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Recitation Week 3 Chapter 23 Problem 21. Two point charges q1 = +2.40 nC and q2 = −6.50 nC are 0.100 m apart. Point A is midway between them; point B is 0.080 m from q1 and 0.060 m from q2 (Fig. 23.31). Take the electric potential to be zero at infinity. Find (a) the potential at point A; (b) the potential at point B; (c) the work done by the electric field on a charge of 2.50 nC that travels from point B to point A.

B 6 0.0

m

0 m

80 0.0

q1 0.050 m A 0.050 m q2 (a) One of the nice features of electric potential is that it is a scalar, so it’s a lot easier to sum up then the vector electric field. kq1 kq2 2.40 · 10−9 C −6.5 · 10−9 C 9 2 2 + VA = V1A + V2A = + = 8.99 · 10 Nm /C · = −737 J/C . (1) r1A r2A 0.050 m 0.050 m (b) VB = V1B + V2B =

kq1 kq2 + = 8.99 · 109 Nm2 /C2 · r1B r2B

2.40 · 10−9 C −6.5 · 10−9 C + 0.080 m 0.060 m

= −704 J/C .

(2)

(c) The electric potential energy change of the moving charge is given by ∆UBA = q∆VBA = q(VA − VB ) = 2.50 · 10−9 C · (−737 + 704) J/C = −82 nJ .

(3)

It makes sense that the charge lost electric potential energy, since it is a positive charge moving to the lower potential point A. The electric potential energy lost went into some other form of energy (kinetic, heat, work, mechanical), but the electric field was giving energy to that other form, so it does positive work: 82 nJ. Problem 22. Two positive point charges, each of magnitude q, are fixed on the y-axis at the points y = +a and y = −a. Take the potential to be zero at an infinite distance from the charges. (a) Show the positions of the charges in a diagram. (b) What is the potential V0 at the origin? (c) Show that the potential at any point on the x-axis is V =

1 2q √ 4πε0 a2 + x2

(4)

(d) Graph the potential on the x-axis as a function of x over the range from x = −4a to x = 4a. (e) What is the potential when x a? Explain why this result is obtained. (a)

ˆj

ˆi

(b) This is particularly easy, since both scalar electric potentials are the same (same q, same r = a, direction doesn’t matter). V0 = V0t + V0b = 2V0t =

2kq . a

(5)

(c) The x-axis is still symmetric (same distance to each charge), so V = 2Vt , and we can use the Pythagorean theorem for r. 2kq 1 2q 1 2q √ V = 2Vt = = = . (6) r 4πε0 r 4πε0 a2 + x2 (c)

V

a

−a

x

(e) As x gets much bigger than a, a2 + x2 → x2 , so V =

2kq k · 2q ≈ . 2 r x2

(7)

In other words, the electric field is pretty much the same as it would be if both charges were sitting at the origin. This makes sense since the distinction that the charge is actually some small distance a off the origin isn’t very important for x a. Problem 25. A positive charge q is fixed at the point x = 0, y = 0, and a negative charge −2q is fixed at the point x = a, y = 0. (a) Show the positions of the charges in a diagram. (b) Derive an expression for the potential V at points on the x-axis as a function of the coordinate x. Take V to be zero at an infinite distance from the charges. (c) At which positions on the x-axis is V = 0? (d) Graph V at points on the x-axis as a function of x in the range from x = −2a to x = +2a. (e) What does the answer to (b) become when x a? Explain why this result is obtained.

ˆj

ˆi

q

−2q

(b) V = V+ + V− =

kq −2kq + = kq |x| |x − a|

1 2 − |x| |x − a|

.

(8)

(c) kq > 0 so V = 0 only when 1 2 − |x| |x − a| 2|x| = |x − a| 0=

−2x = a − x

(9) (10)

2x = a − x a x= 3 a x= 3

x = −a x = −a

2x = x − a x = −a < 0 contradiction

(11) (12) (13) (14)

Where the split to three bands comes from taking all possible absolute value conditions (x < 0 < a; 0 < x < a; 0 < a < xand respectively). (d)

V a

−a

x

Note that I’ve cropped the graph, since the potential explodes to ±∞ at x = 0 and a respectively. (e) As x gets much larger than a, |x − a| → |x|, so V ≈ kq

1 2 − |x| |x|

=

−kq . |x|

(15)

In other words, the electric field is pretty much the same as it would be if both charges were sitting at the origin. This makes sense since the distinction that one charge is actually some small distance a off the origin isn’t very important for x a.

