CALC A semicircle of radius a is in the first and second quadrants, with the center of curvature

CALC A semicircle of radius a is in the first and second...

Key Concepts

Symmetry in Electrostatics

Symmetry arguments are used to simplify the evaluation of electric fields by cancelling out the components that are equal in magnitude but opposite in direction. In problems with symmetric charge distributions, such as semicircular arcs positioned in specific quadrants, symmetry can drastically reduce the complexity of the integration by indicating which components of the field cancel or add constructively.

Superposition Principle

The principle of superposition states that the net electric field created by multiple charges is the vector sum of the electric fields due to each individual charge or charge element. This principle is critical when combining the contributions from different parts of the distribution—such as the positive and negative halves—to obtain the overall field.

Integration of Continuous Charge Distributions

Integrating over a continuous charge distribution involves breaking the distribution into infinitesimal elements that each contribute a small amount to the overall electric field. Parameterizing the curve (for example, by an angle for a semicircle) and integrating these contributions with respect to the parameter allows one to account for the vector nature of the field and determine both its magnitude and direction.

Coulomb's Law

Coulomb's law describes the electric force (and thus the field) created by a point charge. In problems involving continuous charge distributions, Coulomb's law is applied to infinitesimal charge elements, and their individual contributions to the electric field are integrated to determine the total field. This concept is fundamental in understanding how individual charges influence the overall electric field in space.

Charge Distribution

A continuous charge distribution refers to a charge spread out over a region, line, surface, or volume. In this context, uniform linear charge density is used to express that the charge is distributed evenly along a curved path, such as a semicircular arc. This concept helps in setting up the correct integrals by defining how charge per unit length (or area/volume) contributes to the field.

Transcript

00:01 In this exercise, we have two quarters of a ring, on which the quarter on the right has a total charge of minus q, and the quarter on the left has a total charge of plus q.

00:14 What we need to do in this exercise is to find what is the resident field that i'll call eres measured by some one that is placed in the oranging of the circle of radius a.

00:31 Okay, so how do we calculate the electric field of such a configuration? well, we know that kulom's law applies for point charges, and here we are not dealing with two point charges.

00:45 Instead, we are dealing with two quarter of rings that are charged, but we can treat one quarter of a ring as a collection of infinitesimal charges that i'll call here the q.

01:00 So here suppose that i'm taking a very small piece of this quarter of ring.

01:06 So instead we can measure the, we can use columbus law to evaluate what is the electric fields produced by a collection of small charges they are placed around along this quarter of ring.

01:24 So by this, i mean that we will use what we call the infinitesimum columns law.

01:32 So first, it is interesting to define what the q is.

01:37 So we know that the definition of total charge for the case of a quarter of a ring, it will be the linear density of charges times a line element, which is basically this length here, this small length.

02:03 And in our case, we can notice that this small length is just comprehends an angle of d - theta and has a radius of a, so we can rewrite it as lambda times a d -theta, on which we can find what lambda is from the definition of, sorry, this is the differential charge, and a on which we can find what lambda is from the total charge definition, on which for a quarter of a ring, is just a perimeter of this quarter of ring.

02:49 So it's pi over two times a times lambda.

02:55 And we find that lambda is equal to 2k over pi a.

03:04 So you can rewrite the q as to q pi over a times a the theta only to cut this is here okay so we have the the expression for the infinitesimal columns law for the infinitesimal fields for this infinitesimal charge the queue so to where will the field to where will these fields will be pointing to so we know that for a quarter of a ring, the field will be pointing in the radial direction.

03:52 And since the charges that are spread along this ring are negative, it will be pointing towards the source charges.

04:04 So here, so i'll write here that the e on the right will be pointing to the positive radial direction.

04:19 And on the left, we have that the field will also be pointing in the radio direction, but instead of a negative charge, we have a positive charge.

04:31 And so the field will be pointing from outwards the source charges towards the arranging of the circle.

04:44 So here we'll have that the field on the left will be pointing.

04:52 To the negative radial direction.

05:02 And we can find the expression for the radial direction as remembering that this has a unit modulus, which means that the radius direction is equal to the radial vector over the modulus of the radio vector, which is x in the x direction, which i'll call a, plus the y coordinate in the j direction which is the y direction over and we know that the modulus of r for circle of radius a is just the radius of the circle over a and working with polar coordinates we know that x is equal to a kiosign of theta and y is equal to equal to a is equal to a sine theta so that the r direction can be expressed as simply, sorry, as cosine theta in the x direction, plus sine theta in the j direction.

06:27 Great.

06:28 Now we can use all these expressions we found.

06:33 So the expression for the q and the expression for the radial direction in terms of the x and y directions, and find that for the right side, we'll have that the total field will be the integral of columbus infinity as law, which is the er.

07:03 And this means that the total field will be the integral of the q on which the q is 2 q over 2 q over pi so 2 q over pi times a d theta this all is the q as we saw previously times 1 .4 pi epsilon 0, a squared, times in the r direction, but we know that the r direction is just this expression.

07:50 So times, consign theta in the i direction, plus sine theta in the j direction.

08:00 And the limits of our integral, notice that will be integrating over the theta, and for this distribution, so we can sum over all the infinitesimal dq to cover all the quarter of this ring, is going to be a sum from zero to pi over two.

08:30 So, integral from zero to pi over two.

08:35 Okay, so we just solved this integral, and first of all, we can notice that we can cut some quantities, here.

08:44 So sorry, there is no a in here.

08:50 We have that these two cancel with this four, making a two, and this pi multiplies with this pi, making a pi squared.

09:01 And we can take these terms here out of the integral, since they're constant in the theta, and we had that er is equal 2 is equal to q over 2 pi squared, epsilon 0, a squared, times the integral of cosine theta in the i direction plus the integral of sine theta in the j direction, in the j direction, in the y direction over the theta from 0 to pi over 2.

10:03 Okay, so let's take the integral on cosine theta first...

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