Show that the maximum magnitude Emax of the electric field along the axis of a uniformly charged

Show that the maximum magnitude Emax of the electric field...

Key Concepts

Uniform Charge Distribution

This concept refers to the arrangement of charge such that it is evenly spread over a given geometry, in this case the ring. A uniform distribution simplifies the integration process when calculating the electric field, as every infinitesimal segment contributes equally in magnitude, allowing symmetry arguments to be applied.

Symmetry in Electrostatics

Symmetry in the charge configuration, such as that in a ring, greatly simplifies the calculation of the resultant electric field. Due to the circular symmetry, the horizontal components of the field due to opposite charge elements cancel out on the axis, leaving only the vertical (axial) component, which can then be integrated more straightforwardly.

Electric Field Due to a Charge Distribution

This concept involves calculating the net electric field at a point from a continuous charge distribution. It requires summing (integrating) the contributions from all charge elements using Coulomb's law, where each element generates a field that depends on its magnitude and the inverse square of its distance from the point of interest.

Application of Coulomb's Law

Coulomb's law provides the basis for calculating the electric field created by a point charge, and when dealing with continuous charge distributions, it is used in an integral form. In the context of a charged ring, each infinitesimal segment is treated as a point charge that contributes to the total field along the axis.

Differential Calculus for Optimization

Differential calculus is applied to determine the point along the axis where the electric field attains its maximum value. By expressing the electric field in terms of the axial coordinate and then differentiating with respect to that coordinate, one sets the derivative equal to zero to find the optimum (or extremum) value, which in this problem occurs at x = a/?2.

Transcript

00:01 Okay, so when you know the electric field can be equal to k -e times q x over x squared plus a square to the power of 3 over 2.

00:08 So if we take differentiation on each side, we have d over d x is equal to d over d x and inside of it is k -ke x over x over x over x squared to the power of 3 over 2.

00:22 So therefore d e over d x is equal to k eq times a square minus 2x over x squared plus a square to the power of a five over two.

00:32 So as we know, if de over d x is equal to 0, which means when a slope is approached to 0, that means the electric field will be at maximum.

00:43 Okay? so we just simply put the equation here equal to 0.

00:49 So then we'll have a square plus 2x square is equal to 0.

00:57 Since the number below the denominator cannot be equal zero.

01:01 Okay? so we have 2x squared is equal to a square.

01:06 Therefore, x squared is equal to a square over 2.

01:13 And then x is equal to square root a square over 2, which is equal to a over square of 2.

01:26 And then we can plug in the x value back into the equation to determine the maximum electric field.

01:32 So e -max will be, um, equal to k e times q x okay so k e times q and n times a over square of two over x square which is a over square of two plus a square to the power of three over two and this will give us k e q times a over square or two over a square over 2 plus a square and then to the power of 3 over 2 and this will give us k -eq a over square 2 over a square over 3a square over 2 and then to the power of 3 over 2 and we can take a square out then we have k -ke x xx2 times a over square of 2 over 3 times square of 3 over 2 times square of 2.

03:32 Because 3 to the power of 3 over 2 is equal to 3 square of 3...

Please give Ace some feedback