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
on charge q due to charge Q?
d. What is the magnitude of the electric force
on 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
on 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
on 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 7 / 29
2. A negatively charged sphere with charge Q = -20 µC
is placed on an insulating table a tiny charge q and mass
of 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
on charge q.
d. What should be the sign and magnitude of the charge q in
order 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
of 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
of 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?
2
E = kQ/r
9
2
2
-6
E = (9x10 Nm /C )(20x10 C)/(0.8m)
E = 2.8x105 N/C
2
Slide 10 / 29
2. A negatively charged sphere with charge Q = -20 µC
is placed on an insulating table a tiny charge q and mass
of 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
on charge q.
FE
mg
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
of 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
order 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
2
kQ1Q2/r = mg
Q2 = mgr2/kQ1
Q2 = (0.0036 kg)(9.8 m/s2)(0.8 m)2/(9x109 Nm2/C2)(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
origin.
d. Calculate the net electric field at
the origin due to two charges Q1 and
Q2.
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
4
E = 1.8X10 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
9
2
2
-6
2
E = (9x10 Nm /C ) (32x10 C) / (3 m)
4
E = 1.8X10 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
origin.
Enet
EQ1
EQ2
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
4 2
4 2
E = (1.8x10 ) + (1.8x10 )
4
E = 2.5x10 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
of 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
of 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 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
of 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
-6
FE = (500 N/C) (90x10 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
of 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
of 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
_________________
FTcosθ = mg
-1
= tanθ = FE / mg
θ = tan (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
of 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
of 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.
i.
ii.
#F = #(FE2 + mg2)
#F = #(.0452 + .0982) = .108N
mg
o
.7
24
2
a = F / m = 10.8 m/s
FE
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 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.
FT
FT
FE
FE
mg
mg
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#
o
= 1.8 sin 25
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°.
2
o 2
o
9
q = √(r FE / K) = √( (L sin 10 ) (mg tan 10 ) / 9x10 ) )
q = 1.22x10- 6 C
Δq = 5.4 μC - 1.2 μC = 4.2 μC