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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’s 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

Potential energy in a one-dimensional crystal * * Calculate the potential energy, per ion, for an infinite 1 D ionic crystal with separation \(a\); that is, a row of equally spaced charges of magnitude \(e\) and alternating sign. Hint: The power-series expansion of \(\ln (1+x)\) may be of use.

Field from a hemisphere ** (a) What is the electric field at the center of a hollow hemispherical shell with radius \(R\) and uniform surface charge density \(\sigma\) ? (This is a special case of Problem \(1.12\), but you can solve the present exercise much more easily from scratch, without going through all the messy integrals of Problem 1.12.) (b) Use your result to show that the electric field at the center of a solid hemisphere with radius \(R\) and uniform volume charge density \(\rho\) equals \(\rho R / 4 \epsilon_{0}\)

\(N\) charges on a circle \(N\) point charges, each with charge \(Q / N\), are evenly distributed around a circle of radius \(R\). What is the electric field at the location of one of the charges, due to all the others? (You can leave, your answer in the form of a sum.) In the \(N \rightarrow \infty\) limit, is the field infinite or finite? In the \(N \rightarrow \infty\) limit, is the force on one of the charges infinite or finite?

Force in a soap bubble ** Like the charged rubber balloon described at the end of Section 1.14, a charged soap bubble experiences an outward electrical force on every bit of its surface. Given the total charge \(Q\) on a bubble of radius \(R\), what is the magnitude of the resultant force tending to pull any hemispherical half of the bubble away from the other half? (Should this force divided by \(2 \pi R\) exceed the surface tension of the soap film, interesting behavior might be expected!)

\(\mathrm{~ U n i f o r m ~ f i e l d ~ i n ~ a ~ c a v i t y ~ \cjkstart ? ? ? ? ? ?}\) A sphere has radius \(R_{1}\) and uniform volume charge density \(\rho . \mathrm{A}\) spherical cavity with radius \(R_{2}\) is carved out at an arbitrary location inside the larger sphere. Show that the electric field inside the cavity is uniform (in both magnitude and direction). Hint: Find a vector expression for the field in the interior of a charged sphere, and then use superposition. What are the analogous statements for the lower-dimensional analogs with cylinders and slabs? Are the statements still true?

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