|QQ′|d2, where K=14πϵ0, and ϵ0=8.854×10−12C2/(N⋅m2) is the permittivity of free space. Consider two point charges located on the x axis: one charge, q1= -20.0 nC , is located at x1 = -1.685 m ; the second charge, q2 = 34.0 nC , is at the origin (x = 0).What is (Fnet3)x, the x-component of the net force exerted by these two charges on a third charge q3 = 46.5 nC placed between q1 and q2 at x3 = -1.200 m? Your answer may be positive or negative, depending on the direction of the force.
(Fnet3)x=−4.54×10−5N
note:coulomb's law (F1 on2=(k|q1|q2|)/r^2)
do this for each force, then Fnet=F1on2+F3on2...
A small metal bead, labeled A, has a charge of 27 nC . It is touched to metal bead B, initially neutral, so that the two beads share the 27 nC charge, but not necessarily equally. When the two beads are then placed 5.0 cm apart, the force between them is 5.4×10−4 N . Assume that A has a greater charge.What is the charge qA and qB on the beads?
+x=19*10^-9 C or 19nC
-x=8*10^9 C or 8nC
note: F=(k|q1|q2|)/r^2), find one of the q's and place in
q1+q2=27nC -> use quadratic equation
A small metal sphere has a mass of 0.16 g and a charge of -23.0 nC. It is 10.0 cm directly above an identical sphere that has the same charge. This lower sphere is fixed and cannot move. If the upper sphere is released, it will begin to fall. What is the magnitude of its initial acceleration?
F1 on 2=4.8*10^-4N
note: use F=(k|q1|q2|)/r^2)
a=-6.6m/s^2
note: Fnet=F1 on2-m2g=m2a
its going down (-)
qa=1 nc and is 1 cm from B. qb=-1nC and is 1 cm from C. qc=4nC. What is the magnitude of the electric force on charge A
0, and the force pulling on A is zero.
note: use F=(k|q1|q2|)/r^2), but draw a picture to see what direction the elements are moving. (a proton and electron are attracted to each other, so B pulls a in a positive direction and C repells A in a negative direction.)
charge q2 experiences no net electric force. q1 is 10 cm away from negative middle charge and another 10 cm away from q2. the middle negative charge is -2nC. frind the charge of q1.
8nC
note: use Fnet 2=F1 on 2+F-2nC on 2=0 (static)
draw a FBD to know if directions of forces are positive or negative.
A 3.0-mm-diameter copper ball is charged to 50 nC . What fraction of its electrons have been removed? The density of copper is 8900 kg/m3. Find the fraction of electrons removed.
.
note: 9.039*10^-12
mball=(density)(volume)
find the moles (n) -> find no of atoms with avagadros (6.02*10^23)->N0
N=electrons removed is N=q/e.
put N/N0
Two equally charged, 1.00 g spheres are placed with 2.00 cm between their centers. When released, each begins to accelerate at 225 m/s2. What is the magnitude of the charge on each sphere?
Q =100 nC
note: F21=F12=kq1q2/r^2=ma
three positive charges make a triangle each 1cm apart. and each angle at 60 degrees. there is 1nC between 2 qs with 2nC. What is the force on the 1.0nC charge in ? Give the magnitude. Give the direcion.
Fon1-3.12*10^-4
theta:90
note:
F21=(k|q1|q2|)/r^2) and for F31=F=(k|q1|q3|)/r^2)
x-component: Fcos60
y-component:Fsin60
*x will cancel
use sqrt(Fnetx-comp^2+Fnety-comp^2)
theta=tan^-1(fy/fx)
*fx is zero-> und. -> tan is 90
lesson of HW #1
draw the FBD of forces. to calculate Fnet
x-comp. use cos, and y-comp use sin
4 charges. q1= 1nC, q2=2nC, q3=-6nC, q4=2nC. q2 is 5cm above and to the left of q1. q4 is 5cm above and to the right of q1. q3 is above q1 and between q2 and q4. the theta from q2 to q4 is 90 degrees. What is the magnitude of the force on the 1.0 nC charge at the bottom of the figure?What is the direction of the force on the 1.0 nC charge at the bottom of the figure?
1.14*10^-5N
90 degrees
note:
find the x and y components of each force on 1. note if the force is away or toward the the point charges. find the Fnet= sqrt(Fnetx-comp^2+Fnety-comp^2)
theta=tan^-1(fy/fx)
a. what must the charge (sign and magnitude) of a particle of mass 1.42 g be for it to remain stationary when placed in a downward-directed electric field of magnitude 700 N/C ?

