A proton traveling at a speed 1.0*106 m/s of enters the gap between the plates of a 2.0-cm-wide parallel-plate capacitor. The surface charge densities on the plates are 1*10^-6 C/m2.
How far has the proton been deflected when it reaches the far edge of the capacitor? Assume the electric field is uniform inside the capacitor and zero outside.
I know the formula is E=symbol eta/epsilon for a parallel plate. But I don't what to do or how to start this.
Well, this seems like a shocking situation! Don't worry, I'm here to assist you with a spark of humor.
To find out how far the proton has been deflected, we need to determine the electric field within the capacitor. As you mentioned, the formula for the electric field between the plates of a parallel-plate capacitor is E = η/ε, where η represents the surface charge density and ε is the permittivity of free space.
Given the surface charge densities on the plates as 1*10^-6 C/m², we can substitute these values into the equation. However, be prepared for a bit of a jolt!
E = (1*10^-6 C/m²)/(ε)
To calculate the distance the proton has been deflected, we need to consider the time it takes for the proton to travel across the 2.0-cm-wide gap. Since we know the proton's speed is 1.0*10⁶ m/s, the time it takes to cross the gap is t = distance/speed, or t = 0.02 m / 1.0*10⁶ m/s.
But wait, there's more! To find the distance the proton has been deflected, we can now use the formula d = (1/2) * E * t². After all, it's a funny thing how these equations come together!
Now, plug in the previously found values and let's get down to business:
d = (1/2) * E * t²
d = (1/2) * [(1*10^-6 C/m²)/(ε)] * [0.02 m / (1.0*10⁶ m/s)]²
Just like that, you'll have the distance the proton has been deflected when it reaches the far edge of the capacitor! Remember, laughter is the best conductor of learning.
To find the distance the proton is deflected when it reaches the far edge of the capacitor, we can use the equation for the force experienced by a charged particle in an electric field.
The force experienced by a charged particle in an electric field is given by the equation:
F = qE
Where:
F is the force experienced by the charged particle,
q is the charge of the particle, and
E is the electric field.
In this case, the charged particle is a proton with a charge of +1.6 x 10^-19 C, and the electric field between the plates of the capacitor is uniform and given by:
E = sigma / epsilon₀
Where:
sigma is the surface charge density on the plates, and
epsilon₀ is the permittivity of free space.
Given that the surface charge density on the plates is 1 x 10^-6 C/m² and epsilon₀ is a constant value (approximately 8.85 x 10^-12 C²/(N·m²)), we can calculate the electric field:
E = (1 x 10^-6 C/m²) / (8.85 x 10^-12 C²/(N·m²))
Now that we have the electric field, we can find the force experienced by the proton using the equation F = qE.
Next, we need to calculate the time it takes for the proton to travel across the gap between the plates. Since the distance between the plates is given as 2.0 cm, we can divide this distance by the speed of the proton to find the time:
t = d / v
Where:
d is the distance between the plates, and
v is the velocity/speed of the proton.
Once we have the time, we can calculate the displacement by multiplying the time by the horizontal component of the velocity. Since the electric field is in the vertical direction, the proton will experience a vertical acceleration. However, there is no horizontal force acting on the proton, so the horizontal velocity will remain constant.
The horizontal component of the velocity can be found using the equation:
v_horizontal = v * cosθ
Where:
v_horizontal is the velocity component in the horizontal direction,
v is the velocity/speed of the proton, and
θ is the angle between the proton's velocity vector and the perpendicular to the plates.
Given that the angle between the proton's velocity and the perpendicular to the plates is 90 degrees (as the electric field is vertical), we can calculate the displacement using the equation:
displacement = v_horizontal * t
Finally, substitute the values into the equations and calculate the displacement.
To start solving this problem, we need to find the electric field between the plates of the parallel-plate capacitor. We can use the formula you mentioned, which calculates the electric field, E, as the ratio of the surface charge density, σ, to the electric constant of free space, ε₀.
The formula is:
E = σ / ε₀
Given that the surface charge densities on the plates are 1 * 10^-6 C/m², and the electric constant of free space is ε₀ = 8.85 * 10^-12 C²/N*m², we can substitute these values into the formula to find the electric field.
E = (1 * 10^-6 C/m²) / (8.85 * 10^-12 C²/N*m²)
Now, let's calculate E:
E ≈ 1.13 * 10^5 N/C
The next step is to determine the force acting on the proton due to the electric field. The electric force experienced by a charged particle in an electric field is given by the equation:
F = q * E
where F is the electric force, q is the charge of the particle, and E is the electric field.
In this case, the charge of a proton is +1.6 * 10^-19 C, and the electric field is given as 1.13 * 10^5 N/C.
F = (1.6 * 10^-19 C) * (1.13 * 10^5 N/C)
Calculating the value of F yields:
F ≈ 1.81 * 10^-14 N
Now, we know that the force acting on the proton due to the electric field is directed perpendicular to its initial velocity, causing the proton to deflect. This force is known as the centripetal force and can be calculated using the formula:
F = m * a
where F is the force, m is the mass, and a is the acceleration.
In this case, the mass of the proton is approximately 1.67 * 10^-27 kg. Since the force and acceleration are directly proportional, we can set these two equations equal to each other:
F (electric force) = F (centripetal force)
q * E = m * a
Substituting the values, we find:
(1.6 * 10^-19 C) * (1.13 * 10^5 N/C) = (1.67 * 10^-27 kg) * a
Now, we can solve for the acceleration, a:
a ≈ (1.6 * 10^-19 C * 1.13 * 10^5 N/C) / (1.67 * 10^-27 kg)
a ≈ 1.09 * 10^7 m/s²
Since the acceleration is constant throughout the path, we can use the equations of accelerated motion for constant acceleration:
v² = u² + 2 * a * s
where v is the final velocity, u is the initial velocity, a is the acceleration, and s is the distance traveled.
In our case, the final velocity v is 0 m/s (the particle stops), the initial velocity u is 1 * 10^6 m/s, and the acceleration a is 1.09 * 10^7 m/s².
Substituting these values, we can solve for s:
0 = (1 * 10^6 m/s)² + 2 * (1.09 * 10^7 m/s²) * s
s ≈ - 5.49 * 10^-3 m
However, since distance cannot be negative, we take the absolute value:
s ≈ 5.49 * 10^-3 m
Therefore, the proton is deflected by approximately 5.49 * 10^-3 meters when it reaches the far edge of the capacitor.
Knowing E, you then know the force on the proton. From the velocity, you know the time the proton is in the plates (2cm wide).
F*time=massproton*velocity change.
the original velocity in the direction of E is zero, so velcity change must be final velocity, or twice average velocity during the transit.
distancedeflected=avgvelocity*time
=velocitychange/2 * time
you take it from here.