two metal spheres touching each other and a positive rod is brought close by
the charge is transfered to the first sphere and then to the next sphere and as the sphere on the right turns positive and moves away the left sphere turns negative
two identical spheres charge is gradually leaking their charges when they lose half their charge the magnitude will be
1/4F - kq1q2/r - q1=1/2 q2=1/2 which is 1/4
when two charges are distance d apart the elecrtic force that one feels from the other has a magnitude F. To make the force twice as strong what would the distance need to be changed to?
d/root2 > set up a ratio
f1/f2 = d2^2/d1^2
change to
f1/2F1 =d2^2/d1^2
then the F's cancel out
d1^2/2=d2^2
remove the squares
d1/root 2 = d2
two large flat plates are parallel distance d apart. half way between the two plates theres an electric field with magnitude E. if the separation of the plates is reduced to d/2 what is the magnitude of the electric field
magnitude does not change so E is still the same in the middle of the field
One very small uniformly charged plastic ball is located directly above another such charge in a test tube as shown in the figure. The balls are in equilibrium a distance d apart. If the charge on each ball is doubled, the distance between the balls in th
set up a ratio kq1q2/y^2=kq1q2/d^2
k2q2q/y^2=kqq/d^2
4kq^2/y^2=kq^2/d^2
4/y^2=1/d^2
2/y=1/d
2d=y
The figure shows two unequal point charges, q and Q, of opposite sign. Charge Q has greater magnitude than charge q. In which of the regions X, Y, Z will there be a point at which the net electric field due to these two charges is zero?
the charge will be 0 next to the smaller charge away from the big charge x y z itll be in x
Two point charges Q1 and Q2 of equal magnitudes and opposite signs are positioned as shown in the figure. Which of the arrows best represents the net electric field at point P due to these two charges?
-kq/16a^2 + kq/49a^2
count the squares
the -kq/16 i^
thats more negative than the other so it shows its pulling in the negative x direction
Three equal negative point charges are placed at three of the corners of a square of side d as shown in the figure. Which of the arrows represents the direction of the net electric field at the center of the square?
easy just cancel out the opposite charges
The figure shows three electric charges labeled Q1, Q2, Q3, and some electric field lines in the region surrounding the charges. What are the signs of the three charges?
easyyy
the arrows point away from a positive towards a negative
An electron is initially moving to the right when it enters a uniform electric field directed upwards. Which trajectory shown below will the electron follow?
the field is directed upwards and because of this the electron will move opposite to the field i guess and accelerate the direction so it will go down
In Fig. 21.12, charge q1 = 2.3 � 10-6 C is placed at the origin and charge q2=6.2e-6 is placed on the x axis at x= -.2m Where along the x-axis can a third charge Q=-8.3e-6 be placed such that the resultant force on this third charge is zero?
kq1Q/d^2 = kq2Q/(d+.2)^2
q1Q/d^2 = q2Q/(d+.2)^2
(2.3e-6)(-8.3e-6)/d^2 = (-6.2e-6)(-8.3e-6)/(d+.2)^2
do the algebra to solve for d which gives .31
In Fig. 21.13, calculate the x- and y-components of the electrical field produced at point P (at the center of the square) by the four charges placed at the corners of the square. The length of each side of the square is 0.80 m. The charges have the follo
q1 E1= kq1/r^2 r =.5*.8/root2
q2,3,4 are all the same just sub in the q values
then add e1 + e2 since same direction, gives Sigma Ea
subtract e2-e4 since opposite directions gives Sigma Eb
Ex= sigma A cos(-45) + sigma B cos(45) = 2.5e5 N/c
Ey = sigma A si
A proton, mass 1.67 � 10^-27 kg, is projected horizontally midway between two parallel plates that are separated by 0.60 cm, with an electrical field with magnitude 940,000 N/C between the plates. If the plates are 5.40 cm long, find the minimum speed of
F = qE
F= 1.602e-19 * 9.4e5 = 1.51e-13
a = f/m >1.51e-13 * 1.672e-27
= 9.03e13
delta y = 1/2at^2
t=root(2delta y/a)
root((2*.003)/9.03e13) =8.15e-9
v = delta x/ delta t
v=.054/8.15e-9
=6.62e6
A block of a solid is known to contain many dipoles aligned in the same direction (such a material is known as an electret). In an electric field of magnitude 300N/C the block tends to orient itself so that the dipoles are aligned along the electric field
w = Ep(cos theta2- cos theta1)
theta 1 = 0 degrees
theta 2 = 180 degrees
there are n number of dipoles
w =nEp(cos theta2-cos theta1)
4.3 = n (300) (4.7e-30)(cos180-cos0)
n = 1.5e27
Three point charges are arranged along the x - axis. Charge q1 = 3.0 �C is at origin, and charge q2 = -5.0 �C is at x = 0.200m. Where is the charge q3 = -8.00 �C located if the net force on q1 is 7.00 N in the negative x - direction?
