Problem 16.12.!a.) What is the electrical potential at origin?!

y1 = .0125 m

y2 = .018 m

q1 = 4.5 µC

q2 = !2.24 µC

x = .015 mx

b.) What is the electrical potential at (1.5 cm, 0)?!

1.!

a.) What is the electrical potential at origin?!y1 = .0125 m

y2 = .018 m

q1 = 4.5 µC

q2 = !2.24 µC

x = .015 mx

2.!

V = Vq1+ Vq2

= k q1

y1

+ k (!q2 )y2

= 9x109( ) 4.5x10!6 C( ).0125 m( ) + 9x109( ) !2.24x10!6 C( )

.018 m( ) = 2.12x106 volts

y1 = .0125 m

y2 = .018 m

q1 = 4.5 µC

q2 = !2.24 µC

x = .015 mx

b.) What is the electrical potential at (1.5 cm, 0)?!

3.!

V = Vq1+ Vq2

= k q1

r1+ k (!q2 )

r2

= 9x109( ) 4.5x10!6 C( ).0125 m( )2 + .015 m( )2( )1/2 + 9x109( ) !2.24x10!6 C( )

.018 m( )2 + .015 m( )2( )1/2

= 3.18x102 volts

12.)! a.) The electrical potential at origin?! y = .0125 m

y = .018 m

q1 = 4.5 µC

q2 = !2.24 µC

x = .015 mxThe important point here is that electrical potentials

add as SCALARS, not vectors. Absolute electrical potential (point voltages) generated by positive charges are positive and absolute electrical potentials generated by negative charges are negative. In either case, the electrical potential function FOR A POINT CHARGE is !

V1 = kqr1

V = Vq1+ Vq2

= k q1

r1+ k (!q2 )

r2

= 9x109( ) 4.5x10!6 C( ).0125 m( ) + 9x109( ) !2.24x10!6 C( )

.018 m( ) = 2.12x106 volts

Using this on our situation, remembering to include the sign of the charge in the expression, we get: !

r1

r2

4.!

12.)! b.) The electrical potential at (.015,0)?! y = .0125 m

y = .018 m

q1 = 4.5 µC

q2 = !2.24 µC

x = .015 mxThe only things that is different is that the distances

(the “r’s” are different. Aside from that, the everything else looks just like part a. !

V = Vq1+ Vq2

= k q1

r3

+ k (!q2 )r4

= 9x109( ) 4.5x10!6 C( ).01252 +.0152( )1/2 + 9x109( ) !2.24x10!6 C( )

.0182 + .0152( )1/2

= whatever

r3

r4

5.!

19.)!

a.) Calculate the electrical potential energy in the system.!

b.) Calculate the electrical potential energy in the system with the alpha particle included.!

x = 6x10!15 m

x = 6x10!15 mx = 3x10!15 m

y = 3x10!15 m

c.) What’s the change of electrical potential energy if alpha particle is removed.!

6.!

19.)!

a.) Calculate the electrical potential energy in the system. We know:!

x = 6x10!15 m

Also, the voltage at infinity is zero. The voltage r units from a field producing point charge is ! . So . . . !V = k q

r

7.!

19.)!

a.) Calculate the electrical potential energy in the system. We know:!

x = 6x10!15 m

!V =!Uq

" U2 # U1( ) = q V2 # V1( )

Also, the voltage at infinity is zero. The voltage r units from a field producing point charge is ! . So . . . !V = k q

r

Ur ! U"( ) = q Vr ! V"( ) # Ur ! 0( ) = q k q

r! 0$

%&'()

# Ur = k q2

r

# Ur = 9x109( ) 1.6x10!19( )2

6x10!15

$

%&&

'

())

# Ur = 3.84x10!14 J

8.!

19.) (con’t.)!

b.) Calculate the electrical potential energy in the system with the alpha particle included.!

x = 6x10!15 mx = 3x10!15 m

y = 3x10!15 m

Added to the energy required to bring the protons together, we have to determine the work required to bring the alpha particle in. As each protons will be the same distance away, all we need to do is double the work done by one, or:!