b. What is the magnitude of an electric field in which the electric force on a proton is equal in magnitude to its weight?
Use 1.67×10−27 kg for the mass of a proton, 1.60×10−19C for the magnitude of the charge on an electron, and 9.80 m/s2 for the magnitude of the free-fall acceleration.
a. −1.99×10^−5C
note: F=qE=mg -> q=mg/E

b. 1.02×10−7 N/C
note: E=mg/q
when is an electric field zero?
at equilibrium when the force is zero if the charges are not zero
Two point charges are placed on the x axis as shown in (Figure 1). The first charge, q1 = 8.00 nC , is placed a distance 16.0 m from the origin along the positive x axis; the second charge, q2 = 6.00 nC , is placed a distance 9.00 m from the origin along the negative x axis.
a. Find the x-component of the electric field at the origin, point O.
Express your answer in newtons per coulomb to three significant figures, keeping in mind that an x component that points to the right is positive.
b. Now, assume that charge q2 is negative; q2=−6nC, as shown in (Figure 2). What is the x-component of the net electric field at the origin, point O?
Express your answer in newtons per coulomb to three significant figures, keeping in mind that an x component that points to the right is positive.
EOx = .385 N/C
note: Enet=kq1/r1^2(i)+kq2/r2^2(i)
(.385, 0)
EOx = -.947 N/C
Enet=kq1/r1^2(-i)+kq2/r2^2(-i)
(-.947,0)
What magnitude charge creates a 1.0 N/C electric field at a point 1.0 m away?
ans: .11nC
note: E=kq/r^2-> q=Er^2/k
A bumblebee can sense electric fields as the fields bend hairs on its body. Bumblebees have been conclusively shown to detect an electric field of 60 N/C. Suppose a bumblebee has a charge of 24 pC. How far away could another bumblebee detect its presence?
.06m
note: r=sqrt(kq/E)
a dot is 5 cm, 0 cm . 2 positive charges each with 1 nC are at points (0,5cm and 0,-5cm).
a. What is the strength of the electric field at the position indicated by the dot in (Figure 1)?
b. What is the direction of the electric field at the position indicated by the dot in the figure? Specify the direction as an angle above the horizontal line.
a. 2545.58 N/C
note: E=kq/r^2 draw the electric feild lines to the dot. notice it is a triangle 5,5,5sqrt2
find Ex-component=Ecos 45 and
Ey-comp=Esin45
b. 0
the x components cancel
A parallel-plate capacitor is constructed of two square plates, size L×L, separated by distance d. The plates are given charge ±Q. Let's consider how the electric field changes if one of these variables is changed while the others are held constant.
a. What is the ratio Ef/Ei of the final electric field strength Ef to the initial electric field strength Ei if Q is doubled?
b. What is the ratio Ef/Ei of the final electric field strength Ef to the initial electric field strength Ei if L is doubled?
c. What is the ratio Ef/Ei of the final electric field strength Ef to the initial electric field strength Ei if d is doubled?
a. 2
b. 4
c. 1
note: use the equation E=Q/(epsilon knot)(Area)
epsilon knot=8.85*10^-12 -dont need it here
Area=L*L
Ef=Qf/Af Ei=Qi/Ai
distance does not effect this equation
A parallel-plate capacitor is formed from two 4.0 cm ×4.0 cm electrodes spaced 2.2 mm apart. The electric field strength inside the capacitor is 1.0×10^6N/C.
a. What is the charge (in nC) on the positive electrode?
b. What is the charge (in nC) on the negative electrode?
a. +14.2nC
b.-14.2nC
note: Q=E(epsilon knot)(Area)
A parallel-plate capacitor is constructed of two horizontal 17.6-cm-diameter circular plates. A 1.1 g plastic bead with a charge of -4.4 nC is suspended between the two plates by the force of the electric field between them.
What is the charge on the positive plate?
Q=5.3*10^-7C
note: E=Q/A(episolon knot)
E=F/q=mg/q; use the beads information
find Q=A(epsilon knot)E(^)
A=pir^2
What is the strength of an electric field that will balance the weight of an electron?(up/down)
5.57*10^-11N/C
electric field down
note: E=mg/q
electron is neg. Charged
What is the strength of an electric field that will balance the weight of a proton? (E up/down)?
1.02*10^-7 N/C
electric feild up
p+ is positive charged
Raindrops acquire an electric charge as they fall. Suppose a 2.4-mm-diameter drop has a charge of +13 pC. In a thunderstorm, the electric field under a cloud can reach 12,000 N/C, directed upward.