i think because the q2 charge is more negative than q1 its directed in the negative x direction
do equation for q1 and q2 f12
f=kqq/r^2
(9e9) (3e-6) (5e-6) / .2^2 =3.38N x^
for f23
=f net - f12
=7N(-x) - (3.38)(x)
=10.38N
since we dont know the distance f
An electron is projected with an initial speed 1.60� 106 m/s. into the uniform field between the parallel plates. Field between the plates is uniform and directed vertically downward, and the field outside the plates is zero. The electron enters the field
time taken to travel the length of plates (horizontal x first)
delta x = v t +.5at^2
a=0 because horizontal
x=vt
t=(2/100)/(1.6e6)
1.25e-8
next is the magnitude of the field
y=d/2
y=1cm/2 =.5cm
now vertical displacement
y=vt+.5at^2
v=0
y=.5at^2
a=2(.5cm/1
In the figure Q = 5.8 nC and all other quantities are accurate to 2 significant figures. What is the magnitude of the force on the charge Q? (k = 1/4??0 = 8.99 � 109 N ? m2/C2)
F1=kqq/d
F1 = (9e9)(2nC)(5.8nC)/(1cm*e-2) = 1.04e-3N
f=2f sin60 = 2(1.04e-3)sin60
F=1.8e-3
What is the minimum magnitude of an electric field that balances the weight of a plastic sphere of mass 6.4g that has been charged to -3.0 nC
E q = mg
E(3e-9) = (6.4e-3)(9.8)
E=20.9e6
Two small insulating spheres are attached to silk threads and aligned vertically as shown in the figure. These spheres have equal masses of 40 g, and carry charges q1 and q2 of equal magnitude 2.0 ?C but opposite sign. The spheres are brought into the pos
gravitational and coulomb force act on the spheres
grav force= mg
(.04kg)(9.8m/s^2)=.3924N
coulomb =
kqq/r^2
(9e9)(2e-6)(-2e-6)/.15^2 =-1.598N
add the forces together and change the sign to positive
=1.205N
A small sphere with a mass of 441 g is moving upward along the vertical +y-axis when it encounters an electric field of 5.00 N/C . If, due to this field, the sphere suddenly acquires a horizontal acceleration of 13.0 m/s2 , what is the charge that it carr
ma = qE
q=ma/E
(.441)(13)/(5)
=1.15C
If the electric flux through a closed surface is zero, the electric field at points on that surface must be zero.
FALSE
Consider a spherical Gaussian surface of radius R centered at the origin. A charge Q is placed inside the sphere. To maximize the magnitude of the flux of the electric field through the Gaussian surface, the charge should be located
e) The flux does not depend on the position of the charge as long as it is inside the sphere
Under electrostatic conditions, the electric field just outside the surface of any charged conductor
is always perpendicular to the surface of the conductor
At a distance D from a very long (essentially infinite) uniform line of charge, the electric field strength is 1000 N/C. At what distance from the line will the field strength to be 2000 N/C?