9.!

19.) (con’t.)!

b.) Calculate the electrical potential energy in the system with the alpha particle included.!

x = 6x10!15 mx = 3x10!15 m

y = 3x10!15 m

Added to the energy required to bring the protons together, we have to determine the work required to bring the alpha particle in. As each protons will be the same distance away, all we need to do is double the work done by one, or:!

Wtotal = Wbring protons together + 2Wbring ! in due to one proton

" Utotal = Ur # 0( )p+ + 2 Ur # 0( )! = q Vr # 0( )p+ + 2 Vr # 0( )!

" Utotal = kq

p+2

rprotons

+ 2kq

p+ q!

rp to !

" Ur = 9x109( ) 1.6x10#19( )2

6x10#15

$

%&&

'

())+ 2 9x109( ) 1.6x10#19( ) 3.2x10#19( )

32 + 32( )1/2x10#15

$

%&&

'

())

" Ur = 2.55x10#13 J

10.!

19.) (con’t.)!

x = 6x10!15 mx = 3x10!15 m

y = 3x10!15 mc.) What’s the change of electrical potential energy if alpha particle is removed.!

With the alpha particle removed, the system energy change is:!

11.!

19.) (con’t.)!

x = 6x10!15 mx = 3x10!15 m

y = 3x10!15 mc.) What’s the change of electrical potential energy if alpha particle is removed.!

With the alpha particle removed, the system energy change is:!

Ufinal = Uafter ! Ubefore

= (3.84x10!14 ) ! (2.55x10!13) = (!2.17x10!13) Joules

12.!

19.) (con’t.)!

d.) To determine the speed of the alpha particle at infinity (with the two protons held stationary), we need the work/energy theorem.!

e.) Determine the speed at infinity of the two protons if the alpha particle held stationary. (By allowing both protons to move, there is no longer any electric potential energy in the system (a single particle doesn’t need any work done to bring it in from infinity). Sooo . . .!

13.!

19.) (con’t.)!

d.) To determine the speed of the alpha particle at infinity (with the two protons held stationary), we need the work/energy theorem.!

e.) Determine the speed at infinity of the two protons if the alpha particle held stationary. (By allowing both protons to move, there is no longer any electric potential energy in the system (a single particle doesn’t need any work done to bring it in from infinity). Sooo . . .!

KE1+ Uin system at time 1+ Wext = KE2 + Uin system at time 2!!!!!" 0 + (2.55x10-13J) + 0 = 1

24 1.67x10-27kg( )#$ %&v2 + (3.84x10'14 J)

" v = 8.08x106 m / s

KE1+ Uin system at time 1+ Wext = KE2 + Uin system at time 2!!!!!" 0 + (2.55x10-13J) + 0 = 1

22 1.67x10-27kg( )#$ %&v2 + (0)

" v = 1.24x107 m / s

14.!

Extra 1:!a.) What’s the speed of an electron accelerated through 120 volts?!

b.) What’s the speed of a proton accelerated through 120 volts?!

+! -!

15.!

Extra 1:!

a.) What’s the speed of an electron accelerated through 120 volts?!

+! -!

16.!

This is a classic conservation of energy problem. It is ALWAYS best to identify voltage values (or, at least, symbols) before starting . As such, we will assume that the electrical potential of the high voltage side is and the electrical potential of the low voltage side is ZERO. Remembering that the charge of an electron is negative and that to accelerate, it must start at the negative plate, we can write:!

KE1+ Uin system at time 1 + Wext = KE2 + Uin system at time 2!!!!!" 0 + ( # q)V# + 0 = 1

2m

e#v2 + ( # q)V+

" 0 + ( #1.6x10#19 ) 0( ) + 0 = 12

9.1x10#31( )v2 + ( #1.6x10#19 ) 120( )" v = whatever

V+ = 120 volts

V+ = 120 volts V! = 0 volts

Extra 1:!

b.) What’s the speed of a proton accelerated through 120 volts?!