Problem 27. Before the advent of solid-state electronics, vacuum tubes were widely used in radios and other devices. A simple type of vacuum tube known as a diode consists of essentially two electrodes within a highly evacuated enclosure. One electrode, the cathode, is maintained at a high temperature and emits electrons from its surface. A potential difference of a few hundred volts is maintained between the cathode and the other electrode, known as the anode, with the anode at the higher potential. Suppose that in a particular vacuum tube the potential of the anode is 295 V higher than that of the cathode. An electron leaves the surface of the cathode with zero initial speed. Find its speed when it strikes the anode. The electric potential is higher at the anode, so the electric potential energy for an electron will be lower there (because q < 0 and U = qV ). This means that the electron is losing electrical potential energy as it moves from the cathode towards the anode. This energy has to go somewhere, and in this case it ends up as kinetic energy, since there are no other forms of energy in the problem besides kinetic and electric potential energy. Conserving energy, we have Ui = UEi = Uf = K + UEf

(16)

1 K = UEi − UEf = −∆UE = q(Vi − Vf ) = mv 2 2 s r −19 2q(Vi − Vf ) 2(−1.60 · 10 C)(0 − 295 J/C) = = 10.2 Mm/s . v= m 9.11 · 10−31 kg

(17) (18)

Which is pretty fast (but not relativistic). Problem 58. (a) Calculate the potential energy of a system of two small spheres, one carrying a charge of 2.00 µC and the other a charge of −3.50 µC, with their centers separated by a distance of 0.250 m. Assume zero potential energy when the charges are infinitely separated. (b) Suppose that one of the spheres is held in place and the other sphere, which has a mass of 1.50 g, is shot away from it. What minimum initial speed would the moving sphere need in order to escape completely from the attraction of the fixed sphere? (To escape, the moving sphere would have to reach a velocity of zero when it was infinitely distant from the fixed sphere.) (a) Imagine assembling this system from scratch (when V = 0 and the particles are all infinitely seperated). Bringing the first particle in from infinitey takes no energy, since the other particles are still infinitely far away. Once the first charge is in place, the second charge is attracted to the first, and we have to resist this attraction as we bring the charge in slowly, so the system is dropping into a lower potential energy state. U = q− V+ = q−

kq+ 8.99 · 109 Nm2 /C2 · 2.00 · 10−6 C = −3.50 · 10−6 C · = −0.252 J r 0.250 m

(19)

(b) The sphere will need K = −U = 0.252 J of energy to escape, since stopping at infinity will yield (conserving energy) Ei = K + U = Ef = (Kf = 0) + (Uf = 0) = 0 .

(20)

This corresponds to a velocity of 1 mv 2 2 s r 2K 2 · 0.252 J v= = = 18.3 m/s . m 1.50 · 10−3 kg

K=

(21) (22)

Problem 60. A small sphere with mass 1.50 g hangs by a thread between two parallel vertical plates 5.00 cm apart (Fig. 23.36). The plates are insulateing and have uniform surface charge densitied +σ and −σ. The charge on the sphere is q = 8.90·10−6 C. What potential difference between the plates will cause the thread to assume an angle of 30.0◦ with the vertical?

θ q 5.00 cm

The sphere is stationary, so we know the forces must balance. X Fy = T cos θ − mg = 0 mg T = cos θ X Fx = qE − T sin θ = 0

(23) (24) (25)

qE = T sin θ = mg tan θ mg tan θ . E= q

(26) (27)

The E field is constant, so the potential difference between the two plates is given by Z

right

Z

right

E · dx = E

∆V =

dx = Ed ,

(28)

left

left

and the potential difference between the plates is ∆V = Ed =

mgd 1.50 · 10−3 kg · 9.80 m/s2 · 5.00 · 10−2 m tan θ = tan(30.0◦ ) = 47.7 V . q 8.90 · 10−6 C