E=lambda/2pi e D
1000N/C= lambda/2pi e D
rearrange
D=lambda/(2pi e)(1000)
then again but with 2000
d1=lambda/2pi e 2000
im not really sure whats going on here but its d/2
just take the 1000/2000
An uncharged conductor has a hollow cavity inside of it. Within this cavity there is a charge of +10 �C that does not touch the conductor. There are no other charges in the vicinity. Which statement about this conductor is true? (There may be more than on
outer = +10uc
inner = -10uc
A charge Q is uniformly spread over one surface of a very large nonconducting square elastic sheet having sides of length d. At a point P that is 1.25 cm outside the sheet, the magnitude of the electric field due to the sheet is E. If the sheet is now str
the area has increased 4x
so it will be e/4
Two concentric spheres are shown in Fig. 22.9. The inner sphere is a solid nonconductor and carries a charge of +5.00 �C uniformly distributed over its outer surface. The outer sphere is a conducting shell that carries a net charge of -8.00 �C. No other c
the inner shell has a charge of +5C and for some reason there has to be a total charge of 0 so its charge is -5
outer shell is +5-8 = -3
9.5cm
because the charge is uniformly distributed the radius is 0. no electric field means no direction
15cm E=2e6 rad
In Fig. 22.10, a circular plate with radius 1.67 m contains 705 ?C of a charge uniformly distributed. Find the magnitude of the electric field near the plate.
rho=Q/pi r^2
E=rho/2 e0
Positive charge is distributed uniformly throughout a large insulating cylinder of radius r=.9m The charge per unit length in the cylindrical volume is
lambda=3e-9 C/m
Calculate the magnitude of the electric field at a distance of 0.200 m from the axis of
E=lambda d/2pi e r^2
In Fig. 22.11, two small concentric conducting spherical shells produce a radially outward electric field of magnitude 49,000N/C a distance of 4.10m from the center of the shells. If the inner shell contains a charge of -5.30uC find the amount of charge o
Q=4pi E e0 r^2
Q=4(49000)(pi)(4.1)^2(8.85e-12)
9.16e-5 C
A cube with 2.50 m edges is oriented in a region of uniform electric field. Find the electric
flux through the right face if the electric field in N/ C is given by:
a. 3.00 i b. What is the total flux through the cube for each of these fields?
L=2.5m
A=6.25
phi=EA
3.0i * 6.25j = 0
close surface is 0 flux
A nonuniform electric field is directed along the x-axis at all points in space. This magnitude of the field varies with x, but not with respect to y or z. The axis of a cylindrical surface, 0.80 m long and 0.20 m in diameter, is aligned parallel to the x
Q=(E2-E1)(pi r^2) eo
(1000-6000)(pi)(.1)^2(8.85e-12)
-160 Nm^2/C
A positive point charge q = 3.0 ?C is surrounded by a sphere with radius 0.20 m centered on the charge. Find the electric flux through the sphere due to this charge.
flux=charge enclosed/e0
3e-6/8.85e-12
=3.4e5
How many excess of electrons must be added to an isolated spherical conductor 32.0 cm in diameter to produce an electric field of 1150 N/C?
Q=E*r^2/k
N=Q/electrons
Electrons 1.6e-19
N is number of electrons i think
A hollow conducting spherical shell has radii of 0.80 m and 1.20 m, as shown in the figure. The sphere carries a net excess charge of -500 nC. A point charge of +300 nC is present at the center. (k = 1/4??0 = 8.99 � 109 N ? m2/C) The radial component of t
E=(q en)/(4 pi e0 r^2)=
(9e9)(-500e-9 + 300e-9)/1.5^2
-800N/C
If the electric field is zero everywhere inside a region of space, the potential must also be zero in that region.
FALSE
If the electrical potential in a region is constant, the electric field must be zero everywhere in that region.
TRUE
If the electric potential at a point in space is zero, then the electric field at that point must also be zero.
