A slab of insulating material has Problem 22.53. thickness 2...

A slab of insulating material has Problem 22.53. thickness $2 d$ and is oriented so that its faces are parallel to the $y z$-plane and given by the planes $x=d$ and $x=-d$. The $y$ - and $z$-dimensions of the slab are very large compared to $d$ and may be treated as essentially infinite. The slab has a uniform positive charge density $\rho$. (a) Explain why the electric field due to the slab is zero at the centre of the slab $(x=0)$. (b) Using Gauss's law, find the electric field due to the slab (magnitude and direction) at all points in space. (FIGURE CANT COPY)

Key Concepts

Gauss's Law

Gauss's Law relates the electric flux through a closed surface to the charge enclosed within that surface. It simplifies the calculation of electric fields, especially for systems with high symmetry, by allowing us to connect the field directly to the total charge without performing complex integrations.

Symmetry in Electrostatics

Symmetry is a fundamental concept in electrostatics that helps simplify the analysis of electric fields. When a charge distribution is symmetric, such as in the case of a uniformly charged slab, contributions to the field from different parts of the distribution can cancel out at certain points, like the center, leading to a zero net field.

Uniform Charge Distribution

A uniform charge distribution means that the charge density is constant throughout the material. This property enables the use of Gauss's Law by providing a consistent value for the charge density, which simplifies the determination of the electric field both inside and outside the charged region.

Gaussian Surface Selection

Selecting an appropriate Gaussian surface is crucial when applying Gauss's Law. For a symmetric charge distribution, the correct surface simplifies the problem by aligning with the geometry of the system, allowing the electric field to either be constant over parts of the surface or to cancel out due to symmetry, thus making the flux calculation more straightforward.

Transcript

00:01 The problem states a slab of insulating material has thickness 2d, right? it has the thickness 2d and is oriented so that its space are parallel to the yz plane and given by the planes x equals to d and x equals to minus d.

00:30 The y and z dimensions of the slab are very large compared to d.

00:37 Treat them as essentially infinite so the slab has a uniform positive charge density row so now it is asked for the first part explain why electric field due to slab is zero at the circle of at the center of the slab right and using goss law find the electric field due to the slab is zero at the circle of the center of the slab right and using goss law find the electric field due to the slab magnitude and direction at all the points in the space so here we can say the formula for this slab the you can say the charge density formula for any point x right it will be row not equals to x by t square right so that is why at the center of this slab e equals to 0 because x is 0 here since x is 0 here so this whole equation becomes 0 and charge density is 0 then e is also 0 here so now for the b part uh we are asked to find the electric field due to the slab so here we need to find the electric field between you can say it is given 0 less than x and this is less than d right so for this we can say dq is given by row dv or you can write it as volume as ad x right this ad x can be written right again this d x will be x squared d x you can say by dx you can say by dx.

02:53 Can we write this yes so this is the basic thing now q enclosed means the charge enclosed will be between 0 to x here so it will be 2 times of 0 to x dq right and we will just integrate it putting the value dq so row a by d square is a constant 2 is also a constant so taken outside, only x is a variable.

03:28 So x squared when integrated, it becomes 2 by 3, row not a x cube by d square.

03:39 Right...

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