An electron enters a parallel plate apparatus 10.0cm long and 2.0cm wide, moving horizontally
at 8.0 × 10^7 m/s. (top plate is positive, electron is 1.0cm from each plate..) The potential difference between the plates is 6.0 × 10^2 V. Calculate...
(a) the vertical deflection of the electron from its original path.
(b) the velocity with which the electron leaves the parallel plate apparatus.
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Im really confused with questions asking for vertical deflection...im not sure what formula to use..should i be using the formula for energy where U=mgh and solve for h? this is the only thing i can think of..any help would be great
The vertical motion will be towards the positively charged plate due to the E field in that direction. The acceleration rate will be
a = F/m = e E/m = e V/(d m)
where d is the plate separation (0.2 m), V is the voltage, and m is the electron mass.
Once you know a, calculate how far it deflects vertically using
y = (1/2) a t^2
t is the time of flight going through the plates, t = 0.10 m/8*10^7 m/s
The horizontal velcoity component will not change. The vertical velocity acquired is a * t
Well, it seems like you're in a bit of an electric pickle! But fear not, Clown Bot is here to help you out with a comedic twist.
To calculate the vertical deflection of the electron, it's important to consider the electric field between the plates. The electron experiences a force due to the electric field, leading to its deflection. The formula you're looking for is actually related to the force.
We can use the equation F = Eq, where F is the force, E is the electric field, and q is the charge of the electron.
Now, the electric field E can be calculated using the formula E = V/d, where V is the potential difference between the plates and d is the distance between the plates.
So, let's start by calculating the electric field: E = (6.0 × 10^2 V) / (2.0 cm) = 3.0 × 10^4 N/C.
Since the electron has a charge of -1.6 × 10^-19 C, we can substitute the values into the equation F = Eq: F = (3.0 × 10^4 N/C) × (-1.6 × 10^-19 C).
Now, we know that the force experienced by the electron is given by F = ma, where m is the mass of the electron and a is the acceleration it experiences due to the force.
Rearranging the equation, we get a = F / m.
The mass of an electron is approximately 9.11 × 10^-31 kg. Plugging in the numbers, we find a = [(3.0 × 10^4 N/C) × (-1.6 × 10^-19 C)] / (9.11 × 10^-31 kg).
Now, here comes the punchline: if we calculate the acceleration, we find it to be approximately -5.3 × 10^13 m/s^2! Negative acceleration indicates that the electron is deflected downwards.
To find the vertical deflection of the electron from its original path, we can use a classic physics equation: d = (1/2)at^2, where d is the displacement, a is the acceleration, and t is the time.
However, we don't have the time involved in this problem. Given that you've provided information only about the electron's speed, we are missing a key piece of information to calculate the time.
Regarding part (b) of the question, to find the velocity with which the electron leaves the parallel plate apparatus, we need the information about the time spent in the apparatus. Unfortunately, without that missing piece, we can't proceed further.
So, I hope I could bring some laughter to your physics dilemma, but unfortunately, without the missing information, the comedy show must come to an end. If you have any other questions or need further clarification, feel free to ask!
To calculate the vertical deflection of the electron, we need to consider the electric force experienced by the electron due to the potential difference between the plates and the magnetic force experienced by the electron due to its motion perpendicular to the magnetic field.
(a) To calculate the vertical deflection of the electron, we can use the equation for the electric force:
F_el = qE
where F_el is the electric force, q is the charge of the electron (-1.6 x 10^-19 C), and E is the electric field strength. The electric field strength between the parallel plates can be calculated using the formula:
E = V/d
where V is the potential difference between the plates (6.0 x 10^2 V) and d is the distance between the plates (2.0 cm = 0.02 m).
Plugging in the values, we get:
E = (6.0 x 10^2 V)/(0.02 m)
E = 3.0 x 10^4 V/m
Now, we can calculate the electric force:
F_el = (q)(E)
F_el = (-1.6 x 10^-19 C)(3.0 x 10^4 V/m)
F_el = -4.8 x 10^-15 N
Since the electric force is acting vertically upward, the vertical deflection can be calculated using the equation:
F_el = m(a)
where m is the mass of the electron (9.11 x 10^-31 kg) and a is the vertical acceleration.
Rearranging the equation, we get:
a = F_el/m
a = (-4.8 x 10^-15 N)/(9.11 x 10^-31 kg)
a = -5.27 x 10^15 m/s^2
Note that the negative sign indicates that the vertical acceleration is downward.
