·24.83 In the figure, a parallel plate capacitor is connected to a 300 .-V battery. With the cap

step 1 For this problem on the topic of capacitors, we are shown a parallel plate capacitor in the figu

·24.83 In the figure, a parallel plate capacitor is...

$\cdot 24.83$ In the figure, a parallel plate capacitor is connected to a $300 .-\mathrm{V}$ battery. With the capacitor connected, a proton is fired with a speed of $2.00 \cdot 10^{5} \mathrm{~m} / \mathrm{s}$ from (through) the negative plate of the capacitor at an angle $\theta$ with the normal to the plate. a) Show that the proton cannot reach the positive plate of the capacitor, regardless of what the angle $\theta$ is. b) Sketch the trajectory of the proton between the plates. c) Assuming that $V=0$ at the negative plate, calculate the potential at the point between the plates where the proton reverses its motion in the $x$ -direction. d) Assuming that the plates are long enough for the proton to stay between them throughout its motion, calculate the speed (magnitude only) of the proton as it collides with the negative plate.

$\cdot 24.83$ In the figure, a parallel plate capacitor is connected to a $300 .-\mathrm{V}$ battery. With the capacitor connected, a proton is fired with a speed of $2.00 \cdot 10^{5} \mathrm{~m} / \mathrm{s}$ from (through) the negative plate of the capacitor at an angle $\theta$ with the normal to the plate.
a) Show that the proton cannot reach the positive plate of the capacitor, regardless of what the angle $\theta$ is.
b) Sketch the trajectory of the proton between the plates.
c) Assuming that $V=0$ at the negative plate, calculate the potential at the point between the plates where the proton reverses its motion in the $x$ -direction.
d) Assuming that the plates are long enough for the proton to stay between them throughout its motion, calculate the speed (magnitude only) of the proton as it collides with the negative plate.

Key Concepts

Kinematics and Decomposition of Motion

When analyzing the trajectory of a projectile launched at an angle, it is crucial to decompose its initial velocity into components parallel and perpendicular to the electric field. The equations of kinematics for constant acceleration are then applied to the component of the velocity along the direction of the electric field, while the perpendicular component remains unaffected. This approach helps in determining turning points, collision points, and overall trajectory shape.

Uniform Electric Field

In a parallel plate capacitor, the electric field is uniform between the plates. This means that the electric field has a constant magnitude and direction, which greatly simplifies the analysis of the force on a charged particle and its subsequent motion. Understanding this concept is key when calculating the acceleration experienced by a charged particle in the region between the capacitor plates.

Motion of Charged Particles in an Electric Field

A charged particle in a uniform electric field experiences a constant force, causing uniform acceleration in the direction of the field (or against it, depending on the sign of the charge). When the particle is launched with an initial velocity that has components both parallel and perpendicular to the field, its motion can be decomposed into independent motions: one, with constant acceleration (along the field), and another, with constant velocity (perpendicular to the field).

Energy Conservation in Electric Fields

In the context of charged particle motion in an electric field, energy conservation involves the conversion of electrical potential energy into kinetic energy, and vice versa. The change in kinetic energy of the particle is related to the work done by the electric field, which is expressed as the product of the charge and the potential difference. This principle is essential for analyzing situations where the particle reverses direction or when determining the final speed after a certain displacement.

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