Two parallel plates having charges of equal magnitude but...

Two parallel plates having charges of equal magnitude but opposite sign are separated by $12.0 \mathrm{~cm}$. Each plate has a surface charge density of $36.0 \mathrm{nC} / \mathrm{m}^{2} . \mathrm{A}$ proton is released from rest at the positive plate. Determine (a) the potential difference between the plates, $(\mathrm{b})$ the energy of the proton when it reaches the negative plate, (c) the speed of the proton just before it strikes the negative plate, (d) the acceleration of the proton, and (e) the force on the proton. (f) From the force, find the magnitude of the electric field and show that it is equal to that found from the charge densities on the plates.

Two parallel plates having charges of equal magnitude but opposite sign are separated by $12.0 \mathrm{~cm}$. Each plate has a surface charge density of $36.0 \mathrm{nC} / \mathrm{m}^{2} . \mathrm{A}$ proton is released from rest at the positive plate. Determine
(a) the potential difference between the plates, $(\mathrm{b})$ the energy of the proton when it reaches the negative plate,
(c) the speed of the proton just before it strikes the negative plate, (d) the acceleration of the proton, and
(e) the force on the proton. (f) From the force, find the magnitude of the electric field and show that it is equal to that found from the charge densities on the plates.

Key Concepts

Electric Field

The electric field is a vector field that represents the force per unit charge exerted on a charged particle. In the context of parallel plates, the field is uniform and directed from the positive plate to the negative plate, and it plays a crucial role in determining the force on a charged particle moving between the plates.

Electric Potential Difference

The electric potential difference, often called voltage, is the work done per unit charge in moving a charge between two points in an electric field. It quantifies the energy difference between two points and is directly related to the electric field through the distance over which the field is applied.

Energy Conversion in Electric Fields

When a charged particle is released in an electric field, its electric potential energy is converted into kinetic energy as it accelerates. This conservation of energy principle allows one to calculate the speed of the particle by equating the change in potential energy to the gain in kinetic energy.

Electric Force

The electric force is the force experienced by a charged particle in an electric field. It is given by the product of the charge and the electric field (F = qE). This concept is essential for understanding the acceleration and motion of the particle under the influence of the field.

Newton's Second Law of Motion

This law states that the acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass. It is used to relate the electric force acting on a charged particle to its acceleration, thereby linking the concepts of force, mass, and motion.

Surface Charge Density and Its Relation to Electric Field

Surface charge density is defined as the charge per unit area on a surface. For parallel plate configurations, the surface charge density on the plates can be used with Gauss's Law to derive the expression for the electric field between the plates, illustrating the fundamental connection between charge distribution and the resulting electric field.

Transcript

00:01 For this problem on the topic of electric potential, we have two parallel plates that have charges of equal magnitude but of opposite signs.

00:08 They are separated by a distance of 12 centimeters and each plate has a surface charge density of 36 newton nanocoloms per square meter.

00:17 We then have a proton that is released from rest at the positive plate and we want to determine various quantities relating to the plates and the proton.

00:28 So firstly, we know that the positive plate by itself creates a field, we'll call it e plus, and this is the charge density sigma over 2 epsilon 0, which is 36 nanocoules per square meter, or 36 times 10 to minus 9 couloms per square meter divided by two times the electric constant 8 .85 times 10 to the minus 12 coulomb squared per newton meter squared.

01:16 And so if we calculate this, we get the field strength to be 2 .03 kiloons per coulom.

01:26 And this is away from the positive.

01:28 Plate.

01:30 Now the negative plate by itself creates a field of the same magnitude between the plates and it is in the same direction.

01:38 So together, the total field produce e is uniform and has a magnitude of 4 .07 kiloons per coulom.

01:52 So the first thing we want to find is the potential difference between the plates.

01:58 Now we'll take v is equal to zero at the negative plate, and so the potential at the positive plate is therefore v minus 0, and we know that this is equal to minus the integral of e .d .s.

02:11 So this is the integral from 0 to 12 centimeters of the electric field minus 4 .07 kiloons per kulum in one dimension so times d x and so if we solve this this becomes 4 .07 kiloons per kulum times a distance of 0 .12 meters and we get the potential difference to be 400 and 88 volts.

03:03 For part b we want to find the energy of the proton when it reaches the negative plate.

03:09 So we know that a half m v squared plus the electric potential energy qv for the proton initially must equal to a half m v squared plus q v finally.

03:34 So the sum of kinetic and electric potential energy is conserved.

03:38 And so the initial electric potential energy qv is equal to the final kinetic energy...

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