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 $36.0 \mathrm{nC} / \mathrm{m}^{2} .$ A proton is released from rest at the positive plate. Determine (a) the potential difference between the plates, (b) the kinetic 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 the electric field found from the charge densities on the plates.

Key Concepts

Uniform Electric Field

In a system with two large, parallel plates of opposite charge, the electric field between the plates is uniform, meaning it has the same magnitude and direction at every point between the plates. This concept is central to understanding many capacitor problems, as the uniformity simplifies the analysis of forces and potential differences in the region between the plates.

Surface Charge Density

Surface charge density is the measure of electric charge per unit area on a surface. It is crucial in determining the electric field produced by a charged plate, as there is a direct relationship between the surface charge density and the electric field in the vicinity of the plate, especially in idealized parallel plate configurations.

Electric Potential Difference

The electric potential difference (voltage) between two points is a measure of the work done per unit charge in moving a charge between those points. In the context of parallel plates, it is directly related to the uniform electric field and the distance between the plates, making it a key parameter in analyzing energy changes for charges moving in such fields.

Energy Conversion in Electric Fields

When a charged particle moves in an electric field, its potential energy is converted into kinetic energy. This concept of energy conservation is used to calculate the kinetic energy of the particle after traversing a known potential difference, providing insights into the speeds and accelerations involved in the motion.

Kinematics under Constant Acceleration

Charged particles in a uniform electric field experience a constant force, which results in constant acceleration. This allows the use of kinematic equations to analyze the motion of the particle, including calculations of displacement, velocity, and time, which are essential for understanding the dynamics of particles in electric fields.

Electric Force on a Charged Particle

The force on a charged particle in an electric field is given by the product of the charge and the electric field strength. This fundamental relationship underlies many phenomena in electromagnetism, and it connects the microscale (force on individual charges) to the macroscale (overall field effects) in systems like capacitors.

Transcript

00:01 Hello friends here it is given that there are two conducting plates each having the charge same but positive and negative separated apart by the distance of 12 centimeters surface charge density of each plate is 36 nanoculum per meter square a proton is released on the positive plate we have to calculate first potential difference between the plates kinetic energy of proton when it reaches to the negative plate, velocity by which it strikes the negative plate.

00:40 By using the force concept, calculate the electric field and justify that it is equal to sigma upon epsilon knot.

00:50 Let us start first.

01:03 Electric field between the plates will be in the direction.

01:09 Portative plate to negative plate and it will be uniform and value is sigma upon epsilon not so potential difference between the plate will be electric field into distance so you will get sigma d upon epsilon not now substitute the value sigma is 36 nanopulam minus 9 per meter square d is 12 centimeter so point 1 2 meter epsilon epsilon epsilon epsilon epsilon 0 8 .85 10 to the power minus square.

01:51 So potential difference between the plates you will get 488 volt.

02:03 This is the answer of part a.

02:19 Excuse me please.

02:24 In the part we have to find the kinetic energy when it is reaching.

02:31 So kinetic energy is called to work done by the potential.

02:36 So charge into potential difference.

02:38 Charge in a 1 .6 and potential difference is 488 jule so it will be 7 .81 into 10 to the power of minus 17 jules see worked done by the potential is called to gain in kinetic energy so half mb square minus 0 is called to 7 .81 into 10 to the power of minus 17 so velocity by which it will reach to the positive plate will be 7 .81 into 10 to the power of minus 17 upon into 2 upon mass of proton.

03:48 So on solving it you will get the velocity to be 3 .06 into 10 to the power 5 meter per second...

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