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00:00:00 – 00:31:42
The video delves into the calculation of electric fields resulting from different charge distributions, primarily focusing on disks, infinite sheets of charge, and between parallel plates. Using Coulomb's law and principles of integration, the presenter meticulously derives formulas for the electric field. They start with a solid disk and break down the steps to find the electric field at a point from its center, taking into account small charge elements and simplifying with trigonometry.
Key concepts include charge per unit area (σ), the permittivity of free space (ε₀), and the use of integral calculus for solving complex expressions. Specifically, for an infinite sheet of charge, the electric field is derived and simplified to σ/2ε₀ with examples provided. Similarly, the electric field around uniformly charged rectangular plates is shown to be distance-independent, while between two parallel plates, the fields add up inside the plates and cancel out outside, leading to practical problem-solving applications.
00:00:00
In this segment, the video focuses on calculating the electric field due to a charge distribution. The process includes deriving a formula for the electric field of an infinite sheet of charge and between two parallel plates containing a sheet of charge. The main example begins with a solid disk, differentiating between the inner and outer radii. The goal is to calculate the electric field at a point P from the center. By considering a small amount of charge and using Coulomb’s law, it explains the electric field components, showing that the net field in the y-direction is zero, and reducing the problem to finding the X component. The segment uses trigonometry and the Pythagorean theorem to further refine the expressions needed for the calculation.
00:05:00
In this part of the video, the speaker discusses a physics problem involving a circle and an equation for ( dE_x ). They derive the area of a circle as ( pi r^2 ) and differentiate both sides to find ( dA ), leading to an expression ( dA = 2pi r , dr ). Then, they talk about ( sigma ) (the charge per unit area) and rearrange an equation to obtain ( dQ = sigma cdot dA ). Substituting ( dA ) gives ( dQ = 2pi r sigma , dr ), which they use in the equation for ( dE_x ).
By integrating both sides, they aim to find the electric field ( E_x ) from the center to the edge of the circle. They simplify constants and set up an integral involving the variable ( r ). Finally, they suggest using U-substitution to integrate the expression correctly, setting ( u = x^2 + r^2 ).
00:10:00
In this part of the video, the speaker explains the process of differentiating and integrating an expression involving the variables R and X. The derivative of ( R^2 ) is calculated as ( 2R ) times ( dR ), and ( dR ) is expressed in terms of ( dU ). After performing manipulations, including moving terms and changing exponents, the anti-derivative of ( u^{3/2} ) is determined. The expression is rewritten with ( U ) substituted by ( x^2 + R^2 ). Subsequently, the integral for the electric field ( E ) is solved, and the final expression for the electric field in the X direction is derived. The speaker concludes by distributing the variable X into the terms inside the brackets.
00:15:00
In this part of the video, the presenter simplifies a complex equation involving an electric field with constants such as K and Epsilon. They then apply this formula to a specific problem involving a disc of radius 3 meters with a total charge of 480 nanocoulombs. The presenter draws a diagram to illustrate the setup and introduces variables including the radius (R), total charge (Q), and distance (X) from the center of the disc to a point P. They calculate the charge per unit area (Sigma) using the formula ( sigma = frac{Q}{A} ) and find it to be approximately ( 1.697 times 10^8 ) coulombs per square meter. Finally, they use the previously derived electric field equation to continue solving the problem.
00:20:00
In this part of the video, the speaker explains the process of calculating the electric field produced by an infinite sheet of charge. They begin by working through a detailed calculation, starting with certain constants, and demonstrate step-by-step how to derive a numerical result of 8795 Newtons per Coulomb for a specific example. The speaker then generalizes the formula by considering the case where the radius (R) of the disc approaches infinity. They show that as R becomes very large, certain terms in the equation become negligible, simplifying the formula. The final simplified equation for the electric field of an infinite sheet of charge is given as Sigma over 2 Epsilon0. The speaker concludes with a problem for the viewers to try, involving an infinite sheet with a charge per unit area of 5 microcoulombs per square meter at specific distances.
00:25:00
In this part, the video explains how to calculate the electric field at a point near a uniformly charged rectangular plate. A large rectangle with a uniformly distributed positive charge is considered, and the electric field at a point 50 cm away is calculated using the formula for an infinite sheet of charge, showing that the electric field is independent of distance. The electric field is given as (sigma / (2 epsilon_0)), resulting in 2.82 x 10^5 N/C. The concept is extended to the electric field between two parallel plates, one positively charged and the other negatively charged. The electric field between the plates is the sum of the fields from each plate (E1 and E2), while outside the plates, the net electric field is zero due to the equal magnitude and opposite direction of E1 and E2.
00:30:00
In this part of the video, the speaker discusses the electric field between two infinite sheets of charge. The net electric field outside the sheets is zero since the fields from each sheet cancel each other out. However, inside the sheets, the electric fields add up. Using the formula for the electric field due to one infinite sheet of charge (charge per unit area divided by 2 times the permittivity of free space), the speaker explains that the net electric field between the sheets is the charge per unit area divided by the permittivity of free space. The video concludes by summarizing how to apply this formula to problems.