(29)

v

screen

2.0 cm

Problem 63. Cathode-ray tubes (CRTs) are often found in oscilloscopes and computer monitors. In Fig. 23.38 an electron with an initial speed of 6.50·106 m/s is projected along the axis midway between the deflection plates of a cathode-ray tube. The uniform electric field between the plates has a magnitude of 1.10 · 103 V/m and is upward. (a) What is the force (magnitude and direction) on the electron when it is between the plates? (b) What is the acceleration of the electron (magnitude and direction) when acted on by the force in (a)? (c) How far below the axis has the electron moved when it reaches the end of the plates? (d) At what angle with the axis is it moving as it leaves the plates? (e) How far below the axis will it strike the fluorescent screen S?

6.0 cm

12.0 cm

(a) F = qE = −1.60 · 10−19 C · 1.10 · 103 N/C ˆj = −1.76 · 10−16 N ˆj .

(30)

(b) F = ma

a=

F −1.76 · 10−16 Nˆ = j = −1.93 · 1014 m/s2 ˆj . m 9.11 · 10−31 kg

(31)

(c) This part is just like the inkjet printer problem from first week (Ch. 21, Prob. 86). Dusting off the constant acceleration equations. . . No acceleration in the x direction means vx is constant, so the time-of-flight is given by. ∆x = vx ∆t

∆t =

∆x 2.00 · 10−2 m = = 9.23 ns . vx 6.50 · 106 m/s

(32)

We can plug this time-of-flight into our constant-acceleration equation for y(t), ay 2 t + vy0 t + y0 2 ay 2 ∆y = ∆t = −8.24 mm , 2

y(t) =

(33) (34)

or 8.24 mm below the axis. (d) Rembering our constant acceleration equations the y velocity upon exitting the plates will be vy = ay t + vy0 = ay ∆t = −1.79 Mm/s ,

(35)

so the direction of departure is given by θ = arctan

vy vx

= arctan

−1.79 6.50

= −15.4◦

(36)

(e) The total drop is a combination of the drop from the plates (calculated in (c)) and the drop afterward (from simple trig) ∆y = ∆yc + 12.0 · 10−2 m · tan(15.4◦ ) = 8.24 · 10−3 m + 12.0 · 10−2 m ·

1.79 = 4.13 cm 6.50

(37)

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B 6 0.0

m

0 m

80 0.0

q1 0.050 m A 0.050 m q2 (a) One of the nice features of electric potential is that it is a scalar, so it’s a lot easier to sum up then the vector electric field. kq1 kq2 2.40 · 10−9 C −6.5 · 10−9 C 9 2 2 + VA = V1A + V2A = + = 8.99 · 10 Nm /C · = −737 J/C . (1) r1A r2A 0.050 m 0.050 m (b) VB = V1B + V2B =

kq1 kq2 + = 8.99 · 109 Nm2 /C2 · r1B r2B

2.40 · 10−9 C −6.5 · 10−9 C + 0.080 m 0.060 m

= −704 J/C .

(2)

(c) The electric potential energy change of the moving charge is given by ∆UBA = q∆VBA = q(VA − VB ) = 2.50 · 10−9 C · (−737 + 704) J/C = −82 nJ .

(3)

It makes sense that the charge lost electric potential energy, since it is a positive charge moving to the lower potential point A. The electric potential energy lost went into some other form of energy (kinetic, heat, work, mechanical), but the electric field was giving energy to that other form, so it does positive work: 82 nJ. Problem 22. Two positive point charges, each of magnitude q, are fixed on the y-axis at the points y = +a and y = −a. Take the potential to be zero at an infinite distance from the charges. (a) Show the positions of the charges in a diagram. (b) What is the potential V0 at the origin? (c) Show that the potential at any point on the x-axis is V =

1 2q √ 4πε0 a2 + x2

(4)

(d) Graph the potential on the x-axis as a function of x over the range from x = −4a to x = 4a. (e) What is the potential when x a? Explain why this result is obtained. (a)

ˆj

ˆi

(b) This is particularly easy, since both scalar electric potentials are the same (same q, same r = a, direction doesn’t matter). V0 = V0t + V0b = 2V0t =