+! -!

17.!

KE1+ Uin system at time 1 + Wext = KE2 + Uin system at time 2!!!!!" 0 + (q)V+ + 0 = 1

2m

p+ v2 + (q)V#

" 0 + (1.6x10#19 ) 120( ) + 0 = 12

6.7x10#27( )v2 + (1.6x10#19 ) 0( )" v = whatever

V+ = 120 volts

The differences between this and the previous problem is that protons will accelerate from the positive to the negative plates (so the “initial” positions will be reversed), the charge being accelerated will be positive and the mass will be that of a proton. Putting it all together yields:!

V! = 0 volts

Extra 2:!

a.) Determine E. !

b.) Determine!VB

c.) Determine!VD

d.) Determine another way.!VD

A

B

C

D

E!

e.) Release an electron at D. In what direction will it accelerate? Will it experience a voltage decrease or increase as it moves? Will it experience a potential energy decrease or increase as it moves?!

dAC = .7 mdAB = 1 mdAD = .5 m

VA = 45 voltsVC = 25 volts! = 30o

!

18.!

Extra 2:!

a.) Determine E. !

A

B

C

D

E!dAC = .7 m dAB = 1 m dAD = .5 mVA = 45 volts VC = 25 volts ! = 60o

19.!

Note that this problem is completely driven by the relationship!E • d = ! Vfinal position ! Vinitial position( )

E • d = ! Vfinal position ! Vinitial position( ) " E d cos # = ! VB ! VA( ) " E 1 m( ) cos 0o = ! 25 volts ! 45 volts( ) " E = 20 N / C

The first thing to notice is that Points B and C are on equipotential lines (they are both as far down stream in the electric field as the other, so both must have the same voltage). That means that . Assume we traversed the distance between points A and B by starting at A, we go a distance “d” WITH the electric field (see sketch). As such:!

VB = VC = 25 volts

d!

Extra 2:!

b.) Determine!VB

c.) Determine!VDA

B

C

D

E!

!

20.!

We’ve already done this. See Part a.!

Knowing E, assuming we traverse from A to D as the arrow in the sketch shows, and noting that the angle between the line of E and the line of d is we can write:!

E • d = ! VD ! VA( ) " E d cos # = ! VD ! VA( ) " 20 N/C( ) .5 m( ) cos 30o = ! VD ! 45 volts( ) " VD = 36.4 volts

d!

! = 90o " # = 90o " 60o = 30o,

Extra 2:!

A D

E!

!

21.!

We still know E, but we could assume that we traverse from D to A as the arrow in the sketch shows. Noting in that case the angle between the line of E and the line of d is , we can write:!

E • d = ! VD ! VA( ) " E d cos # = ! VA ! VD( ) " 20 N/C( ) .5 m( ) cos 150o = ! 45 volts ! VD( ) " VD = 36.4 volts

d!

d.) Determine another way.!VD!

! = 90o + " = 90o + 60o = 150o

!

!

SAME ANSWER AS PART C . . . !

Note: It doesn’t matter which way you traverse as long as you get the angle right and be consistent with your voltage difference.!

Extra 2:!

A

B

C

D

E!

e.) Release an electron at D. In what direction will it accelerate? !

22.!

If allowed to freely accelerate, electrons always accelerate opposite the direction of the E-fld. In this case, it will take off parallel to the E-fld but moving upstream (physically downward and to the left). (Note: That doesn’t mean you can’t force an electron to go down stream. It just isn’t something an electron would do naturally.!

Will it experience a potential energy decrease or increase as it moves?!

Will it experience a voltage decrease or increase as it moves? !

If protons move from higher voltage to lower voltage, electrons will move from lower voltage to higher voltage.!

All object naturally tend from higher to lower potential energy. What’s tricky here is that a lower voltage multiplied by a negative number (remember, is bigger than a higher voltage multiplied by a negative number. Electrons flow from points at which they have higher potential energy to points where they have lower potential energy.!

Ue!= !q( )V)