The accelerating plates have area A and are a distance d apart. (a) What should the charges on t

The accelerating plates have area A and are a distance d...

Key Concepts

Uniform Electric Field Between Plates

Between two parallel plates, the electric field is approximately uniform and is directly related to the potential difference and the distance between them by E = V/d. This understanding is essential when analyzing the force that the field exerts on a charged particle and ensuring that it achieves the correct acceleration.

Capacitance of a Parallel Plate Capacitor

A parallel plate capacitor’s capacitance is given by C = ??A/d, linking the physical dimensions of the plates (area A and separation d) with how much charge Q can be stored at a given voltage V (since Q = CV). This relationship is key in determining the required charge on the accelerating plates.

Conservation of Energy in Electric Fields

This concept explains how a charged particle gains kinetic energy when accelerated by an electric field. When an ion moves through a potential difference, the work done on it by the field (qV) is converted into kinetic energy (½mv²), allowing one to relate the applied voltage to the desired final speed.

Charge Assignment and Electric Field Direction

The direction of the electric field is determined by the sign of the charges on the plates, with field lines running from the positive plate to the negative plate. Correct assignment of the positive and negative plates is crucial, as it ensures that a positively charged ion, for example, will experience a force in the intended direction to achieve acceleration.

Transcript

00:01 This given problem the uranium ions should come out of the setup of these two plates here these are two parallel plates uranium ions are moving like this and then they should come out of the plates with the velocity v if we consider their initial kinetic energy to be zero when they enter between the plates now mass of the uranium ions is given as m so in the first part of the problem electrostatic potential energy which will be gained by the ions within this region between the plates electrostatic potential energy gained by the the electrons sorry ions between the charged plates could be converted into their kinetic energy when they comes out of the plate like this so if potential difference between the plates is v and charge over the ions is charge over which is equal to the charge over a proton so we can say electrostatic potential energy which is given by the product of charge with the potential difference that should be equal to kinetic energy gained by the ions now if electric field between the region in the region or between these plates is e the gap between the plates is given as d and area of the plates is given as a.

02:43 So as we know potential is equal to line integral of electric fields.

02:50 So for we we will use e into d if electric field is e between the plates and that is equal to half mv square.

03:02 Now for the electric field between the plates using gossess theorem we know electric field between two equally and oppositely charged plates is given as sigma by epsilon not where sigma is the charge density surface charge density of the plates so we can say e times of u by epsilon not a for electric field into d would be equal to half mv square hence charge over the plates charges stored over the plates will be given by epsilon not a times mass of the ions multiplied by speed gained by the ions squared and divided by eb so this is a charge over the plates and this is the answer for the first part of this problem and so that the ions may be energized as moving towards right so it is clear the left plate should be positively charged and right plate should be negatively charged and left plate should be positive so that this positivity of the left plate may repel the positively charged ion towards right so left plate should be positive and right plate should be negative...

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