physics 2 exam#1
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
A small metal sphere of mass 3.6 g and charge 6.3 μC 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 close do they get to each other? Treat the spheres as point charges.
.5 m v^2 = kqq/d .5(3.6e-3)(8.9^2)=(9e9)(6.3e-6)(7.8e-6)/d rearrange for d
A cylindrical wire has a resistance R and resistivity ρ. 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=d/2 So New resistance is =>RN=4*L*(p/2)/pi*(d/2)2 RN=2(4pL/pid2) RN=2R
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
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=2*90 = 180 the potential is Q=C4*V4 V4=Q/C4=180uC/6uF = 30V C3=Q=180uC
A heating element of resistance 185 Ω 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 generated in the internal resistance of the battery?
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 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
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 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 = 1/4πε0 = 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
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 a dielectric material, with dielectric constant k=5.07 What must the magnitude of the charge on each capacitor plate now be, to produce the same potential difference between the plates as before?
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
The capacitors in the network shown in the figure all have a capacitance of 5.0 µF. 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)
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 the same diameter wire is needed so that the same current flows when the same voltage is applied at temperature 400°C?
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 formula for resistence , R = ρ L / A where ρ resistivity of the wire L is the length of the wire A be the area of the cros-section of the wire since , Ro = Rt ρo Lo / Ao = ρt Lt / At ---(1) Calculation: from graph: ρo = 2 x 10-8 Ω -m (at 0o ) ρt = 6 x 10-8 Ω - m (at 400o) since , diameter of the wire is the same for both the cases , the area of the cross- section of the wire at both the cases also same so, Ao = At Given : Lenght of the wire at 0o = 79 cm Lenght of the wire ar 400o = ? from eqn: (1) ρo Lo = ρt Lt ( 2 x 10-8 Ω -m)(79 ) = (6 x 10-8 Ω - m)(Lt ) Lenght of the wire required when thw wire is at 400o : Lt = 26.33 cm
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
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 = 1/4πε0 = 8.99 × 109 N ∙ m2/C) The radial component of the electric field at a point that is 1.50 m from the center is closest to
E=(q en)/(4 pi e0 r^2)= (9e9)(-500e-9 + 300e-9)/1.5^2 -800N/C
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 in the capacitor is U0. If the separation between the plates is doubled, how much energy is stored in the capacitor?
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
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 compared to the dimensions of the plates. The magnitude of the electric field between the plates is now equal to
E=V/d E prime/E = (V/2d)/(V/d) =(V/2d)*(d/v) =1/2 E prime = E/2
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 the cylinder. e0=8.854e-12 C^2/N^-1
E=lambda d/2pi e r^2
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
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 the proton that just misses the lower plate as it emerges from the field.
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
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 = 1/4πε0 = 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
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 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
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) =
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 energy density between the plates was u0, what is the new energy density?
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
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 the plates. The amount of charge on the plates is now equal to
Q/2 Q=CV v is constant C=epsilon a / d C=e/2 q is halved since d has increased 2 fold
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-axis, as shown in the figure. The electric fields E1 and E2, at the ends of the cylindrical surface, have magnitudes of 6000 N/C and 1000 N/C respectively, and are directed as shown. What is the net electric flux passing through the cylindrical surface?
Q=(E2-E1)(pi r^2) eo (1000-6000)(pi)(.1)^2(8.85e-12) -160 Nm^2/C
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 on the outer surface of the larger shell.
Q=4pi E e0 r^2 Q=4(49000)(pi)(4.1)^2(8.85e-12) 9.16e-5 C
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 cylindrical wire has a resistance R and resistivity ρ. 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
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
What length of a certain metal wire of diameter 0.15 mm is needed for the wire to have a resistance of 15 Ω? The resistivity of this metal is 1.68 × 10-8 Ω ∙ m.
R=pl/a l=RA/p l=(15)(pi)(.075e-6)/1.68e-8 l=15.7m
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
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 = 1/4πε0 = 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)
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 infinity?
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
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 electrons?
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
Two point charges of +1.0 μC and -2.0 μC 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
Two thin-walled concentric conducting spheres of radii 5.0 cm and 10 cm have a potential difference of 100 V between them. (k = 1/4πε0 = 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
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 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
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= electronic charge The current through both the wires is same since the wires are connected in series. The number of electrons per atom is same for both the wires since the wires are made of same material. The electronic charge is a unique value. The drift velocity of the electron through the left wire is, Vd=i/neA1 vdr=i/neA2 Vdr/Vd =A1/A2 >>1/4 Vdr/Vd=Vd/4 =vd/4
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 capacitance was C0, what is the new capacitance?
c0=e0(pi)(d^2)/L new capi =e0(pi(2d)^2)/2L =2c0
A 2.0 mm diameter wire of length 20 m has a resistance of 0.25 Ω. 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 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 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 narrow wire is E, the electric field in the wide wire is
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 =E/4
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 equal to zero.
