Homework

Chapter 17: Electrostatics and Potential Energy ## Ch 17: 1, 2, 11, 12, 18, 27, 29, 35, 58, 59, 63, 67

I am notorious for stupid arithmetic errors

so, please, always check my arithmetic!

17.1 How many electrons are there in a charge of -1 C?We know the charge of one electron is

1 e = 1.6 x 10 ^{ - 19}CDivide both sides of that equation by 1.6 x 10

^{ - 19}to get1 C = 6.25 x 10 ^{+ 18}e

17.2 A mole of anything is Avogadro's number (N_{A}= 6.02 x 10^{23}) of those things. How many coulombs are there in a mole of electrons?Q = N _{A}e = (6.02 x 10^{23}) (1.6 x 10^{ - 19}C) = 9.63 x 10^{4}C

17.11 A classical model of the hydrogen atom pictures an electron moving in a circular orbit of radius 0.52 x 10^{ - 10}m about the proton. Based on this model, determine

(a) the Coulomb force on the electron,

(b) the orbital speed of the eelectron, and

(c) the kinetic energy of the electron.(a) The Coulomb force -- or electric force -- is given by

F _{el }= k Q q / r^{2}For the hydrogen atom,

Q = q = e = 1.6 x 10 ^{ - 19}CTherefore,

F _{el}= k Q q / r^{2}F

_{el}= (9 x 10^{9})(1.6 x 10^{ - 19})(1.6 x 10^{ - 19}) / (0.52 x 10^{ - 10})^{2}F

_{el}= 8.52 x 10^{ - 8}N

(b) This Coulumb force is also the centripetal force which keeps the electron in a circular orbit.

F _{c}= m v^{2}/rv

^{2}= F_{c}r / mv

^{2}= F_{el}r / mv

^{2}= (8.52 x 10^{ - 8}N)(0.52 x 10^{ - 10}m) / (9.11 x 10^{ - 31}kg)v

^{2}= [(8.52 x 10^{ - 8})(0.52 x 10^{ - 10}) / (9.11 x 10^{ - 31})] (m/s)^{2}v

^{2}= [4.86 x 10^{12}] (m/s)^{2}v = 2.21 x 10

^{6}m/s

(We will be studying Relativity later this semester. There, the speed of light is quite important. Just for fun, how does this electron's speed compare to the speed of light?

c = 3.0 x 10 ^{8}m/sv/c = (2.21 x 10

^{6}) / (3.0 x 10^{8})v/c = 0.00737

That is just under

one percentthe speed of light ).(c) The kinetic energy is given by

KE = ( ^{1}/_{2}) m v^{2}KE = (

^{1}/_{2}) (9.11 x 10^{ - 31}) (2.21 x 10^{6})^{2 }JKE = 2.22 x 10

^{ - 18}JKE = 2.22 x 10

^{ - 18}J [1 eV / 1.6 x 10^{ - 19}J]KE = 13.9 eV

17.12 Both spheres in Figure 17.51 have a mass of 5.0 g and carry the same amount of positive charge. If the threads on which the spheres are suspended are 20.0 cm long, what is the charge on the spheres?Diagrams are important -- even vital!

F

_{y}= 0T

_{y}= w = mg = (0.005 kg) ( 9.8 m/s^{2})T

_{y}= 4.9 x 10^{ - 2}NT

_{y}= T cos 15° = 0.966 TT = T

_{y}/ 0.966T = (4.9 x 10

^{ - 2}N)/0.966)T = 5.07 x 10

^{ - 2}NT

_{x}= T sin 15° = 0.259 T = (0.259)(5.07 x 10^{ - 2}N)T

_{x}= 1.31 x 10^{ - 2}NFx = 0

F

_{el}= T_{x}F

_{el}= k Q q / r^{2}= T_{x}Now, it's back to the diagram for some geometry, so we can determine r, the distance between the charges.

(

^{1}/_{2}) r = (20 cm) sin 15^{o}= (20 cm) (0.259) = 5.18 cmr = 10.36 cm

r = 0.104 m

(be careful of the units!)

F _{el}= k Q q / r^{2}= T_{x}Q q = T

_{x}r^{2}/ kRemember, Q = q

Q ^{2}= (1.31 x 10^{ - 2}N)(0.104 m)^{2}/ (9 x 10^{9}N m^{2}/C^{2})Q

^{2}= 1.57 x 10^{ - 14}C^{2}Q = 1.25 x 10

^{ - 7 }C

17.18 Charge q = 20 C is placed at the origin. Charges Q_{1}= - 10 C and Q_{2}= 30 C are placed at two corners of a square, 10.0 cm on a side, as shown in Figure 17.55. What are the magnitude and direction of the force on q?1 C = 1 x 10 ^{ - 6}CRemember,

force is a vector!Coulomb's Law allows us to find the force on q due to Q_{1}and it allows us to find the force on q due to Q_{2}. Then we mustaddthose two forcesas vectors.Let us begin with F

_{1}, the forceonqdue toQ_{1},F

_{1}= k Q_{1}q / r2F

_{1}= (9 x 10^{9}) (10 x 10^{ - 6}) (20 x 10^{ - 6}) / (0.1)^{2}

Be very **careful** of the
units!

