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Dec 04, 2023 399 likes •457 views

Slide 1 / 29 Slide 2 / 29 1. Sphere 1 carries a positive charge 1. Sphere 1 carries a positive charge Q = +6 C and located at the origin. Q = +6 C and located at the origin. a. What is the direction of the electric field a. What is

slide-1
SLIDE 1

Slide 1 / 29

1. Sphere 1 carries a positive charge Q = +6 µC and located at the origin.

  • a. What is the direction of the electric field at

point P 0.6 m away from the origin?

  • b. What is the magnitude of the electric

field at point P? A test charge q = +1 µC and mass m = 2.5 g is brought from infinity and placed at point P

  • c. What is the direction of the electric force
  • n charge q due to charge Q?
  • d. What is the magnitude of the electric force
  • n charge q due to charge Q?
  • e. What is the acceleration of charge q at the

instant when it is released from point P?

Slide 2 / 29

  • 1. Sphere 1 carries a positive charge

Q = +6 µC and located at the origin.

  • a. What is the direction of the electric field

at point P 0.6 m away from the origin? to the right

Slide 3 / 29

1. Sphere 1 carries a positive charge Q = +6 µC and located at the origin.

  • b. What is the magnitude of the electric

field at point P?

E = kq/r2 E = (9x109 Nm2/C2)(6x10-6C)/(0.6m)2 E = 1.5x105 N/C

Slide 4 / 29

1. Sphere 1 carries a positive charge Q = +6 µC and located at the origin. A test charge q = +1 µC and mass m = 2.5 g is brought from infinity and placed at point P

  • c. What is the direction of the electric force
  • n charge q due to charge Q?

to the right

Slide 5 / 29

1. Sphere 1 carries a positive charge Q = +6 µC and located at the origin. A test charge q = +1 µC and mass m = 2.5 g is brought from infinity and placed at point P

  • d. What is the magnitude of the electric force
  • n charge q due to charge Q?

FE = kQq/r2 or FE = qE FE = (1x10-6C)(1.5x105 N/C) FE = 0.15N

Slide 6 / 29

1. Sphere 1 carries a positive charge Q = +6 µC and located at the origin. A test charge q = +1 µC and mass m = 2.5 g is brought from infinity and placed at point P

  • e. What is the acceleration of charge q at the

instant when it is released from point P?

ΣF = ma a = ΣF/m a = (0.15N)/(0.025kg) a = 60 m/s2

slide-2
SLIDE 2

Slide 7 / 29

  • 2. A negatively charged sphere with charge Q = -20 µC

is placed on an insulating table a tiny charge q and mass

  • f m = 3.6 g is suspended at rest above charge Q. The

distance between the charges is 0.8 m.

  • a. What is the direction of the electric field due to

charge Q at the distance d above charge Q?

  • b. What is the magnitude of the electric field due to

charge Q at the distance d above charge Q?

  • c. On the diagram below show all the forces applied
  • n charge q.
  • d. What should be the sign and magnitude of the charge q in
  • rder to keep it at equilibrium?

Slide 8 / 29

  • 2. A negatively charged sphere with charge Q = -20 µC

is placed on an insulating table a tiny charge q and mass

  • f m = 3.6 g is suspended at rest above charge Q. The

distance between the charges is 0.8 m.

  • a. What is the direction of the electric field due to

charge Q at the distance d above charge Q?

down

Slide 9 / 29

  • 2. A negatively charged sphere with charge Q = -20 µC

is placed on an insulating table a tiny charge q and mass

  • f m = 3.6 g is suspended at rest above charge Q. The

distance between the charges is 0.8 m.

  • b. What is the magnitude of the electric field due to

charge Q at the distance d above charge Q?

E = kQ/r2 E = (9x109 Nm2/C2)(20x10-6C)/(0.8m)2 E = 2.8x105 N/C

Slide 10 / 29

FE mg

  • 2. A negatively charged sphere with charge Q = -20 µC

is placed on an insulating table a tiny charge q and mass

  • f m = 3.6 g is suspended at rest above charge Q. The

distance between the charges is 0.8 m.

  • c. On the diagram below show all the forces applied
  • n charge q.

Slide 11 / 29

  • 2. A negatively charged sphere with charge Q = -20 µC

is placed on an insulating table a tiny charge q and mass

  • f m = 3.6 g is suspended at rest above charge Q. The

distance between the charges is 0.8 m.

  • d. What should be the sign and magnitude of the charge q in
  • rder to keep it at equilibrium?

