How strong an electric field is needed to accelerate electrons in a TV tube from rest to one-tenth the speed of light in a distance of 4.9 cm ?
Require that
Work done by E-field = Kinetic Energy gain
e*E*X = (1/2) m V^2 = (1/200) m c^2
E = (1/200)(m/e)c^2/X
where X = 0.049 m
m = 9.11*10^-31 kg
e = 1.602*10^-19 C
c = 3.00*10^8 m/s
E = 52200 V/m
The electrons move with a velocity of 3.0x10^6 m/s. What electric field strength is needed to accelerate electrons from rest to this velocity in a distance of 9.0 mm?
To determine the strength of the electric field needed to accelerate electrons in a TV tube, we can use the equations of motion and the principles of electromagnetism.
The equation of motion we can use is given by:
v^2 = u^2 + 2as
Where:
v = final velocity (one-tenth the speed of light)
u = initial velocity (zero, since the electrons are at rest)
a = acceleration (which we need to find)
s = distance (4.9 cm)
From the equation of motion, we can rearrange it to solve for acceleration:
a = (v^2 - u^2) / (2s)
Next, let's calculate the values:
v = 1/10 * c (speed of light)
= (1/10) * 3 * 10^8 m/s
= 3 * 10^7 m/s
u = 0 m/s (since the electrons are at rest)
s = 4.9 cm
= 4.9 * 10^(-2) m
Plugging the values into the equation:
a = ((3 * 10^7)^2 - 0^2) / (2 * 4.9 * 10^(-2))
a = (9 * 10^14 - 0) / (9.8 * 10^(-1))
a = 9 * 10^14 / 9.8 * 10^(-1)
a = 9.18 * 10^15 m/s^2
So, the acceleration required is approximately 9.18 * 10^15 m/s^2.
To find the strength of the electric field, we can use the equation:
E = F / q
Where:
E = electric field strength
F = force experienced by the electron
q = charge of electron
The force experienced by the electron can be calculated using Newton's second law:
F = ma
Where:
m = mass of the electron
The mass of an electron is approximately 9.11 * 10^(-31) kg.
Plugging in the values:
F = (9.11 * 10^(-31)) * (9.18 * 10^15)
F = 8.36 * 10^-15 N
The charge of an electron is approximately -1.6 * 10^(-19) C.
Plugging in the values:
E = (8.36 * 10^-15) / (1.6 * 10^(-19))
E = 5.22 * 10^4 N/C
Therefore, the strength of the electric field needed to accelerate electrons in a TV tube from rest to one-tenth the speed of light in a distance of 4.9 cm is approximately 5.22 * 10^4 N/C.
To determine the strength of the electric field needed to accelerate electrons from rest to a certain speed in a given distance, we can use the equation for the acceleration of an electron in an electric field.
The equation for the acceleration (a) of an electron in an electric field is given by:
a = eE / m
where e is the charge of an electron (-1.6 x 10^-19 C), E is the strength of the electric field, and m is the mass of an electron (9.11 x 10^-31 kg).
Given that the initial velocity of the electrons is zero, we can use the kinematic equation to calculate the final velocity (v) of the electrons. The kinematic equation is:
v^2 = u^2 + 2as
where u is the initial velocity, s is the distance, and a is the acceleration.
In this case, we want to find the electric field strength (E), so we rearrange the first equation to solve for E:
E = am / e
Now, let's plug in the given values and solve for the electric field strength (E):
e = -1.6 x 10^-19 C
m = 9.11 x 10^-31 kg
u = 0 (initial velocity)
v = 0.1c (final velocity)
s = 4.9 cm = 0.049 m
First, let's calculate the acceleration (a):
a = (v^2 - u^2) / (2s)
a = (0.1c)^2 / (2 * 0.049)
a = 0.01c^2 / 0.098
a = 0.102c^2
Now, substitute the values of a, e, and m into the equation to find E:
E = am / e
E = (0.102c^2 * 9.11 x 10^-31 kg) / (-1.6 x 10^-19 C)
Calculating this expression will give us the strength of the electric field needed to accelerate the electrons.