**Don't** write something
like

F_{1} = (9 x
10^{9})(10)(20)/(10)^{2}

The units are **very
important!**

Also, note that 10 cm = 0.1 m

F _{1}= 1.8 x 10^{2 }N = 180 NRemember,

force is a vector, so we can write this asF _{1x}= 0F

_{1y}= 1.8 x 10^{2}N

Now, we continue with F

_{2}, the forcedue tocharge Q_{2}as it actsoncharge q ;F _{2}= k Q_{2}q / r^{2}Again, remember that r = 10 cm = 0.1 m

F _{2}= (9 x 10^{9}) (30 x 10^{ - 6}) (20 x 10^{ - 6}) / (0.1)^{2}F

_{2}= 5.4 x 10^{2}N = - 540 NRemember, force is a vector, so we can write this as

F _{2x}= - 5.4 x 10^{2}NThe minus sign on F

_{2x}indicates that it points to the left.F _{2y}= 0Now we add the two forces--

as vectors!F _{net,x}= F_{1x}+ F_{2x}= 0 - 5.4 x 10^{2}N = - 5.4 x 10^{2}NF

_{net,y}= F_{1y}+ F_{2y}= 1.8 x 10^{2}N + 0 = 1.8 x 10^{2}NNow that we have the components, we need the magnitude and direction.

F _{net}^{2}= F_{net,x}^{2}+ F_{net,y}^{2}F

_{net}^{2}= [(-5.4)^{2}+ (1.8)^{2}] [x 10^{2}N ]^{2}F

_{net}= 5.69 x 10^{2}N = 569 Ntan =

^{opp}/_{adj}= (1.8) / (5.4) = 0.33= 18.4

^{o}

17.27 A proton is placed in a uniform electric field that has a magnitude of 2.3 x 10^{5}N/C and is directed up. What are the magnitude and direction of the force exerted on the proton?F = E q F = (2.3 x 10

^{5}N/C)(1.6 x 10^{ - 19}C)F = 3.68 x 10

^{ - 14 }NThe charge of a proton is positive so the force is in the same direction as the electric field; that is, the force points up.

17.29 Find the magnitude of the electric field 1.0 x 10^{ - 13}m from a proton.E = k Q / r ^{2}E = (9 x 10

^{9})(1.6 x 10^{ - 19}) / (1 x 10^{ - 13})2E = 1.44 x 10

^{17}N/C

17.35 Two 1.0 C charges are 1.0 m apart at two vertices of an equilateral triangle. Find the magnitude and direction of the electric field at the third vertex.As always, start with a diagram.

Due to symmetry,

E _{1}= E_{2}= E = k q / r^{2}= (9 x 10^{9})(1 x 10^{ - 6})/(1)^{2}= 9 x 10^{3}N/CBy symmetry,

E _{1x}= - E_{2x}so that

E _{net,x}= 0Also, by symmetry,

E _{1y}= E_{2y}= E cos 30^{o}= (9 x 10^{3}N/C ) 0.866 = 7.79 x 10^{3}N/CE

_{net,y}= E_{1y}+ E_{2y}= 15.6 x 10^{3}N/CE

_{net}= 15.6 x 10^{3}N/Cdirected

upward.

17.58 Between two charged, parallel plates, the electric field strength is 1500 N/C. How much work is done in moving a proton 1.5 cm in the direction of the field?Is this work done

onorbythe proton?Has the proton's electrical potential energy

increasedordecreased?F = E q

F = (1500 N/C )(1.6 x 10

^{ - 19}C ) = 2.4 x 10^{ - 16}NW = F d cos

W = (2.4 x 10

^{ - 16}N)(0.015 m)W = 3.6 x 10

^{ - 18}JThis is work done

onortothe proton since the electric force is in the direction the proton moves.Doing work

onthe proton -- the work of the electric fieldonthe proton --decreasesits potential energy. If we release the proton at its final position, itwillnotreturn to its initial position.

17.59 Between two charged, parallel plates, the electric field strength is 3000 N/C. How much work is done in moving an electron 2.5 cm in the direction of the field?Is this work done

onorbythe electron?Has the electron's electrical potential energy

increasedordecreased?F = E q

F = (3000 N/C )(1.6 x 10

^{ - 19}C ) = 4.8 x 10^{ - 16}NW = F d cos

W = (4.8 x 10

^{ - 16}N)(0.025 m)W = 1.2 x 10

^{ - 17}JW = 12 x 10

^{ - 18}JThis is work done

tothe electron since the electric force isoppositeto the direction the electron moves.Doing work

tothe electron -- the work of the electric fieldonortothe electron --increasesits potential energy. If we release the electron at its final position, itwillreturn to its initial position.

17.63 How much work is required to move a 2 C charge from 1.0 m away from a 10 C charge to 0.1 m away?By definition,

W = PE PE = k Q q [ 1 / r

_{f}- 1 / r_{i}]PE = ( 9 x 10

^{9}) ( 10 x 10^{ - 6}) ( 2 x 10^{ - 6 }) [^{1}/_{0.1}-^{1 }/_{1.0}]That is,

r _{i}= 1.0 m and r_{f}= 0.1 m.PE = 1.62 J

The potential energy has increased by 1.62 J so 1.62 J of work has been done

onortothe charge q by forces from the outside as it moves from a distance of 1.0 m to only 0.1 m from the charge Q.

17.67 Charges of 2.0 C and - 1.5 C are 80 cm apart. How much work is required to bring a 3.0 C charge from infinity to a point that is on the line connecting the two original charges and is 20 cm from the negative charge and 100 cm from the positive charge?Recall that potential energy is a

scalar; that means there are no directions involved!PE _{tot}= PE_{1}+ PE_{2}PE = k Q q / r

PE

_{1}= ( 9 x 10^{9})( 2.0 x 10^{ - 6})( 3.0 x 10^{ - 6}) / (1.0)PE

_{1}= 0.0540 JPE

_{2}= ( 9 x 10^{9})( - 1.5 x 10^{ - 6})( 3.0 x 10^{ - 6}) / (0.2)PE

_{2}= - 0.2025 JPE

_{tot}= PE_{1}+ PE_{2}PE

_{tot }= (0.0540 - 0.2025) JPE

_{tot}= - 0.1485 J

W = 0.1485 J(c) Doug Davis, 2002; all rights reserved

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