The summary of 'Electric Field from a Ring and a Disk'

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In the video, Physics Ninja explores calculating electric fields for two types of geometries: a uniformly charged ring and a disk. The analysis begins with a uniformly charged ring of radius ( r ) and total charge ( Q ), placed at a point along the ring's axis at distance ( x ). The vertical components of the electric field from elements of the ring cancel out due to symmetry, so only the horizontal components need to be integrated. This leads to an expression for the total electric field ( frac{kxQ}{(x^2 + R^2)^{3/2}} ). The result is verified through symmetry, confirming the field is zero at the center and behaves like a point charge far away.

Next, the process extends to a uniformly charged disk with surface charge density ( sigma = frac{Q}{pi R^2} ). By treating the disk as composed of many infinitesimal rings, the presenter derives the electric field contribution from each ring. Integration over all rings from the center to the disk’s outer radius provides an expression for the total electric field. After changing variables and integrating, the derived field near the disk surface aligns with the expected result ( frac{sigma}{2 epsilon_0} ), confirming the conclusions with Gauss's law. Throughout, the analysis effectively uses symmetry and integration techniques to solve complex electromagnetic problems.

00:00:00

In this part of the video, Physics Ninja aims to calculate the electric field due to two geometries: a uniformly charged ring and a disk made from a series of rings. He starts by explaining the uniformly charged ring with a radius ( r ) and a point on the ring’s axis at a distance ( x ) from its center. The ring has a total charge ( Q ). To determine the electric field, he considers an infinitesimal charge element ( dq ) and evaluates the field at a specific point using standard electric field formulas. He breaks the electric field vector into its components and discusses the symmetry in the problem, noting that each charge element on the ring has a counterpart directly opposite it, simplifying the calculation of the total electric field.

00:03:00

In this segment, the focus is on determining the electric field produced by a small charge element ( dq ) on a ring. The vertical components of the electric field from opposite sides of the ring cancel out, so only the horizontal (x) components need to be considered. To find the total electric field, you need to integrate the x components from all charges around the ring. The x component of the electric field due to one charge ( dq ) is given by an expression involving Coulomb’s constant ( k ), the charge ( dq ), the distance variables ( x ) and ( r ), and the cosine of the angle ( theta ). The final step involves substituting these values into the integral to compute the total electric field.

00:06:00

In this segment of the video, the speaker discusses the process of summing electric fields to find the total electric field resulting from a ring of charge. The integration involves taking out constant terms such as ( k ) and ( x ), simplifying the integral to account for the total charge ( Q ) of the ring, leading to the expression ( frac{kxQ}{(x^2 + R^2)^{3/2}} ). The direction of the field is confirmed to be in the ( x )-direction.

To verify the result, the speaker suggests testing it in two limits: at the center of the ring and far away from the ring. At the center (where ( x = 0 )), the electric field should be zero due to symmetry, which the derived expression confirms. When far away from the ring, the field should resemble that of a point charge, and the speaker begins to demonstrate this by analyzing the expression under those conditions.

00:09:00

In this part of the video, the presenter explores the electric field resulting from different charge distributions. Initially, the focus is on a scenario where ( x ) is much larger than ( r ), leading to the simplification of the denominator. This results in an approximate formula for the electric field in the x direction, showing that it is proportional to ( frac{kq}{x^2} ), akin to the field from a point charge.

The presenter then transitions to discussing a uniformly charged disk, explaining that the disk has a uniform charge density ( sigma ) derived from the total charge ( Q ) and the area ( pi R^2 ). Using symmetry, it is explained that the total electric field must point away from the disk. The presenter introduces the concept of constructing the disk’s field by summing the fields from a series of infinitesimal rings.

To analyze this, the presenter focuses on the electric field contribution from a single ring, noting that the total charge of the disk ( Q ) can be broken into elements ( dQ ) for the rings. The expression for the electric field from one ring is given, considering a series of rings of varying radii, ultimately intending to sum these contributions to determine the total field.

00:12:00

In this part of the video, the speaker discusses the distribution of charge over a ring and how to express elemental charge ( dq ). The key point is that the charge density is uniform, and a ring further out will have more charge due to a larger area. The speaker then explains how to write ( dq ) as the product of charge density and the area of the ring, defining the area as the circumference of the ring (( 2pi r )) multiplied by its thickness (( dr )). They derive an expression for the electric field, indicating that to find the total field from a disk, one must integrate over all rings from the center to the outer radius. The integration setup involves extracting constants and changing variables to simplify the integral. The explanation ends with setting up the change of variable, preparing to solve the integral in the next segment.

00:15:00

In this part of the video, the speaker explains how to modify the integration limits and perform a change of variable from (r) to (u). Initially, (r) has a lower limit of zero and an upper limit of (R), which translates to (u) having corresponding limits of (x^2) and (x^2 + r^2). The constants (kx sigma 2pi) are factored out, and the integral’s limits are updated to integrate (u) from (x^2) to (x^2 + r^2). The (r dr) term is replaced with (frac{d u}{2}), and the integral simplifies to a straightforward form. The (u) integral is then evaluated, yielding a result involving ( frac{1}{ sqrt{x^2 + r^2} } ) and ( frac{1}{ sqrt{x^2} }). The final step involves simplifying the expression by redistributing constants and adjusting signs to obtain the final result for the field produced by the disc.

00:18:00

In this segment, the video derives the final expression for the electric field of a disk. The k value is identified as 1 over 4 pi epsilon zero, and the charge density is defined as the total charge divided by pi r squared. The video then examines the case where x is very small, i.e., close to the disk’s surface. It simplifies the expression by considering limits and cancellations, resulting in the field being the charge density divided by 2 epsilon 0. This result aligns with expectations from Gauss’s law, confirming the approach. The segment concludes by noting the successful integration of the ring and disk problems.

The summary of ‘Electric Field from a Ring and a Disk’