For the magnitude: ΣF = ma since the charge q is at rest, a = 0... ΣF = 0 FE - W = 0 FE = W kQ1Q2/r

2 = mg

Q2 = mgr

2/kQ1

Q2 = (0.0036 kg)(9.8 m/s

2)(0.8 m) 2/(9x10 9 Nm 2/C 2)(20x10

  • 6 C)

Q2 = 1.25x10

  • 7 C

Since the charge on the sphere is negative, charge q must be repelled and therefore, Q2 is negative

Slide 12 / 29

  • 3. A charge Q1 = -32 µC is fixed on the y axis at

y = 4 m, and a charge Q2 = +18 µC is fixed on the x axis at x = 3 m.

  • a. Calculate the magnitude of the electric field

E1 at the origin due to charge Q1

  • b. Calculate the magnitude of the electric field

E2 at the origin due to charge Q2. c. On the diagram (to the right), draw and label the electric fields E1, E2 and the net electric field at the

  • rigin.
  • d. Calculate the net electric field at

the origin due to two charges Q1 and Q2.

slide-3
SLIDE 3

Slide 13 / 29

  • 3. A charge Q1 = -32 µC is fixed on the y axis at

y = 4 m, and a charge Q2 = +18 µC is fixed on the x axis at x = 3 m.

  • a. Calculate the magnitude of the electric field

E1 at the origin due to charge Q1 E = KQ1 / r1

2

E = (9x109 Nm2/C2) (18x10-6 C2) / (4 m)2 E = 1.8X104 N/C

Slide 14 / 29

  • 3. A charge Q1 = -32 µC is fixed on the y axis at

y = 4 m, and a charge Q2 = +18 µC is fixed on the x axis at x = 3 m.

  • b. Calculate the magnitude of the electric field

E2 at the origin due to charge Q2. E = KQ1 / r1

2

E = (9x109 Nm2/C2) (32x10-6 C) / (3 m)2 E = 1.8X104 N/C

Slide 15 / 29

  • 3. A charge Q1 = -32 µC is fixed on the y axis at

y = 4 m, and a charge Q2 = +18 µC is fixed on the x axis at x = 3 m. c. On the diagram (to the right), draw and label the electric fields E1, E2 and the net electric field at the

  • rigin.

EQ2 EQ1 Enet

Slide 16 / 29

  • 3. A charge Q1 = -32 µC is fixed on the y axis at

y = 4 m, and a charge Q2 = +18 µC is fixed on the x axis at x = 3 m.

  • d. Calculate the net electric field at

the origin due to two charges Q1 and Q2. Enet = EQ1 + EQ2 E = (1.8x104)2 + (1.8x104)2 E = 2.5x104 N/C

Slide 17 / 29

  • 4. A wall has a negative charge

distribution producing a uniform field. A small dielectric sphere of mass 10 g and charge of -90 µC is attached to one end of insulating string 0.6 m long. The other end

  • f the string is attached to the wall. The

uniform electric field has a magnitude of 500 N/C.

  • a. What is the direction of the uniform electric field?
  • b. What is the direction and magnitude of the electric field on charge q due to uniform

electric filed?

  • c. On the diagram below draw and label all the applied forces on charge q.
  • d. Calculate the angle θ between the wall and the string.
  • e. Calculate the shortest distance between the wall and charged sphere.
  • f. The string is cut:
  • i. Calculate the magnitude of the net acceleration of the charged sphere q.
  • ii. Describe the resulting path of the charged sphere.

Slide 18 / 29

  • 4. A wall has a negative charge

distribution producing a uniform field. A small dielectric sphere of mass 10 g and charge of -90 µC is attached to one end of insulating string 0.6 m long. The other end

  • f the string is attached to the wall. The

uniform electric field has a magnitude of 500 N/C.

  • a. What is the direction of the uniform electric field?

left

slide-4
SLIDE 4

Slide 19 / 29

  • 4. A wall has a negative charge

distribution producing a uniform field. A small dielectric sphere of mass 10 g and charge of -90 µC is attached to one end of insulating string 0.6 m long. The other end

  • f the string is attached to the wall. The

uniform electric field has a magnitude of 500 N/C.

  • b. What is the direction and magnitude of the electric field on charge q due to uniform

electric filed?

FE = Eq FE = (500 N/C) (90x10-6 C) = 0.045 N

Slide 20 / 29

  • 4. A wall has a negative charge

distribution producing a uniform field. A small dielectric sphere of mass 10 g and charge of -90 µC is attached to one end of insulating string 0.6 m long. The other end

  • f the string is attached to the wall. The

uniform electric field has a magnitude of 500 N/C.

  • c. On the diagram below draw and label all the applied forces on charge q.

θ FT FE mg

Slide 21 / 29

  • 4. A wall has a negative charge

distribution producing a uniform field. A small dielectric sphere of mass 10 g and charge of -90 µC is attached to one end of insulating string 0.6 m long. The other end

  • f the string is attached to the wall. The

uniform electric field has a magnitude of 500 N/C.

