Question

A proton traveling along the \(x\) -axis at a speed of \(1.00 \cdot 10^{6} \mathrm{~m} / \mathrm{s}\) enters the gap between the plates of a \(2.00-\mathrm{cm}\) -wide parallel plate capacitor. The surface charge distributions on the plates are given by \(\sigma=\pm 1.00 \cdot 10^{-6} \mathrm{C} / \mathrm{m}^{2} .\) How far has the proton been deflected sideways \((\Delta y)\) when it reaches the far edge of the capacitor? Assume that the electric field is uniform inside the capacitor and zero outside.

Short answer

Answer: The proton is deflected sideways by 4.32 x 10⁻³ m (or 4.32 mm) when it reaches the far edge of the capacitor.

Step-by-step solution

Calculate the electric field inside the capacitor

To find the electric field (E) inside the capacitor, we can use the relation: E = σ/ε₀ where σ is the surface charge density (1.00 x 10⁻⁶ C/m²) and ε₀ is the vacuum permittivity (8.85 x 10⁻¹² C²/N·m²). Plugging in the values, we get: E = (1.00 x 10⁻⁶ C/m²) / (8.85 x 10⁻¹² C²/N·m²) = 1.13 x 10⁵ N/C

Calculate the force acting on the proton

The electric force (F) acting on the proton can be calculated using its charge (q) and the electric field (E) found earlier: F = qE The charge q of the proton is 1.60 x 10⁻¹⁹ C, so: F = (1.60 x 10⁻¹⁹ C) x (1.13 x 10⁵ N/C) = 1.81 x 10⁻¹⁴ N

Determine the time it takes for the proton to cross the capacitor

The time (t) it takes for the proton to cross the width of the capacitor (d) can be found using the proton's speed along the x-axis (vₓ) and the width of the capacitor (2.00 cm): t = d/vₓ Converting the capacitor width from cm to meters, we have: d = 2.00 cm × (1 m / 100 cm) = 0.020 m Now, we can find the time: t = 0.020 m / (1.00 x 10⁶ m/s) = 2.00 x 10⁻⁸ s

Calculate the acceleration of the proton in the y-direction

The acceleration (a) of the proton in the y-direction can be found using Newton's second law (F = ma) and the mass of the proton (mₚ = 1.67 x 10⁻²⁷ kg): a = F/mₚ a = (1.81 x 10⁻¹⁴ N) / (1.67 x 10⁻²⁷ kg) = 1.08 x 10¹³ m/s²

Calculate the distance traveled in the y-axis

Lastly, we will calculate the sideways deflection (∆y) of the proton by using the equation of motion: ∆y = 0.5 × a × t² Plugging in the values obtained earlier: ∆y = 0.5 × (1.08 x 10¹³ m/s²) × (2.00 x 10⁻⁸ s)² = 4.32 x 10⁻³ m So, the proton has been deflected sideways by 4.32 x 10⁻³ m (or 4.32 mm) when it reaches the far edge of the capacitor.

Key concepts

Electric Field Inside Capacitor

Understanding the electric field within a parallel plate capacitor is fundamental when studying how charged particles behave in electric fields. In our problem, the capacitor has a uniform electric field generated by equal and opposite charges on the two plates. This field is defined as the force per unit charge and can be calculated using the equation
\( E = \frac{\sigma}{\varepsilon_0} \).
Here, \(\sigma\) is the charge density on the plates and \(\varepsilon_0\) is the vacuum permittivity. The product of these values gives us the electric field strength in Newtons per Coulomb (N/C). In a parallel plate capacitor, the electric field is uniform except near the edges where fringe effects occur, but these are negligible for our discussion. This uniformity means that any charged particle within the capacitor will experience a constant force and hence a constant acceleration perpendicular to the plates.

Force on a Proton in Electric Field

Now let's delve into how a charged particle like a proton is affected by this electric field. The force acting on a proton is determined by Coulomb’s law, which states that force is the product of the electric field and the charge of the proton. Mathematically, it is given by
\( F = qE \).
The proton has a charge \( q = 1.60 \times 10^{-19} \) Coulombs, and when it's multiplied by the electric field strength, E, we get the force in Newtons. This force is directed perpendicular to the motion of the proton and towards the oppositely charged plate, causing the proton to accelerate sideward as it moves between the capacitor plates.

Proton Deflection in Capacitor

Moving on, let's analyze what happens when a proton is deflected within a capacitor. The force we previously calculated causes the proton to deviate from its initial path; it's accelerated in a direction perpendicular to its velocity. This deviation or deflection, \( \Delta y \), happens due to the electric force imparting a sideways acceleration on the proton as it travels through the capacitor. We use the kinematic equations of motion to calculate this deflection, considering that the proton's initial velocity in the y-direction is zero and it undergoes constant acceleration within the electric field.

Uniform Electric Field Effects

The beauty of a uniform electric field is that it simplifies a lot of the physics involved. Since the field is constant between the plates, the acceleration of the proton is also constant. This simple yet accurate model allows us to use basic motion equations to predict future positions. In this uniform field, we computed the proton's deflection by using the equation
\( \Delta y = 0.5 \times a \times t^2 \).
This equation is only valid under constant acceleration, a condition met by the uniform electric field inside a parallel plate capacitor, giving us a straightforward way to determine the effect of this field on the proton.

Motion of Charged Particle in Electric Field

Exploring the motion of a charged particle like a proton when it's subjected to an electric field reveals a lot about basic principles of electromagnetism. Charged particles accelerate when exposed to an electric field, following Newton's second law, \( F = ma \). Here, the proton's mass plays a key role in determining the acceleration; a lighter particle would have experienced greater acceleration. As the proton moves through the field, its velocity vector changes due to this acceleration, causing it to follow a curved path. The time it spends within the field, combined with its acceleration, allows us to determine its final displacement, reflecting a fundamental understanding of dynamics and electric field interactions.