Consider the electric dipole of Example 21.14 . (a) Derive...

Consider the electric dipole of Example 21.14 . (a) Derive an expression for the magnitude of the electric field produced by the dipole at a point on the $x$ -axis in Fig. $21.33 .$ What is the direction of this electric field? (b) How does the electric field at points on the $x$ -axis depend on $x$ when $x$ is very large?

Key Concepts

Electric Dipole

An electric dipole consists of two equal and opposite charges separated by a small distance. It is characterized by a dipole moment, which is the product of one of the charges and the separation vector. This concept lays the groundwork for understanding how the arrangement of charges can produce a specific spatial distribution of the electric field.

Superposition Principle in Electric Field Calculation

The electric field due to a collection of charges is determined by adding the vector contributions from each individual charge. In the case of a dipole, the total field at any point is the sum of the fields produced by each of the two charges, taking into account both magnitude and direction. This principle is essential in deriving expressions for the field along specific lines, such as the axis of the dipole.

Electric Field along the Dipole Axis

For points located along the line that passes through both charges (the dipole axis), the vector nature of the electric field allows the individual contributions to be added directly in terms of magnitude and direction. This approach leads to an expression that not only quantifies the field's strength but also confirms its direction along the axis, typically pointing from the negative to the positive charge or vice versa depending on the sign convention.

Far-Field Approximation

When the distance from the dipole is much larger than the separation between the charges, the dipole’s electric field can be approximated through a series expansion or limiting process. In this far-field region, the field’s dependence on distance simplifies, showing a characteristic 1/x^3 decay, which results from the cancellation of lower-order terms in the expansion. This approximation is a crucial concept in understanding how dipole fields behave at large distances.

Transcript

00:02 Okay, so this is the graph i drew for this question.

00:05 So the point along the x -axis should experience two electric field.

00:09 One's from the positive charge, another one's from a negative charge.

00:13 So e -1 here is the electrophil due to positive charge.

00:15 And e -2 here is the electric view due to a negative charge.

00:20 And as you can tell, the x component for each electric field should be canceled because the magnitude of both electric fuel is equal.

00:28 So that the electric field is having a sense.

00:32 Same magnitude by opposite direction.

00:35 Since the amount of charge on each one is the same, and the distance is the same.

00:40 So therefore, since we consider the positive direction is along positive x -axis and positive direction around a vertical direction is the positive y -axis here.

00:51 So therefore, e -e should be equal to e1 -y plus e2y, which is 2e1 -1.

00:56 You can use e2 as one because e1 and e2 is equal.

01:00 Okay, so therefore we have negative 2e1 sine theta -j.

01:02 The reason is negative because it's pointing downward, which means that it's a negative direction.

01:08 And j here just means it's a y components.

01:11 And we know e1y1 should be equal to e1 sine theta because there's angle theta here and the y components of the depth field is the sine theta.

01:19 And we know e1 should be could kk over the distance square, which is the distance between the point and the charge, which is d over two square plus x square, okay? as you can tell is here.

01:29 And then sine theta should be equal to d over two over square root, d over two square plus x square, which is d over two over two square plus x squared to the power of one half.

01:41 So now we can combine everything together.

01:43 So we have negative 2 kq over d over 2 square plus x squared plus x squared to the power of 1 half and the j vector...

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