Consider the electric dipole of Example 21.14. (a) Derive an expression for the magnitude of the

step 1 Okay, so this is the graph I drew for this question. So the point along the x -axis should exper

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

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 e1 here is the electric view due to positive charge.

00:15 And e2 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 out because the magnitude of both the electric field 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 the vertical direction is the positive y -axis here.

00:51 So therefore, e is equal to e1y plus e2y, which is 2e1 -1.

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

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

01:02 The reason why it's negative because you can use e2 as well, okay? you can use e2 as well, because e2 is equal.

01:03 Because it's pointing downward, which means that it's a negative direction.

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

01:11 And we know e1y should be equal to e1 sine theta because there's angle theta here.

01:15 And the y components of the depth field is the sine theta.

01:19 And we know e1 should be equal to kq over the distance square, which is the distance between the point and the charge, which is d over 2, plus x squared, 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 squared plus which is d over 2 over 2 square plus x square to the power of 1 half so now we can combine everything together 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 and as you can tell i can saw the two terms so therefore we have e s equal negative k q d over d over 2 square plus x squared to the power of 3 over 2 j so therefore it's just um you do some arrangement here which is um is equal to negative kqd over x squared plus d over 2 square to the power of the 3 over 2 j vector...

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