Plot the variation of electric field intensities in case of a cylindrically symmetric charge dis

Plot the variation of electric field intensities in case of...

Key Concepts

Electric Field Variation

The variation of the electric field intensity with distance is a critical aspect of understanding electric fields in symmetric charge distributions. By applying Gauss's Law in a cylindrical coordinate system, the electric field can be determined for regions both inside and outside the charged object, highlighting how the field strength changes with radial distance.

Gauss's Law

Gauss's Law is a fundamental principle in electromagnetism that relates the net electric flux through a closed surface to the charge enclosed within that surface. It is particularly useful in situations with high symmetry, such as cylindrical symmetry, allowing for the determination of electric field magnitudes without complex integration calculations.

Cylindrical Symmetry

Cylindrical symmetry implies that the physical properties of a system, such as the charge distribution and the electric field, are invariant under rotations about, and translations along, a central axis. This symmetry simplifies the analysis of the electric field, as the field magnitude depends only on the radial distance from the axis.

Uniform Charge Density

Uniform charge density indicates that the charge is distributed evenly throughout the volume or surface of an object. In problems involving cylindrical symmetry, a uniform charge density ensures that the resulting electric field can be derived using symmetry arguments, leading to simplified expressions depending on the radial position.

Transcript

00:01 Okay, so we have a spherically symmetric charge distribution and we have to calculate the electric flux in this charge distribution.

00:11 We know that using the electric flux we can say that the close integral of the field, the electric field da is going to be equal the charge inside the gaussian surface divided by absolute so on zero.

00:33 Okay, so since the electric field is constant, we can say that this is going to be e that multiplies the integral of the area, which is the surface area of this sphere.

00:51 So it's 4 pi r square.

00:55 This is the gaussian surface.

00:56 Remember, this is going to be equal the charge inside the gaussian surface and the charge inside the gaussian surface can be described as the density dv actually the integral of the density dv divided by epsilon zero so what is this integral this is the integral of the volume therefore we can say this is going to be one moment let's put in here we have e that multiplies four r square that is equal 1 divided by epsilon 0 the close the integral not close integral of the density and the problem says that the density is just a divided by r and the volume the volume is just the area let's put in here the volume dv is just the area multiply by the r therefore the surface area this is going to be just 4 pi r square d r and the radius goes from 0 to r therefore we just need to calculate this so let's continue here we have that the electric field is 4 pi r square is going to be equal 1 divided by epsilon 0 a goes out of the integral so this is just a divided by epsilon 0 we have 4 pi in here integral of r d r from 0 to r therefore let me see this is going to to be let's continue here it is going to be e 4 pi r square it's going to be equal 4 pi a epsilon 0 r squared divided by 2 from r to 0 so this is just 4 pi a divided by 2, absalom 0, r...

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