2kq . a

(5)

(c) The x-axis is still symmetric (same distance to each charge), so V = 2Vt , and we can use the Pythagorean theorem for r. 2kq 1 2q 1 2q √ V = 2Vt = = = . (6) r 4πε0 r 4πε0 a2 + x2 (c)

V

a

−a

x

(e) As x gets much bigger than a, a2 + x2 → x2 , so V =

2kq k · 2q ≈ . 2 r x2

(7)

In other words, the electric field is pretty much the same as it would be if both charges were sitting at the origin. This makes sense since the distinction that the charge is actually some small distance a off the origin isn’t very important for x a. Problem 25. A positive charge q is fixed at the point x = 0, y = 0, and a negative charge −2q is fixed at the point x = a, y = 0. (a) Show the positions of the charges in a diagram. (b) Derive an expression for the potential V at points on the x-axis as a function of the coordinate x. Take V to be zero at an infinite distance from the charges. (c) At which positions on the x-axis is V = 0? (d) Graph V at points on the x-axis as a function of x in the range from x = −2a to x = +2a. (e) What does the answer to (b) become when x a? Explain why this result is obtained.

ˆj

ˆi

q

−2q

(b) V = V+ + V− =

kq −2kq + = kq |x| |x − a|

1 2 − |x| |x − a|

.

(8)

(c) kq > 0 so V = 0 only when 1 2 − |x| |x − a| 2|x| = |x − a| 0=

−2x = a − x

(9) (10)

2x = a − x a x= 3 a x= 3

x = −a x = −a

2x = x − a x = −a < 0 contradiction

(11) (12) (13) (14)

Where the split to three bands comes from taking all possible absolute value conditions (x < 0 < a; 0 < x < a; 0 < a < xand respectively). (d)

V a

−a

x

Note that I’ve cropped the graph, since the potential explodes to ±∞ at x = 0 and a respectively. (e) As x gets much larger than a, |x − a| → |x|, so V ≈ kq

1 2 − |x| |x|

=

−kq . |x|

(15)

In other words, the electric field is pretty much the same as it would be if both charges were sitting at the origin. This makes sense since the distinction that one charge is actually some small distance a off the origin isn’t very important for x a.

Problem 27. Before the advent of solid-state electronics, vacuum tubes were widely used in radios and other devices. A simple type of vacuum tube known as a diode consists of essentially two electrodes within a highly evacuated enclosure. One electrode, the cathode, is maintained at a high temperature and emits electrons from its surface. A potential difference of a few hundred volts is maintained between the cathode and the other electrode, known as the anode, with the anode at the higher potential. Suppose that in a particular vacuum tube the potential of the anode is 295 V higher than that of the cathode. An electron leaves the surface of the cathode with zero initial speed. Find its speed when it strikes the anode. The electric potential is higher at the anode, so the electric potential energy for an electron will be lower there (because q < 0 and U = qV ). This means that the electron is losing electrical potential energy as it moves from the cathode towards the anode. This energy has to go somewhere, and in this case it ends up as kinetic energy, since there are no other forms of energy in the problem besides kinetic and electric potential energy. Conserving energy, we have Ui = UEi = Uf = K + UEf

(16)

1 K = UEi − UEf = −∆UE = q(Vi − Vf ) = mv 2 2 s r −19 2q(Vi − Vf ) 2(−1.60 · 10 C)(0 − 295 J/C) = = 10.2 Mm/s . v= m 9.11 · 10−31 kg

(17) (18)

Which is pretty fast (but not relativistic). Problem 58. (a) Calculate the potential energy of a system of two small spheres, one carrying a charge of 2.00 µC and the other a charge of −3.50 µC, with their centers separated by a distance of 0.250 m. Assume zero potential energy when the charges are infinitely separated. (b) Suppose that one of the spheres is held in place and the other sphere, which has a mass of 1.50 g, is shot away from it. What minimum initial speed would the moving sphere need in order to escape completely from the attraction of the fixed sphere? (To escape, the moving sphere would have to reach a velocity of zero when it was infinitely distant from the fixed sphere.) (a) Imagine assembling this system from scratch (when V = 0 and the particles are all infinitely seperated). Bringing the first particle in from infinitey takes no energy, since the other particles are still infinitely far away. Once the first charge is in place, the second charge is attracted to the first, and we have to resist this attraction as we bring the charge in slowly, so the system is dropping into a lower potential energy state. U = q− V+ = q−

kq+ 8.99 · 109 Nm2 /C2 · 2.00 · 10−6 C = −3.50 · 10−6 C · = −0.252 J r 0.250 m

(19)