FALSE
A negative charge, if free, will tend to move from
low potential to high potential
Suppose you have two point charges of opposite sign. As you move them farther and farther apart, the potential energy of this system relative to infinity
increases
Suppose you have two negative point charges. As you move them farther and farther apart, the potential energy of this system relative to infinity
decreases
A nonconducting sphere contains positive charge distributed uniformly throughout its volume. Which statements about the potential due to this sphere are true? All potentials are measured relative to infinity. (There may be more than one correct choice.)
the potential is highest at the center of the sphere
A conducting sphere contains positive charge distributed uniformly over its surface. Which statements about the potential due to this sphere are true? All potentials are measured relative to infinity. (There may be more than one correct choice.)
the potential at the center of the sphere is the same as the potential at the surface
Suppose a region of space has a uniform electric field, directed towards the right, as shown in the figure. Which statement about the electric potential is true?
the potential at points A and B are equal, and the potential at point C is lower than the potential at point A
The figure shows two arcs of a circle on which charges +Q and -Q have been spread uniformly. What is the value of the electric potential at the center of the circle?
the points all cancel each other out
Three point charges of -2.00 ?C, +4.00 ?C, and +6.00 ?C are placed along the x-axis as shown in the figure. What is the electrical potential at point P (relative to infinity) due to these charges? (k = 1/4??0 = 8.99 � 109 N ? m2/C2)
v1=kq/r
9e9(-2e-6)/.2root2
=-6.37e4 V
v2=kq/r
(9e9)(6e-6)/.2
=1.8e5 V
v3=(9e9)(6e-6)/.2root2
=1.91e5 V
total potential
Vp= V1 + V2 + V3
add all those up
A small metal sphere of mass 3.6 g and charge 6.3 ?C is fired with an initial speed of directly toward the center of a second metal sphere carrying charge 7.8uC This second sphere is held fixed. If the spheres are initially a large distance apart, how clo
.5 m v^2 = kqq/d
.5(3.6e-3)(8.9^2)=(9e9)(6.3e-6)(7.8e-6)/d
rearrange for d
In Fig. 23.7, two point charges, q1 = +18.0 nC and q2 = -41.0 nC, are separated by .5m A third charge of 41.0nC is placed at the point A, 0.18 m to the left of q2. Find the work needed to move the third charge to point B, 0.40 m to the left of q1.
Potential at A
VA=k(q1/(d-a) +q2/a)
potential at B
VB=k(q1/b +q2/d-b)
work
w=Q(VB - VA) =
Two point charges, Q and -3Q, are located on the x-axis a distance d apart, with -3Q to the right of Q. Find the location of ALL the points on the x-axis (not counting infinity) at which the potential (relative to infinity) due to this pair of charges is
d/4 to the right of Q
and
d/2 to the left of Q
Four point charges of magnitude 6.00 ?C and of varying signs are placed at the corners of a square 2.00 m on each side, as shown in the figure. (k = 1/4??0 = 8.99 � 109 N ? m2/C2)
(a) What is the electric potential (relative to infinity) at the center of
v=Kq/r
v=k(6m/r + 6m/r + -6m/r + -6m/r)
4kqcos45/r^2
(4)(9e9)(6e-6) cos45/(root(2^2+2^2)/2)^2
solve.
Two point charges of +2.0 ?C and -6.0 ?C are located on the x-axis at x = -1.0 cm and x = +2.0 cm respectively. Where should a third charge of +3.0-?C be placed on the +x-axis so that the potential at the origin is equal to zero? (k = 1/4??0 = 8.99 � 109
v=kq/r
(9e9)(2e-6)/1e-2 = 1.8e6 VA
(-9e9)(6e-6)/2e-2 = -2.7e6 VB
(1.8e6-2.7e6)=(9e9)(3e-6)/x
rearrange and solve
gives 3cm
A +4.0 ?C-point charge and a -4.0-?C point charge are placed as shown in the figure. What is the potential difference, VA - VB, between points A and B? (k = 1/4??0 = 8.99 � 109 N ? m2/C2)
do pythag theorem to find distance between the points and the charge it isnt directly attached to
b to +4.0 is .5 m same as a
potential at point a
Va=kq/r
V=9e9(-4e-6/.5 + 4e-6/.3)
=48000v 48kv
Vb=kq/r
9e9(-4e-6/.3 + 4e-6/.5)
=-48kv
Va-Vb=96kV
Two point charges of +1.0 ?C and -2.0 ?C are located 0.50 m apart. What is the minimum amount of work needed to move the charges apart to double the distance between them?