a. For a droplet exposed to this field, how do the magnitude of the electric force compare to those of the weight force?

b. What is the direction of the electric force?
a. 2.2*10^-3
b. up
note:
Fe=Eq
Fg=mg=density(volume)(gravity)
Felec<Fg upward E
The nucleus of a 125Xe atom (an isotope of the element xenon with mass 125 u) is 6.0 fm in diameter. It has 54 protons and charge q=+54e (1 fm = 1 femtometer = 1×10−15m.) Hint: Treat the spherical nucleus as a point charge.

a. What is the electric force on a proton 2.0 fm from the surface of the nucleus?

b. What is the proton's acceleration?
F=498N
a=2.98*10^29 m/s^2
note: Fnuc on p+=kqq/r^2
a=F/m
four charges at the corners of a square (L*L). top left corner is -10nC, top right is +q, bottom left is Q, bottom right is -10nC.
What magnitude and sign of charge Q will make the force on charge q zero?
Q=28nC
note: call left corner q1, bottom right corner q 3
make a FBD (x comp of Q and 3 is cos)
find x comp (each F...on q=k...q/r^2)
Q to q is sqrt (2)L
use Newton 2nd Law: F (1 on q)=-F(3 on q)
In a simple model of the hydrogen atom, the electron moves in a circular orbit of radius 0.053nm around a stationary proton.How many revolutions per second does the electron make? Hint: What must be true for a force that causes circular motion?
f=6.6*10^15revs/sec
F1on2=kq1q2/r^2
centripetal acceleration= mv^2/r =mw^2r w=2pif
f=(1/2pi)sqrt(ke^2/mr^3)
The electron gun in a television tube uses a uniform electric field to accelerate electrons from rest to 5.0×10^7 m/s in a distance of 1.2 cm. What is the electric field strength?
E=5.9*10^5N/C
note: uniform feild=e- will be constant
F=qE=ma
vf^2=vi^2+2a(deltax)
A 0.024 g plastic bead hangs from a lightweight thread. Another bead is fixed in position beneath the point where the thread is tied. One bead swings out 5 cm at a 45 degree angle. If both beads have charge q, the moveable bead swings out to the position.What is q?
q=7.4nC
note:F=mg=T
find Y comp: F=Tcos45=mg -> T=mg/cos45
find x- comp:sum of Fnetx=(kqq/r^2)-(Tsin45)=0
q=r sqrt(mgtan45/k)
An electric field E⃗ =( 2.30×105 N/C , right) causes the 2.0 g ball in to hang at an angle. q=25 nC. What is θ?
14.297 degrees
note: F=qE
F=mg
theta=tan^-1(y/X) -> tan^-1(qE/mg)
A parallel-plate capacitor consists of two plates, each with an area of 25 cm2 separated by 3.0 mm. The charge on the capacitor is 9.8 nC . A proton is released from rest next to the positive plate. How long does it take for the proton to reach the negative plate?
t=1.14*10^-8s
note:
E=Q/(epsilon knot)(Area)
F=Eq
a=F/m
v=sqrt(2ax)
t=v/a
Hw #2 highlights
using variables of speed with charge, area, distance, diameter, time, force, tension, gravity, x and y components, pendulum problems, electric feild
It takes 3.5 μJ of work to move a 14 nC charge from point A to B. It takes -5.0 μJ of work to move the charge from C to B.
What is the potential difference VC−VA ?
607V
note: V=w/q
find the V from A to B and from B to C
(Vb-Va)-(Vb-Vc)=(Vc-Va)
What potential difference is needed to accelerate a He+ ion (charge +e, mass 4u) from rest to a speed of 1.6×10^6 m/s ?
-53440V
note: KE=-qV=1/2mv^2 ->V=-mv^2/2q
An electron with an initial speed of 460,000 m/s is brought to rest by an electric field.

