If an electron travels 0.250 m from an electron gun to a TV screen in 32.0 ns, what voltage was used to accelerate it? (Note that the voltage you obtain here is lower than actually used in TVs to avoid the necessity of relativistic corrections.)
Well, before we zap away with the answer, let's buckle up for a little physics fun, shall we?
In order to find the voltage used to accelerate our electron, we'll take a look at the good ol' kinematic equations. Specifically, we'll use the equation relating distance, time, and acceleration:
d = (1/2) * a * t^2
But since we're dealing with constant acceleration, we can simplify this equation to:
d = (1/2) * a * t^2
Now, we need to find the acceleration (a) of the electron. We know that the initial velocity (vi) is zero, and the final velocity (vf) is the velocity of light (3.00 x 10^8 m/s) because the electron is traveling at a speed very close to the speed of light. So we can use the equation:
vf^2 = vi^2 + 2*a*d
Since vi is zero, this equation becomes:
vf^2 = 2*a*d
Rearranging for acceleration:
a = vf^2 / (2*d)
Now, we know that our electron traveled a distance of 0.250 m in 32.0 ns (or 32.0 x 10^-9 s). Plugging in the values, we get:
a = (3.00 x 10^8 m/s)^2 / (2 * 0.250 m)
a ≈ 7.20 x 10^16 m/s^2
Now, let's move on to finding the voltage (V). We can use the equation:
V = a * q / e
Where q is the charge of the electron (approximately 1.60 x 10^-19 C) and e is the elementary charge (also approximately 1.60 x 10^-19 C). Plugging in the values:
V = (7.20 x 10^16 m/s^2) * (1.60 x 10^-19 C) / (1.60 x 10^-19 C)
V ≈ 7.20 x 10^16 V
So, the voltage used to accelerate our electron is approximately 7.20 x 10^16 volts. Now let's get ready for liftoff and "electrify" some TV screens!
To calculate the voltage used to accelerate the electron, we can use the equation for the displacement of a uniformly accelerated particle:
d = (1/2)at^2 + vt
Where:
d = displacement (0.250 m)
a = acceleration (constant)
t = time (32.0 ns = 32.0 × 10^-9 s)
v = initial velocity (0, as the electron starts from rest)
Let's solve for the acceleration first:
d = (1/2)at^2
Rearranging the equation:
2d = at^2
a = (2d) / t^2
Substituting the given values:
a = (2 * 0.250 m) / (32.0 × 10^-9 s)^2
a = 1.953 × 10^15 m/s^2
Now, we can calculate the voltage using the formula for the electric field:
E = V / d
Where:
E = electric field (acceleration due to the applied voltage)
V = voltage
d = displacement (0.250 m)
Rearranging the equation:
V = Ed
Substituting the known values:
V = (1.953 × 10^15 m/s^2) * 0.250 m
V = 4.8825 × 10^14 volts
Therefore, the voltage used to accelerate the electron is approximately 4.8825 × 10^14 volts.
To find the voltage used to accelerate the electron, we need to use the equation for the acceleration of a charged particle in an electric field. The equation is:
acceleration = voltage / distance
In this case, the acceleration of the electron is caused by the electric field created by the voltage. We know the distance traveled by the electron (0.250 m) and the time taken (32.0 ns).
First, let's convert the time to seconds:
32.0 ns = 32.0 × 10^(-9) s
Next, we need to calculate the average velocity of the electron. We can use the formula:
average velocity = distance / time
Substituting the values:
average velocity = 0.250 m / 32.0 × 10^(-9) s
Now, let's convert the average velocity to meters per second:
average velocity = 0.250 m / (32.0 × 10^(-9) s) = 7.8125 × 10^6 m/s
The acceleration of the electron is given by:
acceleration = change in velocity / time
Since the electron is initially at rest, the change in velocity is simply the average velocity we calculated. Substituting the values into the equation:
acceleration = (7.8125 × 10^6 m/s) / (32.0 × 10^(-9) s)
Now, we can rearrange the equation to solve for voltage:
voltage = acceleration * distance
Substituting the values:
voltage = (7.8125 × 10^6 m/s) × (0.250 m)
Finally, we can calculate the voltage:
voltage = 1.953 × 10^6 volts
So, the voltage used to accelerate the electron is approximately 1.953 × 10^6 volts.