W = -(u2 -u1)
-(kq1q2/r2 -kq1q2/r1)
-(9e9)(2e-6)(6e-6)(1/(2(.5))-1/.5)
=solve
A conducting sphere is charged up such that the potential on its surface is 100 V (relative to infinity). If the sphere's radius were twice as large, but the charge on the sphere were the same, what would be the potential on the surface relative to infini
V1 = 100v
Q1=Q2 since the charge stays the same
r1=2r2 since the readius is doubled
gaus law
v=kq/r
V1=kq1/r1
V2=kq2/2r2
V1/V2=q1/q2*2r2/r1
100/v2=1*2
v2=50V
Consider the group of three+2.4 nC point charges shown in the figure. What is the electric potential energy of this system of charges relative to infinity? (k = 1/4??0 = 8.99 � 109 N ? m2/C2)
U=KQ^2/r1 + KQ^2/r2 + KQ^2/r3
plug in
r1=.03 r2=.04 r3=root(.03^2 + .04^2)
The charge on the square plates of a parallel-plate capacitor is Q. The potential across the plates is maintained with constant voltage by a battery as they are pulled apart to twice their original separation, which is small compared to the dimensions of
Q/2
Q=CV
v is constant
C=epsilon a / d
C=e/2
q is halved since d has increased 2 fold
The electric field between square the plates of a parallel-plate capacitor has magnitude E. The potential across the plates is maintained with constant voltage by a battery as they are pulled apart to twice their original separation, which is small compar
E=V/d
E prime/E = (V/2d)/(V/d)
=(V/2d)*(d/v)
=1/2
E prime = E/2
When two or more capacitors are connected in series across a potential difference
each capacitor carries the same amount of charge
the potential difference across the combination is the algebraic sum of the potential differences across the idividual capacitors
the equivalent capacitance of the combination is less that the capacitance o
When two or more capacitors are connected in parallel across a potential difference
the potential difference across each capacitor is the same
An ideal parallel-plate capacitor consists of a set of two parallel plates of area A separated by a very small distance d. When this capacitor is connected to a battery that maintains a constant potential difference between the plates, the energy stored i
E=.5((e0)(A)/d)*V^2
sub in 2d for d
pull the 2 from the bottom and put at the front as 1/2
then you have the same thing thats equal to E
if it said triple the distance it would be E/3
An ideal parallel-plate capacitor consists of a set of two parallel plates of area A separated by a very small distance d. When the capacitor plates carry charges +Q and -Q, the capacitor stores energy U0. If the separation between the plates is doubled,
...
An ideal air-filled parallel-plate capacitor has round plates and carries a fixed amount of equal but opposite charge on its plates. All the geometric parameters of the capacitor (plate diameter and plate separation) are now DOUBLED. If the original energ
Q fixed L=separation d=diameter
C=((e0)(pi d^2))/L
energy U=Q^2/2C
=Q^2L/(2pi(e0)(d^2))
energy density u0=U/pi d^2 L
=Q^2/(2pi(e0)d^4)
new energy density
Q^2/2pi^2(e0)(2d)^4
u0/16
youre putting equations inside each other
An ideal air-filled parallel-plate capacitor has round plates and carries a fixed amount of equal but opposite charge on its plates. All the geometric parameters of the capacitor (plate diameter and plate separation) are now DOUBLED. If the original capac
c0=e0(pi)(d^2)/L
new capi
=e0(pi(2d)^2)/2L
=2c0
An ideal air-filled parallel-plate capacitor has round plates and carries a fixed amount of equal but opposite charge on its plates. All the geometric parameters of the capacitor (plate diameter and plate separation) are now DOUBLED. If the original energ
u0=.5q^2/c
.5Q^2 d/((e0)(A))
.5Q^2(2d)/((e0)(pi)(r^2))
=
u0/2
stays the same
An air-filled parallel-plate capacitor is connected to a battery and allowed to charge up. Now a slab of dielectric material is placed between the plates of the capacitor while the capacitor is still connected to the battery. After this is done, we find t
the charge on the capacitor had increased
Two capacitors of capacitance 6.00 ?F and 8.00 ?F are connected in parallel. The combination is then connected in series with a 12.0-V voltage source and a 14.0-?F capacitor, as shown in the figure.
