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00:00:00 – 00:20:49
Physics Ninja's video tutorial deals with the calculation of the electric field produced by a uniformly charged ring and extends this to a uniformly charged disk. The speaker begins by calculating the electric field at a point along the axis of a uniformly charged ring and methodically breaks down the contribution from each small charge element, ( dq ). Emphasis is placed on breaking the electric field vector into horizontal and vertical components, and leveraging symmetry to simplify calculations, ultimately leading to an integral expression for the field in the x-direction.
Moving forward, the video meticulously derives the final formula for the electric field of the ring, performing checks at various limits (center of the ring and far away) to validate consistency. The derived electric field expression is ( k cdot x cdot Q / (x^2 + r^2)^{3/2} ).
Next, the concept is extended to a uniformly charged disk, approached by considering the disk as a series of concentric rings. The speaker discusses calculating the elemental charge ( dq ) for each ring and uses integration to sum the contributions from all such concentric rings. The integral simplifies using a change of variables, leading to the final expression after integration.
Towards the end, the focus shifts to the specific case where the distance ( x ) from the disk approaches zero. Simplifying the expression reveals that the electric field close to the disk is the charge density divided by ( 2 epsilon_0 ), which is consistent with Gauss's Law. The video concludes by summarizing the integration process for deriving the electric field of both the ring and the disk, illustrating the logical progression from a ring to a disk.
00:00:00
In this segment of the video, Physics Ninja explains how to calculate the electric field due to a uniformly charged ring and how this result can be extended to a disk composed of such rings. He begins by discussing the uniformly charged ring of radius ( r ) with a total charge ( Q ). A point at a distance ( x ) from the center on the axis is considered for the calculation. The process starts by evaluating the contribution from a small charge element ( dq ) on the ring, using basic principles of electrostatics.
He defines the geometry, including the distance from the charge element to the point of interest, denoted by ( sqrt{x^2 + r^2} ), and the angle ( theta ). Next, he describes the vector nature of the electric field ( dE ) produced by ( dq ), emphasizing its magnitude as ( dE = k frac{dq}{x^2 + r^2} ) where ( k = frac{1}{4 pi epsilon_0} ).
Key steps involve breaking down ( dE ) into its horizontal (( dE_x )) and vertical (( dE_y )) components and utilizing symmetry, as each charge on one side of the ring has an equivalent charge directly opposite. This aids in simplifying the computation of the total electric field by accounting for the cancellation and addition of these component vectors.
00:03:00
In this segment of the video, the speaker explains how to determine the electric field produced by a small charge element, ( dq ), positioned symmetrically on a ring. They discuss breaking down the resultant electric field vector into vertical and horizontal components. Key points include:
– The vertical components of the electric field cancel out due to symmetry, meaning the total electric field only has a horizontal (x) component.
– The direction of this electric field depends on the charge’s sign, pointing away from the ring if positively charged.
– To find the total electric field, one must integrate all the x components produced by the individual charges around the ring.
– The magnitude of the electric field vector for each charge is integrated using the cosine of the angle ( theta ), which is consistent for all charges on the ring, with ( cos(theta) ) being ( frac{x}{sqrt{x^2 + r^2}} ).
– Finally, an expression for the x component of the electric field, ( dE_x ), is derived and simplified using the integral and substitution of relevant values.
00:06:00
In this segment of the video, the speaker focuses on integrating the components of an electric field to find the total electric field produced by a ring of charge. They explain the process of simplifying the integral by removing constant terms and constants related to the charge distribution on the ring. The resulting expression for the electric field is given by ( k cdot x cdot Q / (x^2 + r^2)^{3/2} ), with the direction specified as the x-direction.
To verify the correctness of this result, the speaker suggests checking specific limits:
1. At the center of the ring (x=0), the electric field should be zero due to symmetry, which is confirmed by the result.
2. Far away from the ring (x >> r), the ring should resemble a point charge, and the field should match the field produced by a point charge.
The speaker intends to further explore these special cases and their implications in the following parts of the discussion.
00:09:00
In this part of the video, the speaker starts by simplifying the electric field equation for cases where distance ( x ) is much larger than the radius ( r ). They demonstrate that the denominator simplifies to ( x^3 ), resulting in the electric field approximating ( frac{kq}{x^2} ), which is the field from a point charge. Then, they move on to discussing a uniformly charged disk. The total charge ( q ) is distributed over the area ( pi R^2 ) of the disk. By considering the disk as a series of concentric rings, the field generated by each ring is summed to determine the total electric field. The charge ( dq ) of each ring is considered, and this incremental charge is used in the field equations, taking into account the varying radius ( r ) of each ring.
00:12:00
In this part of the video, the speaker discusses how to calculate the elemental charge, (dq), for a uniformly charged ring and extends the concept to a disk. The charge density, multiplied by the ring’s area (determined by its circumference, (2pi r), and thickness, (dr)), yields (dq). The speaker then substitutes this into the expression for the electric field and sets up an integral to sum the contributions from all rings. The constants are factored out to simplify the integral. The integral from (0) to the disk’s radius, (R), is set up with respect to (r), transforming to a new variable, (u = x^2 + r^2), for easier computation. The process of variable substitution is explained for solving the integral, preparing to finalize the expression for the field’s magnitude and direction.
00:15:00
In this part of the video, the speaker discusses changing the variable of integration from ( dr ) to ( du ), which requires modifying the integral limits accordingly. Initially, the lower limit for ( r ) is zero, translating to ( u = x^2 ). The upper limit, where ( r ) equals ( R ), translates to another value for ( u ). They make the change of variables, rewriting the integral in terms of ( du ), with all constants such as ( k ), ( sigma ), and ( 2pi ) factored in. The integral becomes straightforward, changing from ( rdr ) to ( du/2 ).
The crucial step involves writing ( (x^2 + r^2) ) as ( u ) and simplifying the integral to ( u^{-3/2} ). After integrating ( u^{-1/2} ), the limits are substituted back to derive the final expression for the field produced by the disk. Simplification follows, including distributing a negative sign and switching the order of terms, leading to the final result which includes factors of ( k ), ( sigma ), and ( pi ).
00:18:00
In this segment of the video, the focus is on deriving the final expression for the electric field of a disk with charge density. The presenter starts by discussing the general expression and then examines the case where the distance from the disk (x) is very small (close to zero). When x tends towards zero, simplifying the expression reveals that the terms involving x cancel out, leaving the result that the electric field is the charge density divided by 2 epsilon 0. This result is verified as consistent with Gauss’s Law, confirming that close to the disk, it appears infinite in size. The segment closes by summarizing the integration process for the ring and disk.