a.What was the potential difference that stopped the electron?

b. What was the initial kinetic energy of the electron, in electron volts?
a. -.602 V
note: KE=-qV
b. .6 eV
note: KEi= (mv^2)/2
A proton with an initial speed of 700,000 m/s is brought to rest by an electric field.
a. What was the potential difference that stopped the proton?

b. What was the initial kinetic energy of the proton, in electron volts?
2557.18 V
note: KE=(mv^2)/2=e(V knot)
V knot=potential diffrence
V= (.5 mv^2)/e

2557.18eV
note: KE=1/2 mv^2 K=W/q
The electric potential at a point that is halfway between two identical charged particles is 210 V .What is the potential at a point that is 25% of the way from one particle to the other?
440 V
note: find V=kq/r with middle point
electric point at the middle v=kq/(r/2)+kq/(r/2)
V=kq/(3r/4)+kq(r/4)
solve for V
A 1.0 cm × 1.0 cm parallel-plate capacitor has a 3.0 mm spacing. The electric field strength inside the capacitor is 1.5×105 V/m .
a. What is the potential difference across the capacitor?

b. How much charge is on each plate?
a. 450 V
note: deltaV=Ed
b. +/- 1.327*10^-10
note: C=A(epsilon knot)/d
Q=CV
Two 2.70 cm × 2.70 cm plates that form a parallel-plate capacitor are charged to ± 0.708 nC .

a. What is the electric field strength inside the capacitor if the spacing between the plates is 1.50 mm ?

b. What is potential difference across the capacitor if the spacing between the plates is 1.50 mm ?

c. What is the electric field strength inside the capacitor if the spacing between the plates is 3.00 mm ?

d. What is the potential difference across the capacitor if the spacing between the plates is 3.00 mm ?
a. 2*10^5N/C
note: E=q/A(epsilon knot)

b. 300V
note: deltaV=Ed

c. 2*10^5 N/V
note: E=q/A(epsilon knot)

d. 600
note: deltaV=Ed
Parallel plate capactior is 3mm apart. the right sided plate( positive is 300V). and the left sided negative plate is 0V.

a. What is the electric field strength inside the capacitor?

b. What is the potential energy of a proton at the midpoint of the capacitor?
a. 10^5 V/m
note:E=V/d

b. 2.4*10^-17J
note: U=qdeltaV
delta V=150
there is a proton with 3 charges around it (A, B, and C).
q=2 nC, da=1cm, db=1cm, dc=2cm
a. What is the potential difference ΔVAB?

