A charge of -6.50 nC is spread uniformly over the surface of one face of a nonconducting disk of

A charge of -6.50 nC is spread uniformly over the surface of...

Key Concepts

Uniform Charge Distribution on a Disk

This concept deals with the situation where electric charge is spread uniformly over the surface of a disk. The electric field at points along the axis of the disk is calculated by integrating the contributions from infinitesimal charge elements, taking advantage of the symmetry of the disk. This method demonstrates how distributed charge creates a resultant field that depends on both the radial distribution and distance from the disk, and it forms the basis for understanding fields produced by extended charged objects.

Charge Distribution on a Ring

In this case, all of the electric charge is concentrated uniformly along the rim of a disk, which effectively forms a ring. The electric field produced by a ring of charge at points along its axis has a distinct dependence on the geometry compared to a uniformly charged disk, and it is derived by summing the contributions of all the charges on the ring. This situation illustrates how moving the charge outward alters the distribution of the field in space.

Point Charge Approximation

This concept concerns the scenario where the entire charge is concentrated at a single point. The electric field of a point charge follows Coulomb’s law, which shows that the field magnitude decreases with the square of the distance from the charge. This simple case serves as a baseline for comparing how different spatial distributions of charge, such as distributed over a disk or ring, affect the electric field at various locations.

Electric Field along the Symmetry Axis

Calculating the electric field along the symmetry axis of symmetric charge distributions, such as disks or rings, utilizes the inherent symmetry to simplify the integration of contributions from each charge element. This approach allows for derivations of field expressions that are functions of axial distance and characteristic dimensions of the charge distribution, providing insight into how geometry influences field magnitude and direction.

Effect of Charge Distribution on Field Strength

The spatial arrangement of charge greatly influences the strength and spatial profile of the resulting electric field. When charge is distributed uniformly over an area, closer proximity of a larger portion of the charge to a point of interest typically enhances the field. Conversely, when the charge is moved to the periphery or concentrated at a point, the resulting field may be weaker or stronger depending on the distance and manner of distribution. This concept illustrates the importance of charge geometry in determining the local electric field intensity.

Transcript

00:01 So we are given the charge capital q that is equal to negative 6 .5 nanoculum.

00:10 So that is 10 to the par minus 9 coulum and radius of the disk that's given to be 1 .25 times 10 to the power minus 2 meter.

00:24 Now in the first part of the problem, we are required to find the electric field at its point x which is 0 .02 meter from the origin.

00:40 Now let's assume the axis of the disk to be the positive x axis.

00:50 Now recall equation 21 .11 that is in example 21 .11 and we will use this equation in order to find the electric field in the first part.

01:06 So according to this equation, the magnitude of the electric field is the x component of the electric field and that is equal to sigma over 2 epsilon not times 1 minus 1 over r over x whole square plus 1 and whole root over of that.

01:32 Now x and r are already known.

01:37 Find a sigma which is the surface charge density and over here since it is a disk the surface charge density is simply charged over the surface area of the disk and we have two epsilon not let me write that as well so yeah now the rest of the values i'm just going to copy paste that now since this is a disk, the area of the disk is just an area of a circle, so that is pi times square of the radius, which is capital r over here.

02:19 And epsilon not is a constant and is equal to 8 .85 times 10 to the par minus 12.

02:31 Now we can simply substitute each of these values over here and doing that define the electric field to be 1 .14 times 10 to the power 5 newton per column.

02:59 So that's the magnitude of the electric field now coming to the direction.

03:05 So because the charge is negative, then the field lines should be oriented towards the disk and the electric field has only one component that is the one in the x direction.

03:18 So the net electric field's direction is negative icap.

03:27 So we can mark the vector sign since we have already included the direction.

03:33 So that's the net electric field in the first part.

03:36 Now in the second part the disk becomes a ring.

03:42 Now if all charges because it becomes a ring so all of the charges that is that was distributed throughout the disk gets pushed away to the outer rim of the ring now that's that is when the disc can be considered as a ring now recall the law of the loop in example 21 .9 so there we got the electric field to be given by the x component of the field only and that is equal to kq x over x square plus a square to the power 3 over 2 and here a is simply the radius of the rink now k is a constant and is equal to 8 .99 times 10 to the power 9 newton meter square for cool on square now we can again substitute all of these values since x has the same value 0 .02 only the disk has become a ring now so our equation has changed so plugging all these variables over here we find the electric field the magnitude of the electric field to be 8 .92 times 10 to the power 4 newton per cool up now coming to the direction again since the charge is negative so the electric field lines are oriented towards the ring and only the x component of the electric field stays and the y component cancel out so the net field is again in the negative x direction so negative i cap and again we can mark a vector sign over here since we have already included the direction in the third part the disc becomes a point charge now if the charge is brought all together, then obviously the disk can be considered as a point charge.

06:00 And we know that electric field of a point charge is simply kq over x square...

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