Using the kinematic equation:
v_f^2 = v_i^2 + 2aΔy
where v_f is the final velocity in the vertical direction, v_i is the initial velocity in the vertical direction (zero in this case), a is the vertical acceleration, and Δy is the vertical displacement, we can calculate the vertical deflection:
0^2 = (8.0 x 10^7 m/s)^2 + 2(-5.27 x 10^15 m/s^2)Δy
Simplifying the equation, we have:
0 = 6.4 x 10^14 m^2/s^2 + (-10.5 x 10^15 m^2/s^2)Δy
Rearranging the equation, we get:
10.5 x 10^15 m^2/s^2 Δy = 6.4 x 10^14 m^2/s^2
Δy = (6.4 x 10^14 m^2/s^2) / (10.5 x 10^15 m^2/s^2)
Δy = 0.061 m
Therefore, the vertical deflection of the electron from its original path is 0.061 m (or 6.1 cm).
(b) To calculate the velocity with which the electron leaves the parallel plate apparatus, we need to consider the horizontal motion of the electron.
The electron experiences a magnetic force due to its motion perpendicular to the magnetic field, given by the equation:
F_mag = qvB
where F_mag is the magnetic force, q is the charge of the electron (-1.6 x 10^-19 C), v is the velocity of the electron, and B is the magnetic field strength.
Since the electron is moving horizontally, the magnetic force acts vertically, exactly opposite to the electric force.
Therefore, we can equate the two forces:
|F_mag| = |F_el|
Using the values:
|qvB| = |-4.8 x 10^-15 N|
Rearranging the equation, we get:
B = (|-4.8 x 10^-15 N|)/(|q|v)
B = (4.8 x 10^-15 N)/(1.6 x 10^-19 C)(8.0 x 10^7 m/s)
B = 0.03 T
The velocity with which the electron leaves the parallel plate apparatus is the same as its initial velocity, which is 8.0 x 10^7 m/s.
Therefore, the velocity with which the electron leaves the parallel plate apparatus is 8.0 x 10^7 m/s.
To calculate the vertical deflection of the electron and its final velocity, we need to consider the forces acting on it.
(a) The vertical deflection of the electron can be determined by considering the electric force acting on it due to the potential difference between the plates.
The electric force (F_electric) can be calculated using the formula:
F_electric = q * E
where q is the charge of the electron and E is the electric field between the plates. The electric field can be calculated using the formula:
E = V / d
where V is the potential difference between the plates and d is the distance between the plates.
First, let's calculate the electric field:
E = (6.0 × 10^2 V) / (2 * 10^(-2) m) = 3.0 × 10^4 N/C
The charge of an electron (q) is approximately -1.6 × 10^(-19) C.
Now, we can find the electric force:
F_electric = (-1.6 × 10^(-19) C) * (3.0 × 10^4 N/C)
= -4.8 × 10^(-15) N
Since the electric force is directed upward and the electron's original path is horizontal, the vertical deflection is the distance the electron is displaced vertically from its original path. We can calculate it using the formula:
Vertical deflection (h) = (1/2) * a * t^2
where a is the acceleration and t is the time taken.
The net force acting on the electron in the vertical direction is due to the electric field and gravity, so we have:
Net Force (F_net) = F_electric + F_gravity
The force due to gravity (F_gravity) can be calculated using the formula:
F_gravity = m * g
where m is the mass of the electron and g is the acceleration due to gravity (approximately 9.8 m/s^2).
The mass of the electron (m) is approximately 9.1 × 10^(-31) kg.
Now, let's calculate the net force:
F_net = (-4.8 × 10^(-15) N) + (9.1 × 10^(-31) kg) * (9.8 m/s^2)
= (-4.8 × 10^(-15) N) + 8.9 × 10^(-30) N (Neglecting the gravitational force compared to the electric force)
The net force is very small compared to the electric force, so the vertical deflection will also be very small. In this case, we can neglect the vertical deflection. Therefore, the electron's path remains predominantly horizontal.
(b) Since the electric force is the only significant force acting on the electron in the horizontal direction, it does not affect the velocity of the electron. Therefore, the velocity with which the electron leaves the parallel plate apparatus will be the same as its initial velocity, which is 8.0 × 10^7 m/s.