b. What is the potential difference ΔVBC?
a. 0V
note: v=kq/r
Find Va and Vb
vb-va
b. -900V
find Vc
vc-vb
3 point charges and a dot, making a box shape. The dot is in the lower right corner, 2nC is in all other corners. The box is 3cm*4cm. What is the electric potential at the point indicated with the dot?
V=14.1*10^2V
note: 345 tirangle
V=kq/r
Vnet=V1+V2+V3...
The electric potential at point A is -400 V. What is the potential at point B, which is 8.5 cm to the right of A? the E field is 1200V/m, and B is moving to the upper right at a 30 degree angle.
Vb=-488.32V
deltaV=Ed
a proton sits at the origin. point A is 5 cm above the proton and point B is 1.5 cm to the right of the proton. the proton is 25nC, and the points are q=3nC. What is the change in electrical potential energy of a 2.0 nC point charge when it is moved from point A to point B in the figure?
delta U=3.15*10^-5 J
note: U=kq1q2/r12
Find UA and UB
UB-UA=deltaU
A -3.0 nC charge is on the x-axis at x = -9.0 cm and a +4.0 nC charge is on the x-axis at x = 19 cm .
At what point or points on the y-axis is the electric potential zero?
+/-16.7 cm
note: V=kq/r
Vnet=V1+V2=0 -> Newtan's 2nd
q-P=sqrt[x^2+(0-d^2)] find both r1 and r2
kq2/r2=-kq1/r1
A proton follows the path shown in (Figure 1). Its initial speed is v0 = 2.4×106 m/s.What is the proton's speed as it passes through point P?
1.2*10^6m/s
conservation of Energy: K2+U2=K1+U1
Ei=Ef
1/2mvi^2+kqq/r1=1/2mvf^2+kqq/r2
A parallel-plate capacitor is charged to 5000 V. A proton is fired into the center of the capacitor at a speed of 2.3×10^5 m/s as shown in (Figure 1). The proton is deflected while inside the capacitor, and the plates are long enough that the proton will hit one of them before emerging from the far side of the capacitor.What is the impact speed of the proton?
7.3*10^5m/s
note: F=qE=ma-> a=qE/m
Vf^2=Vi^2+2ax
1. Two 4.0 cm ×4.0 cm square aluminum electrodes, spaced 0.70 mm apart, are connected to a 300 V battery.
a. What is the capacitance?
b. a. What is the charge on the positive electrode?
a. 20.2pF or 20.2*10^-12F
note: C=A(epsilon knot)/d
b. 6*10^-9C
note: q=CV
An uncharged capacitor is connected to the terminals of a 4.0 V battery, and 6.0 μC flows to the positive plate. The 4.0 V battery is then disconnected and replaced with a 8.0 V battery, with the positive and negative terminals connected in the same manner as before.
6*10^-6C
note: C=Q1/V1=Q2/V2
deltaQ=Q2-Q1
You need to construct a 300 pF capacitor for a science project. You plan to cut two L×L metal squares and place spacers between them. The thinnest spacers you have are 0.30 mm thick.
The capacitance of a parallel plate capacitor is given byC= k e0 A /dwhere k is the relative permittivity of the dielectric between the plates , assume air=1
10.082 cm
note:A= (C d) / (k e0)
L=sqrt(A)
A 2.7 nF parallel-plate capacitor has an air gap between its plates. Its capacitance increases by 3.5 nF when the gap is filled by a dielectric.
2.296
note:C=epsilon*A/d
electric permittivity=dielectric constant*epsilon
(capicetance increase+parallel plate capacitance)/parallel plate capacitance
*dilectric constant usually found on a table
A parallel-plate capacitor is charged by a 20.0 V battery, then the battery is removed.

a. What is the potential difference between the plates after the battery is disconnected?

b. What is the potential difference between the plates after a sheet of Teflon is inserted between them? (teflon k=2)
a. 20V
note: Assuming an ideal capacitor. When the battery is removed after charging the capacitor to a potential difference of 20 V. The potential difference will still remain at 20V.
b. 10V
note:
V'=Q'/C' -> Q=kCV'-> V'=Q/kC->V/k=Q/C
20V/2
1. Capacitor 2 has half the capacitance and twice the potential difference as capacitor 1.
a. What is the ratio (UC)1/(UC)2.
1/2
note:
Uc1=1/2C1V1^2
Uc2=1/2C2V2^2
C2=1/2C1
100 pJ of energy is stored in a 1.0 cm × 1.0 cm × 1.0 cm region of uniform electric field.
a. What is the electric field strength?
4753.82 V/m
note:E/Volume=1/2(epsilon)E^2
a triangle is formed by q1=2nC on the top, 3cm above two qs each with the charge -1nC and 3cm away from each other. a dot is in the middle of the triangle.
a. What is the electric potential at the point indicated with the dot?
0V
note: Vdot=kq/d
the distance sqrt(3)
Two 2.1-cm-diameter-disks spaced 1.8 mm apart form a parallel-plate capacitor. The electric field between the disks is 5.0×105 V/m .
a. What is the voltage across the capacitor?
b. How much charge is on each disk?
c. An electron is launched from the negative plate. It strikes the positive plate at a speed of 2.5×107 m/s . What was the electron's speed as it left the negative plate?
a. 1000V
note:
V=Ed