  • d. Calculate the angle θ between the wall and the string.

FTsinθ = FE

_________________ = tanθ = FE / mg

FTcosθ = mg θ = tan-1 (FE / mg) θ = 24.7o

Slide 22 / 29

  • 4. A wall has a negative charge

distribution producing a uniform field. A small dielectric sphere of mass 10 g and charge of -90 µC is attached to one end of insulating string 0.6 m long. The other end

  • f the string is attached to the wall. The

uniform electric field has a magnitude of 500 N/C.

  • e. Calculate the shortest distance between the wall and charged sphere.

r = Lsin# r = .6m sin(24) r = .25m

Slide 23 / 29

  • 4. A wall has a negative charge

distribution producing a uniform field. A small dielectric sphere of mass 10 g and charge of -90 µC is attached to one end of insulating string 0.6 m long. The other end

  • f the string is attached to the wall. The

uniform electric field has a magnitude of 500 N/C.

  • f. The string is cut:
  • i. Calculate the magnitude of the net acceleration of the charged sphere q.
  • ii. Describe the resulting path of the charged sphere.

24.7

  • FE

mg i. ii.

#F = #(FE

2 + mg2)

#F = #(.0452 + .0982) = .108N a = F / m = 10.8 m/s2

Slide 24 / 29

  • 5. Two small spheres with masses m1 = m2 = m = 25 g are hung by

two silk threads of length L = 1.8 m from a common point. The spheres are charged with equal charges q1 = q2 = q, each thread makes an angle # = 25° with the vertical line.

  • a. On the diagram (to the left) draw and label all the applied

forces on each sphere.

  • b. Calculate the distance between the spheres.
  • c. Calculate the magnitude of the electric force between the

spheres.

  • d. Calculate the magnitude of the charge on spheres.

Some of the electric charge has left the spheres due to humidity in the room. Assuming the electric charge leaves the spheres at the same rate and after some time the new angle between the string and the vertical line is 10°.

  • e. Calculate the change in the electric charge on each sphere when the angle

changes from 25° to 10°.

slide-5
SLIDE 5

Slide 25 / 29

  • 5. Two small spheres with masses m1 = m2 = m = 25 g are hung by

two silk threads of length L = 1.8 m from a common point. The spheres are charged with equal charges q1 = q2 = q, each thread makes an angle # = 25° with the vertical line.

  • a. On the diagram (to the left) draw and label all the applied

forces on each sphere.

mg mg FE FE FT FT

Slide 26 / 29

  • 5. Two small spheres with masses m1 = m2 = m = 25 g are hung by

two silk threads of length L = 1.8 m from a common point. The spheres are charged with equal charges q1 = q2 = q, each thread makes an angle # = 25° with the vertical line.

  • b. Calculate the distance between the spheres.

r1 = r2 = L sin# = 1.8 sin 25 o r1 = r2 = .76 m r = 1.52 m

Slide 27 / 29

  • 5. Two small spheres with masses m1 = m2 = m = 25 g are hung by

two silk threads of length L = 1.8 m from a common point. The spheres are charged with equal charges q1 = q2 = q, each thread makes an angle # = 25° with the vertical line.

  • c. Calculate the magnitude of the electric force between the

spheres.

X: FE = FT Sin# Y: mg = FT Cos#

tanθ = FE/mg FE = mg tanθ FE = (0.025 kg)(9.8 m/s2)(tan 25) = 0.114 N

Slide 28 / 29

  • 5. Two small spheres with masses m1 = m2 = m = 25 g are hung by

two silk threads of length L = 1.8 m from a common point. The spheres are charged with equal charges q1 = q2 = q, each thread makes an angle # = 25° with the vertical line.

  • d. Calculate the magnitude of the charge on spheres.

FE = K q2 / r2 q = #(r2 FE / K) = #( (1.52 m)2 (.114 N) / 9x109

Nm2/C2) )

q = 5.4x10- 4 C

Slide 29 / 29

  • 5. Two small spheres with masses m1 = m2 = m = 25 g are hung by

two silk threads of length L = 1.8 m from a common point. The spheres are charged with equal charges q1 = q2 = q, each thread makes an angle # = 25° with the vertical line.

Some of the electric charge has left the spheres due to humidity in the room. Assuming the electric charge leaves the spheres at the same rate and after some time the new angle between the string and the vertical line is 10°.

  • e. Calculate the change in the electric charge on each sphere when the angle

changes from 25° to 10°.

q = √(r2 FE / K) = √( (L sin 10o )2 (mg tan 10o ) / 9x109) ) q = 1.22x10- 6 C Δq = 5.4 μC - 1.2 μC = 4.2 μC