(b) The sphere will need K = −U = 0.252 J of energy to escape, since stopping at infinity will yield (conserving energy) Ei = K + U = Ef = (Kf = 0) + (Uf = 0) = 0 .

(20)

This corresponds to a velocity of 1 mv 2 2 s r 2K 2 · 0.252 J v= = = 18.3 m/s . m 1.50 · 10−3 kg

K=

(21) (22)

Problem 60. A small sphere with mass 1.50 g hangs by a thread between two parallel vertical plates 5.00 cm apart (Fig. 23.36). The plates are insulateing and have uniform surface charge densitied +σ and −σ. The charge on the sphere is q = 8.90·10−6 C. What potential difference between the plates will cause the thread to assume an angle of 30.0◦ with the vertical?

θ q 5.00 cm

The sphere is stationary, so we know the forces must balance. X Fy = T cos θ − mg = 0 mg T = cos θ X Fx = qE − T sin θ = 0

(23) (24) (25)

qE = T sin θ = mg tan θ mg tan θ . E= q

(26) (27)

The E field is constant, so the potential difference between the two plates is given by Z

right

Z

right

E · dx = E

∆V =

dx = Ed ,

(28)

left

left

and the potential difference between the plates is ∆V = Ed =

mgd 1.50 · 10−3 kg · 9.80 m/s2 · 5.00 · 10−2 m tan θ = tan(30.0◦ ) = 47.7 V . q 8.90 · 10−6 C

(29)

v

screen

2.0 cm

Problem 63. Cathode-ray tubes (CRTs) are often found in oscilloscopes and computer monitors. In Fig. 23.38 an electron with an initial speed of 6.50·106 m/s is projected along the axis midway between the deflection plates of a cathode-ray tube. The uniform electric field between the plates has a magnitude of 1.10 · 103 V/m and is upward. (a) What is the force (magnitude and direction) on the electron when it is between the plates? (b) What is the acceleration of the electron (magnitude and direction) when acted on by the force in (a)? (c) How far below the axis has the electron moved when it reaches the end of the plates? (d) At what angle with the axis is it moving as it leaves the plates? (e) How far below the axis will it strike the fluorescent screen S?

6.0 cm

12.0 cm

(a) F = qE = −1.60 · 10−19 C · 1.10 · 103 N/C ˆj = −1.76 · 10−16 N ˆj .

(30)

(b) F = ma

a=

F −1.76 · 10−16 Nˆ = j = −1.93 · 1014 m/s2 ˆj . m 9.11 · 10−31 kg

(31)

(c) This part is just like the inkjet printer problem from first week (Ch. 21, Prob. 86). Dusting off the constant acceleration equations. . . No acceleration in the x direction means vx is constant, so the time-of-flight is given by. ∆x = vx ∆t

∆t =

∆x 2.00 · 10−2 m = = 9.23 ns . vx 6.50 · 106 m/s

(32)

We can plug this time-of-flight into our constant-acceleration equation for y(t), ay 2 t + vy0 t + y0 2 ay 2 ∆y = ∆t = −8.24 mm , 2

y(t) =

(33) (34)

or 8.24 mm below the axis. (d) Rembering our constant acceleration equations the y velocity upon exitting the plates will be vy = ay t + vy0 = ay ∆t = −1.79 Mm/s ,

(35)

so the direction of departure is given by θ = arctan

vy vx

= arctan

−1.79 6.50

= −15.4◦

(36)

(e) The total drop is a combination of the drop from the plates (calculated in (c)) and the drop afterward (from simple trig) ∆y = ∆yc + 12.0 · 10−2 m · tan(15.4◦ ) = 8.24 · 10−3 m + 12.0 · 10−2 m ·

1.79 = 4.13 cm 6.50

(37)

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