b.1.4*10^-9C
C=(epsilon knot)(A)/d
A=pir^2
Q=CV

c. 7*10^6 m/s
note: a=Ee/m
vf^2-Vi^2=2ax
vi=sqrt(vf^2-2ax)
A parallel-plate capacitor is connected to a battery and stores 3.5 nC of charge. Then, while the battery remains connected, a sheet of Teflon is inserted between the plates. For the dielectric constant, use the value from Table 21.3 (teflon is 2)
By how much does the charge change?
3.5nC
note: Cplate=k(C knot)
7-3.5=3.5
A proton is fired from far away toward the nucleus of an iron atom. Iron is element number 26, and the diameter of the nucleus is 9.0 fm. (1fm=10−15m.) Assume the nucleus remains at rest.
a. What initial speed does the proton need to just reach the surface of the nucleus?
note: use conservation of energy
1/2(mv^2)=(qkZe)/r
Z=element atomic number
e=1.6*10^-19
use mass,q, of proton
find v
Suppose that a = 1.5 cm . point one 50V and is 2a away from 0V. point 2 is 50 V and 1a away from 0V.
a. Determine the magnitude of the electric field at point 1
b. Determine the magnitude of the electric field at point 2.
a. |E|=862.07V/m down
note: E=-deltaV/deltax

b. |E|=1724.17 V/m up
note: E=-deltaV/deltax
A capacitor consists of two 7.0-cm-diameter circular plates separated by 1.0 mm. The plates are charged to 130 V , then the battery is removed.

a. How much energy is stored in the capacitor?
4.352*10^-7J
note:
d= total distance travelled
Conservation; Ei+PEi=KEf+PEf
C=A(epsilon knot)/d
U=1/2CV^2

Q=CV
C'=A(epsilon knot)/d'
U'=Q^2/2C'

W=U'-U
concept: which way does and e- travel through a battery with a wire connected to both terminals
e- travel from postive to neg in a battery to increase its potentail energy. This movement causes thermal energy in a wire where potentil energy is lost and chemical energy to be the expense.
Rank in order, from largest to smallest, the resistivities ρ1 to ρ5 of these wires. Three wires meet at a junction. Wire 1 has a current of 0.40 A into the junction. The current of wire 2 is 0.67 A out of the junction
ρ1: r, L
ρ2: 2r, L
ρ3: 2r,2L
ρ4: r, 2L
ρ5: 2r, 4L
ρ2>ρ3>ρ1=ρ5>ρ4
note: ρ=AR/L
*R=resistance
ρ=resistivity
ex.ρ1=(Rpir^2)/L
In proton beam therapy, a beam of high-energy protons is used to deliver radiation to a tumor, killing its cancerous cells. In one session, the radiologist calls for a dose of 4.9×108 protons to be delivered at a beam current of 120 nA.
t=6.53*10^-4 sec
q= i*t
A 9.5 V battery supplies a 1.5 mA current to a circuit for 7.0 h.

a. How much charge has been transferred from the negative to the positive terminal?
Express your answer with the appropriate units.


b. How much electric potential energy has been gained by the charges that passed through the battery?
Express your answer with the appropriate units.
359.1J
note: Q=It
W=QV
A 9.0 V potential difference is applied between the ends of a 0.60-mm-diameter, 65-cm-long nichrome wire.
What is the current in the wire?
I=2.6A
note:
Area=(pid^2)/4
R=(resistivity of chrome)(length)/Area
Resistivity of chrome=1.5´10^-6
V=IR->I=V/R
The length of a 60 W, 240 Ω light bulb filament is 60 cm.
If the potential difference across the filament is 120 V, what is the strength of the electric field inside the filament?
Suppose the length of the bulb's filament were doubled without changing its diameter or the potential difference across it. What would the electric field strength be in this case?

Remembering that the current in the filament is proportional to the electric field, what is the current in the filament following the doubling of its length?

