## 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 C

Divide both sides of that equation by 1.6 x 10 - 19 to get

1 C = 6.25 x 10 + 18 e

17.2 A mole of anything is Avogadro's number (NA = 6.02 x 1023) of those things. How many coulombs are there in a mole of electrons?

Q = NA e = (6.02 x 1023) (1.6 x 10 - 19 C) = 9.63 x 104 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

Fel = k Q q / r2

For the hydrogen atom,

Q = q = e = 1.6 x 10 - 19 C

Therefore,

Fel = k Q q / r2

Fel = (9 x 109)(1.6 x 10 - 19)(1.6 x 10 - 19) / (0.52 x 10 - 10)2

Fel = 8.52 x 10 - 8 N

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

Fc = m v2/r

v2 = Fc r / m

v2 = Fel r / m

v2 = (8.52 x 10 - 8 N)(0.52 x 10 - 10 m) / (9.11 x 10 - 31 kg)

v2 = [(8.52 x 10 - 8)(0.52 x 10 - 10) / (9.11 x 10 - 31)] (m/s)2

v2 = [4.86 x 1012] (m/s)2

v = 2.21 x 106 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 108 m/s

v/c = (2.21 x 106) / (3.0 x 108)

v/c = 0.00737

That is just under one percent the speed of light ).

(c) The kinetic energy is given by

KE = (1/2) m v2

KE = (1/2) (9.11 x 10 - 31) (2.21 x 106)2 J

KE = 2.22 x 10 - 18 J

KE = 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!

Fy = 0

Ty = w = mg = (0.005 kg) ( 9.8 m/s2)

Ty = 4.9 x 10 - 2 N

Ty = T cos 15° = 0.966 T

T = Ty / 0.966

T = (4.9 x 10 - 2 N)/0.966)

T = 5.07 x 10 - 2 N

Tx = T sin 15° = 0.259 T = (0.259)(5.07 x 10 - 2 N)

Tx = 1.31 x 10 - 2 N

Fx = 0

Fel = Tx

Fel = k Q q / r2 = Tx

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 15o = (20 cm) (0.259) = 5.18 cm

r = 10.36 cm

r = 0.104 m

(be careful of the units!)

Fel = k Q q / r2 = Tx

Q q = Tx r2 / k

Remember, Q = q

Q2 = (1.31 x 10 - 2 N)(0.104 m)2 / (9 x 109 N m2/C2)

Q2 = 1.57 x 10 - 14 C2

Q = 1.25 x 10 - 7 C

17.18 Charge q = 20  C is placed at the origin. Charges Q1 = - 10 C and Q2 = 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 C

Remember, force is a vector! Coulomb's Law allows us to find the force on q due to Q1 and it allows us to find the force on q due to Q2. Then we must add those two forces as vectors.

Let us begin with F1, the force on q due to Q1,

F1 = k Q1 q / r2

F1 = (9 x 109) (10 x 10 - 6) (20 x 10 - 6) / (0.1)2

Be very careful of the units!

Don't write something like

F1 = (9 x 109)(10)(20)/(10)2

The units are very important!

Also, note that 10 cm = 0.1 m

F1 = 1.8 x 102 N = 180 N

Remember, force is a vector, so we can write this as

F1x = 0

F1y = 1.8 x 102 N

Now, we continue with F2, the force due to charge Q2 as it acts on charge q ;

F2 = k Q2 q / r2

Again, remember that r = 10 cm = 0.1 m

F2 = (9 x 109) (30 x 10 - 6) (20 x 10 - 6) / (0.1)2

F2 = 5.4 x 102 N = - 540 N

Remember, force is a vector, so we can write this as

F2x = - 5.4 x 102 N

The minus sign on F2x indicates that it points to the left.

F2y = 0

Now we add the two forces--as vectors!

Fnet,x = F1x + F2x = 0 - 5.4 x 102 N = - 5.4 x 102 N

Fnet,y = F1y + F2y = 1.8 x 102 N + 0 = 1.8 x 102 N

Now that we have the components, we need the magnitude and direction.

Fnet2 = Fnet,x2 + Fnet,y2

Fnet2 = [(-5.4)2 + (1.8)2] [x 102 N ]2

Fnet = 5.69 x 102 N = 569 N

tan = opp/adj = (1.8) / (5.4) = 0.33

= 18.4o

17.27 A proton is placed in a uniform electric field that has a magnitude of 2.3 x 105 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 105 N/C)(1.6 x 10 - 19 C)

F = 3.68 x 10 - 14 N

The 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 / r2

E = (9 x 109)(1.6 x 10 - 19) / (1 x 10 - 13)2

E = 1.44 x 1017 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.

Due to symmetry,

E1 = E2 = E = k q / r2 = (9 x 109)(1 x 10 - 6)/(1)2 = 9 x 103 N/C

By symmetry,

E1x = - E2x

so that

Enet,x = 0

Also, by symmetry,

E1y = E2y = E cos 30o = (9 x 103 N/C ) 0.866 = 7.79 x 103 N/C

Enet,y = E1y + E2y = 15.6 x 103 N/C

Enet = 15.6 x 103 N/C

directed 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 on or by the proton?

Has the proton's electrical potential energy increased or decreased?

F = E q

F = (1500 N/C )(1.6 x 10 - 19 C ) = 2.4 x 10 - 16 N

W = F d cos

W = (2.4 x 10 - 16 N)(0.015 m)

W = 3.6 x 10 - 18 J

This is work done on or to the proton since the electric force is in the direction the proton moves.

Doing work on the proton -- the work of the electric field on the proton -- decreases its potential energy. If we release the proton at its final position, it will not return 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 on or by the electron?

Has the electron's electrical potential energy increased or decreased?

F = E q

F = (3000 N/C )(1.6 x 10 - 19 C ) = 4.8 x 10 - 16 N

W = F d cos

W = (4.8 x 10 - 16 N)(0.025 m)

W = 1.2 x 10 - 17 J

W = 12 x 10 - 18 J

This is work done to the electron since the electric force is opposite to the direction the electron moves.

Doing work to the electron -- the work of the electric field on or to the electron -- increases its potential energy. If we release the electron at its final position, it will return 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 / rf - 1 / ri ]

PE = ( 9 x 109 ) ( 10 x 10 - 6 ) ( 2 x 10 - 6 ) [ 1/0.1 - 1 /1.0 ]

That is,

ri = 1.0 m and rf = 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 on or to the 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!

PEtot = PE1 + PE2

PE = k Q q / r

PE1 = ( 9 x 109 )( 2.0 x 10 - 6 )( 3.0 x 10 - 6 ) / (1.0)

PE1 = 0.0540 J

PE2 = ( 9 x 109 )( - 1.5 x 10 - 6 )( 3.0 x 10 - 6 ) / (0.2)

PE2 = - 0.2025 J

PEtot = PE1 + PE2

PEtot = (0.0540 - 0.2025) J

PEtot = - 0.1485 J

W = 0.1485 J