equivalent capacitance
1/Ce = 1/(c1 + c2) + 1/c3
=1/6+8 + 1/14 =1/7 =7uF
to find the charge
v1/v2=C3/Cp = 1/1 = v1=v2 v1=v/2
v1=Q1/C1
V/2=Q1/C1
Q1=v/2C1
= 12/2 *6 = 36uC
potential difference
v1/v2=c3/cp therefore v=1/c
v1/v2=14/14
v1=v2
v1=v/2
12/2 =6V
Four capacitors are connected across a 90-V voltage source as shown in the figure.
1/Ceq=1/C1 +1/c2
1/2 + 1/4 = 6/8
ceq=8/6uF
Q=CV
Q=8/6*90 = 120uC
since its a series Q1=Q2=Q=120
c1=c2=120
potential difference across c4
find the charges on c3&c4
since its in series c3=c4
Q=CV
1/Ceq = 1/c3 + 1/c4 = 1/3 + 1/6 = 9/18
Ceq=18/9uF=2uF
V=90
Q=
A parallel-plate air capacitor is made from two plates 0.070 m square, spaced 6.3 mm apart. What must the potential difference between the plates be to produce an energy density of .037 J/m^3
A=.07 m^2
d=6.3=-3
E=.037 J/m^3
energy density is
u=1/2 (epsilon)( | E |)^2
.037=.5
8.85e-12
| E |^2
electric field between the plates
E^2=83.61e8
E=9.14e4
potential difference between the plates
V=Ed
(9.14e4)(6.3e-3)
=575.82V
A parallel-plate capacitor, with air between the plates, is connected to a battery. The battery establishes a potential difference between the plates by placing charge of magnitude
3.47e-6 C
on each plate. The space between the plates is then filled with
C=KA/d
K=ek
e=permitivity of free space
k=relative permitivity (given)
a=area of the plates
d=distance between the plates
c1/c2 = (ea/d)/(eka/d)
c1/c2=1/k
Q=CV
v=Q/C
v1=v2
Q1/C1=Q2/C2 =1/k
Q2= K Q1
Q2=5.07*Q2 = kQ1
Q2=5.07*3.47e-6
Q2=17.6e-6
Two capacitors C1=20 ?C , and C2=25 ?C are connected in series. This combination is
then connected in parallel with a third capacitor C3=15 ?C .
Find the equivalent capacitance.
...
Two thin-walled concentric conducting spheres of radii 5.0 cm and 10 cm have a potential difference of 100 V between them. (k = 1/4??0 = 8.99 � 109 N ? m2/C2)
(a) What is the capacitance of this combination?
(b) What is the charge carried by each sphere?
a=.05m
b=.1m
k=9e9
capacitance=4pi(eo)ab/(b-a)
(1/9e9)(.05)(.1)/(.1-.05)
=11.1 PF
charge carried
Q=CV
11.1 pf gets changed to 1.11nC multiply by 100
A metal cylinder of radius 2.0 mm is concentric with another metal cylinder of radius 5.0 mm. If the space between the cylinders is filled with air and the length of the cylinders is 50 cm, what is the capacitance of this arrangement? (k = 1/4??0 = 8.99 �
radius of inner cylinder=2mm
radius of outer cyl=5mm
length of cyl=50cm
capcitance is
C=l/(2k ln(b/a))
.5/2(9e9)ln(5mm/2mm)
solve
30pF
The capacitance per unit length of a very long coaxial cable, made of two concentric cylinders, is 50 pF/m. What is the radius of the outer cylinder if the radius of the inner one is 1.0 mm? (k = 1/4??0 = 8.99 � 109 N ? m2/C2)
C/L=(2pi)(e0)/ln (rb/ra)
ln(rb/ra)=2
pi
8.87e-12/50e-12
rb/ra=e^1.114
=3.0465
rb=3.0465*1
=3mm
Three capacitors are connected as shown in the figure. What is the equivalent capacitance between points a and b?
