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Chapter 1: Problem 45

Zero field \(?\) Four charges, \(q,-q, q\), and \(-q\), are located at equally spaced intervals on the \(x\) axis. Their \(x\) values are \(-3 a,-a, a\), and \(3 a\), respectively. Does there exist a point on the \(y\) axis for which the electric field is zero? If so, find the \(y\) value.

Short Answer

Expert verified
[y=2a, y= -2a] represent the two possible points on the y-axis where the electric field will be zero.

Step by step solution

01

Recognize the symmetries

Because of symmetry, the x components of the electric fields created by the charges at \(-a,-3a,\) and \(a, 3a\) mutually cancel each other. Therefore, we can ignore them and focus on the y-components. We will calculate the electric fields produced by the individual charges at any point \((0, y)\) on the y-axis.
02

Apply Coulomb's Law

We apply Coulomb鈥檚 law to calculate the magnitude of every charge's electric field at a point on the y-axis. For any charge \(Q\), the electric field at a point P, given distance \(r\), can be found using Coulomb's law as \(E= k|Q| / r^2\), where \(k\) is Coulomb's constant.
03

Calculate the Field for Each Charge

We now calculate the y-component of electric field created by each charge at location (0, y) on the y-axis. Let's choose a positive point \(+y\) on the y-axis for our convenience. For the charges at \(-3a\) and \(3a\), the distance from the point is \(\sqrt{y^2+9a^2}\), while for the charges at \(-a\) and \(a\), the distance is \(\sqrt{y^2+a^2}\). Also, the fields by each charges \(q\) are upwards and by each \(-q\) are downwards. We now sum them to set it possibly equal to zero.\n Let \(E_{3a}\) represent the electric field by charges at \(-3a, 3a\) and \(E_a\) represent the electric field by charges at \(-a, a\), then we have:\n\(E_{3a} = kq/(y^2+9a^2)^{3/2}\) and \(E_a= kq/(y^2+a^2)^{3/2}\)\n Summing the y-components gives: \(E_{3a} - E_a = 0\) since \(E_{3a}\) is downwards and \(E_a\) is upwards.
04

Solve for \(y\)

The equation from Step 3 gives: \(kq/(y^2+9a^2)^{3/2} = kq/(y^2+a^2)^{3/2}\)\nNote that \(k\), \(q\), cancel out, we are left with: \(1/(y^2+9a^2)^{3/2} = 1/(y^2+a^2)^{3/2}\)\n We simply now solve this equation for \(y\) which finally gives \(y=\pm 2a\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Coulomb's Law
Understanding Coulomb's Law is crucial when dealing with electric forces and fields. It describes the force between two static charges, stating that the magnitude of the electrostatic force of interaction between two point charges is directly proportional to the scalar multiplication of the magnitudes of the charges and inversely proportional to the square of the distance between them.

The formula encapsulating this principle is usually written as: \[ F = k \frac{|q_1 q_2|}{r^2} \]where \( F \) is the magnitude of the force, \( q_1 \) and \( q_2 \) are the amounts of the charges, \( r \) is the distance between the charges, and \( k \) is Coulomb's constant, approximately \( 8.987 \times 10^9 \) N m^2/C^2. When dealing with the electric field, a related expression \[ E = k \frac{|Q|}{r^2} \] is used, where \( E \) denotes the electric field and \( Q \) is the charge that creates the field at a given point. By applying Coulomb's Law, one can determine the electric field induced by any charge distribution.
Electric Charge Symmetry
The principle of electric charge symmetry plays a significant role in simplifying the calculation of electric fields in many scenarios. Symmetry in charge distribution allows for predictions about the resulting electric field without complicated calculations.

In the context of the given exercise, we saw how the charges placed at symmetric intervals along the x-axis produced fields that canceled each other's x-components at points along the y-axis. Due to this symmetry, the resulting electric field at any point on the y-axis only has a y-component. This is because the horizontal components from charges at equal distances but opposite sides cancel out. Thus, the symmetry significantly reduces the complexity of the problem, allowing us to focus on the vertical components of the electric field. In general, recognizing symmetrical arrangements can facilitate the process of solving many electrostatic problems.
Electric Field Components
The electric field components of a charge distribution can be understood by breaking down the electric field vector into its constituent x, y, and z components. This approach is especially helpful in configurations where symmetry allows us to focus on a specific axis.

In our example, the problem is simplified by recognizing that only the y-components of the electric field need to be considered due to charge symmetry. To calculate these components, we apply Coulomb's Law to each individual charge, considering the distances from the charges to a point on the y-axis. The y-component of the field from each charge is calculated by multiplying the total field by the cosine of the angle between the field vector and the y-axis. By summing these components, we determine the net electric field at that point. This method of component-wise analysis is a fundamental technique in understanding electric fields in vector form.

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Most popular questions from this chapter

Building a sheet from rods ** An infinite uniform sheet of charge can be thought of as consisting of an infinite number of adjacent uniformly charged rods. Using the fact that the electric field from an infinite rod is \(\lambda / 2 \pi \epsilon_{0} r\), integrate over these rods to show that the field from an infinite sheet with charge density \(\sigma\) is \(\sigma / 2 \epsilon_{0}\)

Field from a spherical shell, right and wrong ** The electric field outside and an infinitesimal distance away from a uniformly charged spherical shell, with radius \(R\) and surface charge density \(\sigma\), is given by Eq. (1.42) as \(\sigma / \epsilon_{0}\). Derive this in the following way. (a) Slice the shell into rings (symmetrically located with respect to the point in question), and then integrate the field contributions from all the rings. You should obtain the incorrect result of \(\sigma / 2 \epsilon_{0}\) (b) Why isn't the result correct? Explain how to modify it to obtain the correct result of \(\sigma / \epsilon_{0} .\) Hint: You could very well have performed the above integral in an effort to obtain the electric field an infinitesimal distance inside the shell, where we know the field is zero. Does the above integration provide a good description of what's going on for points on the shell that are very close to the point in question?

Hydrogen atom \(* *\) The neutral hydrogen atom in its normal state behaves, in some respects, like an electric charge distribution that consists of a point charge of magnitude \(e\) surrounded by a distribution of negative charge whose density is given by \(\rho(r)=-C e^{-2 r / a_{0}} .\) Here \(a_{0}\) is the Bohr radius, \(0.53 \cdot 10^{-10} \mathrm{~m}\), and \(C\) is a constant with the value required to make the total amount of negative charge exactly \(e\). What is the net electric charge inside a sphere of radius \(a_{0} ?\) What is the electric field strength at this distance from the nucleus?

Escaping field lines * Charges \(2 q\) and \(-q\) are located on the \(x\) axis at \(x=0\) and \(x=a\), respectively. (a) Find the point on the \(x\) axis where the electric field is zero, and make a rough sketch of some field lines. (b) You should find that some of the field lines that start on the \(2 q\) charge end up on the \(-q\) charge, while others head off to infinity. Consider the field lines that form the cutoff between, these two cases. At what angle (with respect to the \(x\) axis) do these lines leave the \(2 q\) charge? Hint: Draw a wisely chosen Gaussian surface that mainly follows these lines.

Oscillating on a line ** Two positive point charges \(Q\) are located at points \((\pm \ell, 0) .\) A particle with positive charge \(q\) and mass \(m\) is initially located midway between them and is then given a tiny kick. If it is constrained to move along the line joining the two charges \(Q\), show that it undergoes simple harmonic motion (for small oscillations), and find the frequency.
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