d/4 to the right of Q and d/2 to the left of Q
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
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 +4.0 μC-point charge and a -4.0-μC point charge are placed as shown in the figure. What is the potential difference, VA - VB, between points A and B? (k = 1/4πε0 = 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
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
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 of any of the capacitors
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
Two capacitors of capacitance 6.00 μF and 8.00 μF are connected in parallel. The combination is then connected in series with a 12.0-V voltage source and a 14.0-μF 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
If the electric field is zero everywhere inside a region of space, the potential must also be zero in that region.
false
If the electric flux through a closed surface is zero, the electric field at points on that surface must be zero.
false
If the electric potential at a point in space is zero, then the electric field at that point must also be zero.
false
A positive point charge q = 3.0 μC 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
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 μC but opposite sign. The spheres are brought into the positions shown in the figure, with a vertical separation of 15 cm between them. Note that you cannot neglect gravity. (k = 1/4πε0 = 8.99 × 109 N ∙ m2/C2) The tension in the lower thread is closest to
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
Three point charges are arranged along the x - axis. Charge q1 = 3.0 µC is at origin, and charge q2 = -5.0 µC is at x = 0.200m. Where is the charge q3 = -8.00 µC 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 for f23 do f23=kqq/x^2 10.38(-x)=(9e9)(3e-6)(8e-6)/x^2 (-x) x=root((9e9)(3e-6)(8e-6)/10.38) = .144
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
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
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
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
Under electrostatic conditions, the electric field just outside the surface of any charged conductor
is always perpendicular to the surface of the conductor
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
A negative charge, if free, will tend to move from
low potential to high potential
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 carries?
ma = qE q=ma/E (.441)(13)/(5) =1.15C
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
An uncharged conductor has a hollow cavity inside of it. Within this cavity there is a charge of +10 µC 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 one correct choice.)
outer = +10uc inner = -10uc
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 following values: q1=3.5e-6 q2=-4.3e-6 q3=-4.9e-6 q4=-.1e-6 and Be sure to indicate whether each component, if nonzero, is positive or negative for the coordinate system.
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 sin(-45) + sigma B sin(45) = -8.35e4 n/c
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 = 1/4πε0 = 8.99 × 109 N ∙ m2/C2)
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
A wire of resistivity ρ 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
In Fig. 22.10, a circular plate with radius 1.67 m contains 705 μC of a charge uniformly distributed. Find the magnitude of the electric field near the plate.
rho=Q/pi r^2 E=rho/2 e0
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 the test tube would become
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
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 stretched so that its sides have length 2d, what is the magnitude of the electric field at P?
the area has increased 4x so it will be e/4
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
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 that
the charge on the capacitor had increased
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
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
Two concentric spheres are shown in Fig. 22.9. The inner sphere is a solid nonconductor and carries a charge of +5.00 µC uniformly distributed over its outer surface. The outer sphere is a conducting shell that carries a net charge of -8.00 µC. No other charges are present. The radii shown in the figure have the values R1 = 10.0 cm, R2 = 20.0 cm, and R3 = 30.0 cm.
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 radially outward 27cm E=0 35cm E=2.2e5 radially inward
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
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
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
When two or more capacitors are connected in parallel across a potential difference
the potential difference across each capacitor is the same
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
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 at a point midway between the plates. If the electron just misses the upper plate as it emerges from the field, find the magnitude of the electric field. Length of the plates is 2.00 cm and distance between them is 1 00 cm.
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/100)/1.25e-8 =6.4e13 m/s^2 electric force = F=qE>>ma=qE>>E=ma/q E=(9.11e-31)(6.4e13)/1.602e-19) =363.94 N/C
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
If the electrical potential in a region is constant, the electric field must be zero everywhere in that region.
true
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 energy stored in the capacitor was U0, how much energy does it now store?
u0=.5q^2/c .5Q^2 d/((e0)(A)) .5Q^2(2d)/((e0)(pi)(r^2)) = u0/2 stays the same
Three point charges of -2.00 μC, +4.00 μC, and +6.00 μC 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 = 1/4πε0 = 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
Four point charges of magnitude 6.00 μC and of varying signs are placed at the corners of a square 2.00 m on each side, as shown in the figure. (k = 1/4πε0 = 8.99 × 109 N ∙ m2/C2) (a) What is the electric potential (relative to infinity) at the center of this square due to these charges? (b) What is the magnitude of the electric field due to these charges at the center of the square?
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 μC and -6.0 μC 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-μC be placed on the +x-axis so that the potential at the origin is equal to zero? (k = 1/4πε0 = 8.99 × 109 N ∙ m2/C2)
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 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 lines. If 4.3 J are required to rotate the solid so that the dipole directions are reversed, and each dipole has dipole moment 4.7e-30 how many dipoles are there in the block?
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