C total= C4uf + C6uf
=10uf
in parallel capacitors the total capacitance is equal to the sum of the capacitance of each individual capacitor.
Cab=1/((1/C2uf)(1/Ctotal))
1/((1/2)+(1/10))
= 1.7
The capacitors in the network shown in the figure all have a capacitance of 5.0 �F. What is the equivalent capacitance, Cab, of this capacitor network?
Ceq of 3 capacitors = 5*5/(5+5) + 5 = 7.5 uF
Cab = 7.5*5/(5+7.5) = 3uF
note ( in series Ceq = C1*C2/(C1+C2) , in parallal Ceq = C1 +C2)
The figure shows a steady electric current passing through a wire with a narrow region. What happens to the drift velocity of the moving charges as they go from region A to region B and then to region C?
The drift velocity increases from A to B and decreases from B to C
The figure shows two connected wires that are made of the same material. The current entering the wire on the left is 2.0 A and in that wire the electron drift speed is vd. What is the electron drift speed in the wire on the right side?
area of cross section of left wire is A1=(pi d1^2)/4
d1=diameter of left wire
A1= (pi(1e-3)^2)/4
=7.85e-7
A2=pi d^2/4
A2=pi(2e-3)^2/4
=31.4e-7
A1/A2=7.85e-7/31.4e-7
=1/4
Vd=I/neA1
i=current through the left wire
n=number of electrons per atom
e= electroni
A narrow copper wire of length L and radius b is attached to a wide copper wire of length L and radius 2b, forming one long wire of length 2L. This long wire is attached to a battery, and a current is flowing through it. If the electric field in the narro
current will be the same in both wires but will have different resistances
R1=PL/a
R2=PL/4a (4since the radius is doubled so crosssectional area will be 4x)
R1=4(R2)
voltage
V1=4(V2) (since v is proportional to R)
E1= 4(E2) (E proportional to V)
E2=E1/4
=
When a potential difference of 10 V is placed across a certain solid cylindrical resistor, the current through it is 2 A. If the diameter of this resistor is now tripled, the current will be
R = 10/2 = 5 ohm
If dia. is triple
R = pL/A
=pL/pi*r^2
r=3r1
s0
R = 5/3^2
and
I = V/R = 10*3^2/5
= 18 A
A cylindrical wire has a resistance R and resistivity ?. If its length and diameter BOTH cut in half, what will be its resistance?
R = (rho) l / A
R1/R2 = L1/L2 * A2/A1
L2 = L1 /2
A2 = pi * d1 ^2/4
R1/R2 = 2 * 1/4
R1/R2 = 1/2
R2 = 2 R1
A cylindrical wire has a resistance R and resistivity ?. If its length and diameter BOTH cut in half, what will be its resistivity?
1)
Answer is C) p
Resistivity of the wire is constant .Does not vary with length and diameter.
2)
B)2R is the answer
Since Area of cylindrical wire is given by
A=pi*d2/4
Resistance of wire is given by
R=pL/A=pL/(pi*d2/4)
R=4pL/pi*d2
Given
L1=(L/2) and d1=
The current in a wire varies with time according to the equation I(t) = 6.00 A + (4.80 A/s)t, where t is in seconds. How many coulombs of charge pass a cross section of the wire in the time period between t = 0.00 s and t=3s
i=DQ/DT
Q=integral DQ = int 0to3 iDT
int 0-3 (6+4.8t) dt
6t + 4.8t^2/2 | 0to3
plug in
39.6C
A silver wire has a cross sectional area A = 2.0 mm2. A total of 9.4 � 1018 electrons pass through the wire in 3.0 s. The conduction electron density in silver is 5.8 � 1028 electrons/m3 and e = 1.60 � 10-19 C. What is the drift velocity of these electron
Vd=J/qn
=I/qnA
=(deltaQ/detaT)/(qnA)
=((9.4e18)e/3secs)/((e)(5.8e28)(2e-6))
=((9.4e18)/(3secs))/((5.8e28)(2e-6))
=2.7e-5
An electric device delivers a current of 5.0 A to a device. How many electrons flow through this device in 10 s? (e = 1.60 � 10-19 C)
i=q/t
q=ne
i=ne/t
n=it/e
n=((5A)(10s))/(1.6e-19)
=3.125e20
What length of a certain metal wire of diameter 0.15 mm is needed for the wire to have a resistance of 15 ?? The resistivity of this metal is 1.68 � 10-8 ? ? m.
