Download It is halfway between the plates, so the potential must be 34 V. Or

5.
(a) What is the potential at point K?
It is halfway between the plates, so the potential must be 34 V.
Or, find the value of the electric field, see (b). Then at point K, the voltage would be
20 + .025(560) = 34 V.
(b) What is the electric field at point K?
E = V/d = (48-20)/.05 = 560 V/m
It is a uniform electric field, so I calculated based on the potential difference between the plates. I could
have plugged in (34-20)/0.025, but it equals the same thing.
(c) If an electron were placed at point K, and released from rest, which plate would it strike? How fast will
it be going when it hits the plate?
qV = 1/2mv2
v = sqrt(2qV/m) = sqrt(2*1.6×10-19)(48-34)/9.11×10-31) = 2.2×106 m/s
6. In a television set, electrons are first accelerated from rest through a potential difference in an electron gun.
They then pass through deflecting plates before striking the screen.
(a) Determine the potential difference through which the electrons must be accelerated in the electron gun in
order to have a speed of 6.0 × 107 m/s when they enter the deflecting plates.
qV = 1/2mv2
V = 1/2mv2/q = ½(9.11×10-31)(6×107)2/(1.6×10-19) = 10248.75 V
Answer key: 10000 V or 104 V
(3 points)
The pair of horizontal plates shown below is used to deflect electrons up or down in the television set by placing
a potential difference across them. The plates have length 0.04 m and separation 0.012 m, and the right edge of
the plates is 0.50 in from the screen. A potential difference of 200 V is applied across the plates, and the
electrons are deflected toward the top of the screen. Assume that the electrons enter horizontally midway
between the plates with a speed of 6.0 × 107 m/s and that fringing effects at the edges of the plates and gravity
are negligible.
(b) Which plate in the pair must be at the higher potential for the electrons to be deflected upward? Check the
appropriate box below.
 Upper plate
 Lower plate
Justify your answer.
The electrons were deflected towards the top. That means the electric field is in the opposite direction
down. The field is drawn from High to Low. So the top plate must have the higher potential.
(2 points)
(c) Considering only an electron's motion as it moves through the space between the plates, compute the
following.
i. The time required for the electron to move through the plates
x = vt
.04 = 6×107 t
t = 6.7 × 10-10 s
(only 1 point)
ii. The vertical displacement of the electron while it is between the plates
So I’m thinking kinematics and horizontal electron tiger.
x = vot + 1/2 at2
the initial upward velocity is 0. We just found t. need to get acceleration.
ma = qE
ma = qV/d
a = qV/md = (1.6×10-19C*200V)/(9.11×10-19kg*0.012m) = 2.93×1015 m/s2
x = 1/2 (2.93×1015) (6.7 × 10-10)2 = 6.5 × 10-4 m
(5 points)
(d) Explain why it is a reasonable assumption to neglect gravity in part (c).
FE >> FG
aE >> g
2.93×1015 m/s2 >> 9.8 m/s2
(2 points)
(e) Still neglecting gravity, describe the path of the electrons from the time they leave the plates until they strike
the screen. State a reason for your answer.
The electron will travel in a parabolic upward shape while between the plates since the only force acting
is the electric force. After it passes through the plates, there are no forces acting on it, so it will travel in a
straight line in the direction it was traveling when it leaves the plates until it hits the screen.
(2 points)
7. A small conducting sphere of mass 5 × 10−3 kilogram, attached to a string of length 2 × 10−1 meter, is at rest
in a uniform electric field E, directed horizontally to the right as shown above. There is a charge of 5 × 10−6
coulomb on the sphere.
The string makes an angle of 30° with the vertical. Assume g = 10 meters second squared.
(a) In the space below, draw and label all the forces acting on the sphere.
(b) Calculate the tension in the string and the magnitude of the electric field.
Equilibrium
Tsin(30) = FE
Tcos(30) = mg
Tcos(30) = mg
T = mg/cos(30) = (5 × 10−3)(10)/cos(30) = 0.058 N
Tsin(30) = F = qE
E = Tsin(30)/q = (0.058)sin(30)/( 5 × 10−6) = 5800 N/C
(c) The string now breaks. Describe the subsequent motion of the sphere and sketch on the following diagram
the path of the sphere while in the electric field.