The electric flux is (4.0 V ·m / s^5) t^5 through a certain area of a dielectric that has diele

The electric flux is (4.0 V ·m / s^5) t^5 through a certain...

Key Concepts

Time Derivative

The time derivative denotes how a physical quantity changes with time. In the context of electromagnetism, taking the time derivative of electric flux allows one to derive the displacement current. This mathematical operation is essential for analyzing dynamic systems where fields or fluxes are not static, thereby linking temporal variations to observable physical effects.

Electric Flux

Electric flux is defined as the measure of the number of electric field lines passing through a given surface. It is a fundamental concept in electromagnetism that provides insights into how electric fields interact with surfaces, and it plays a crucial role in Gauss's law. Understanding electric flux is essential for analyzing how fields change over time and space.

Displacement Current

Displacement current is a term introduced by Maxwell to complement Ampère’s circuital law. It represents the rate of change of electric flux through a surface and serves as a source of the magnetic field in situations where conduction current is absent. Recognizing displacement current is key for understanding electromagnetic waves and the continuity of current in circuits with capacitive elements.

Dielectric Materials

Dielectric materials are insulating substances characterized by their ability to reduce the internal electric field when an external electric field is applied. The dielectric constant quantifies this effect by specifying how much the material polarizes, thereby influencing the overall electric flux and related quantities in the medium. This concept is important for understanding capacitors, the propagation of electromagnetic fields in media, and modifications to displacement current in dielectric contexts.

Transcript

00:01 Hi there, so for this problem, we are given the value for the electric flops and that value is equal to four balls times meters per second to the five and this times the time elevated to the five.

00:25 Okay, and through a certain area of a electric that has an electric constant and the electric constant and the electric is also given and that is equal to 2 .5.

00:39 So for part a of this problem, the question is about to find the displacement current through that area at the time that is equal to 1 .5 seconds.

00:57 Okay, now in these problems we are dealing with the displacement current.

01:03 So the displacement current, so the displacement current we know that that is equal to the f times the rate of change of the electric loops with respect to time.

01:16 And also we know that the electric field in this dielectric is equal to the dielectric times epsilon sub zero.

01:29 Now for power a of this problem we want the displacement current at a equals to 1 .5 seconds.

01:38 So what we need to do is to substitute the expression that we are given for the electric loops in here into this expression along with the above along with the emph in here so that will give us the electric constant times epsilon sub zero this times the rate of change of the expression that we are given that is 4 times the time to the 5 and this could respect the time now, the solution for this is just the constant, epsilon sub 0.

02:14 The derivative takes down the exponent, so we will have 4 times 5, so that will be 20 in here, and this times the time to the 4.

02:25 Now, we just need to simply substitute the values in there, so that will be 2 .5, the value of epsilon sub 0 .0.

02:39 You can search for that value, okay? and that is a constant.

02:45 You can search for it.

02:47 And this times 20 times the time that is 1 .5 and that to the 4.

02:52 Now, after we substitute all of the values in here, we will obtain a value of 2 .2 times 10 to the minus 9 ampers, which is the same as 2 .2 nano -ampers.

03:05 So that's a solution for part a.

03:09 Of this problem...

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