R=pl/a
l=RA/p
l=(15)(pi)(.075e-6)/1.68e-8
l=15.7m
A 110-V hair dryer is rated at 1200 W. What current will it draw when operating from a 110-V electrical outlet?
i=1200/110 = 11A
P=VI
A 400-W computer (including the monitor) is turned on for 8.0 hours per day. If electricity costs 10� per kWh, how much does it cost to run the computer annually for a typical 365-day year?
to to find Watt* hrs you just multiply your time and your wattage
400 * 8 = 3200 Wh or 3.2 kWh / day
now find out the total used in a year
3.2 * 365 = 1168 kWh /year
finally multiply that by the cost
1168 * .1 = $116.80
The emf and the internal resistance of a battery are as shown in the figure. If a current of 8.3 A is drawn from the battery when a resistor R is connected across the terminals ab of the battery, what is the power dissipated by the resistor R?
apply E = V-iR
E = 95-(8.3*5)
E = 53.5 V
now Power P = Vi
P = 53.5* 8,3
P = 444.5 Volts
option 2 it is
A piece of wire 58.4 cm long carries a current I when a voltage V is applied across its ends at a temperature of 0�C. If the resistivity of the material of which the wire is made varies with temperature as shown in the graph in the figure, what length of
Concept:
As current I and voltage V passing at
both the temparatures are same ,
The resistence offred by the wire
at bothe these temparatures is the same
Let ,resistence of the wire be R
Now, consider two cases
at 0o and at 400o
But, it is known by the fo
A certain fuse "blows" if the current in it exceeds 1.0 A, at which instant the fuse melts with a current density of 620 A/cm^2 What is the diameter of the wire in the fuse?
apply current densty J = current /area
so area = pir^2 = 1/(620/10000)
pi r^2 = 0.161 um
r^2 = 0.0512 um
r = 0.226 mm
diamter = 2r = 0.45 mm
If a current of 2.4 A is flowing in a cylindrical wire of diameter 2.0 mm, what is the average current density in this wire?
I=J/A
=J/((pi)(d/2)^2)
2.4/((pi)(2e-3/2)^2)
=7.6e5
A silver wire with resistivity 1.59e-8 omega*m carries a current density of 4.0 A/mm^2 What is the magnitude of the electric field inside the wire?
current density
J=i/a
formula for specific resistance of the wire
p=RA/L
IR/L = j*P
V/L=J*P
E=J*P
the electric field in the wire is
E=(1.59e-8)(4.0/e-6)
0.063 V/m
A 2.0 mm diameter wire of length 20 m has a resistance of 0.25 ?. What is the resistivity of the wire?
cross section area of the wire is
a=pi r^2
pi(d/2)^2
pi(2e-3/2)^2
3.14e-6
resistance
R=pl/a
p=RA/l
((.25)(3.14e-6)/20
=3.9e-8
A heating element of resistance 185 ? is connected to a battery of emf 425 V and unknown internal resistance "r". It is found that heat energy is being generated in the heater element at a rate of 68.0 W. What is the rate at which heat energy is being gen
68W =I^2R=I^2(185 ohm)
i-current in the circuit .606A
i=EMF/R+r = 425V/185ohm +r
425/185+r = .606A
solve for r
=516.32ohm
heat energy generated on internal resistance=i^2r
=189.6W
A wire of resistivity ? must be replaced in a circuit by a wire of the same material but 4 times as long. If, however, the resistance of the new wire is to be the same as the resistance of the original wire, the diameter of the new wire must be
resistance is
R1=pl1/pi(d1/2)^2
resistance in new wire
R2=pl2/pi(d2/2)^2
set them equal to each other
but throw a 4 on the l
solve not too hard