What is the resistance of the filament following the doubling of its length?
Find the total charge that flows during the pulse.
a. 345 W
note: Power = V*i

b.0.345 J
note: Energy = Power*time

c. Q=7.5´10^-4C
When running on its 11.4 V battery, a laptop computer uses 8.3 W. The computer can run on battery power for 5.0 h before the battery is depleted.

a. What is the current delivered by the battery to the computer?
b. How much energy, in joules, is this battery capable of supplying?
c. How high off the ground could a 75 kg person be raised using the energy from this battery?
a. I=.728 A
note: I=P/V
b. R=15.6J
note:R=V/I
c.H=204 m
note:E=I^2(t)R
E=mgh->h= E/mg
The biochemistry that takes place inside cells depends on various elements, such as sodium, potassium, and calcium, that are dissolved in water as ions. These ions enter cells through narrow pores in the cell membrane known as ion channels. Each ion channel, which is formed from a specialized protein molecule, is selective for one type of ion. When an ion channel opens in a cell wall, monovalent (charge e) ions flow through the channel at a rate of 1.0×107ions/s.

What is the current through the channel?

The potential difference across the ion channel is 70 mV. What is the power dissipation in the channel?
a. 1.6e-12
note: I=Q/t
b. P=1.12*10^-13
note: P=IV
The total charge a battery can supply is rated in mA⋅h, the product of the current (in mA) and the time (in h) that the battery can provide this current. A battery rated at 1000 mA⋅hr can supply a current of 1000 mA for 1.0 h, 500 mA current for 2.0 h, and so on. A typical AA rechargeable battery has a voltage of 1.2 V and a rating of 1800 mA⋅h.
For how long could this battery drive current through a long, thin wire of resistance 22 Ω?
Express your answer in hours.
T=33h
note: I=V/R
Rate=1800mAh->1800´10^-3 Ah
T=(1800´10^-3)/ .0545A
1. You've made the finals of the Science Olympics! As one of your tasks, you're given 1.0 g of copper and asked to make a wire, using all the metal, with a resistance of 9.0 Ω . Copper has a density of 8900 kg/m3.
What length will you choose for your wire?

What diameter will you choose for your wire?
a. 7.75
note: R=rho(L)/A->rho(L^2)/A(l)->(rho)(L^2)/V
b. 1.35*10^-4
note:
A=pi(d/2)^2
A=L(rho)/R
d^2=4(L)(rho)/piR
Three wires meet at a junction. Wire 1 has a current of 0.40 A into the junction. The current of wire 2 is 0.67 A out of the junction.
Part A
What is the magnitude of the current in wire 3?
I3=.27A
note: kirchhoff's current law
I1+I2+I3=0-> -I1+I2=I3
current flows out of junction
note:(+ into junction and - is out of junction)
An electric eel develops a potential difference of 460 V , driving a current of 0.75 A for a 1.0 ms pulse.
Find the power of this pulse.
Find the total energy of this pulse.
Find the total charge that flows during the pulse.
a.345W
note: P=VI

b. .345 J
note: E=Pt

c. 7.5*10^-4C
note: Q=It
What is the time constant for the discharge of the capacitors in the figure (Figure 1)?
2
note:
A 20 μFμF capacitor initially charged to 30 μCμC is discharged through a 1.5 kΩkΩ resistor.
How long does it take to reduce the capacitor's charge to 10 μCμC?
Express your answer with the appropriate units.
33 ms
Q=Qie(-t/RC)
take ln of both sides
Initially, the switch in the figure is in position A and capacitors C2C2C_2 and C3C3C_3 are uncharged. Then the switch is flipped to position B.
a. Afterward, what is the charge on C1C1 capacitor?
Express your answer in microcoulombs.

b. Afterward, what is the potential difference across C1C1 capacitor?
Express your answer in volts.

c. Afterward, what is the charge on C2C2 capacitor?
Express your answer in microcoulombs.

d. Afterward, what is the potential difference across C2C2 capacitor?
Express your answer in volts.

e. Afterward, what is the charge on C3C3 capacitor?
Express your answer in microcoulombs.

f. Afterward, what is the potential difference across C3C3 capacitor?
Express your answer in volts.
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