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A point charge is placed at each corner of a square with side length a. All charges have magnitude \(q\). Two of the charges are positive and two are negative (Fig. E21.38). What is the direction of the net electric field at the center of the square due to the four charges, and what is its magnitude in terms of \(q\) and \(a\) ?

Short Answer

Expert verified
The net electric field at the center of the square is directed along the line joining the negative charges or opposite to the line joining the positive charges. Its magnitude is \(2k\frac{q}{a^2}\)

Step by step solution

01

Determine individual electric fields

Compute the electric field due to each charge at the center of the square using Coulomb’s law. The electric field \(E\) due to a point charge \(q\) at a distance \(r\) is given by \(E=k\frac{q}{r^2}\), where \(k\) is Coulomb's constant.
02

Find the electric field due to each charge in x and y direction

Since the charges are located at the corners of a square, the net electric field due to each charge has both x and y components. The distance from each charge to the center of the square is \(r = a/ \sqrt{2}\) and the angle each electric field vector makes with x or y axis is 45 degrees. Therefore, we can express the electric field components \$E_x = E \cos (45°) = E/\sqrt{2}\$ and \$E_y = E \sin (45°) = E/\sqrt{2}\$.
03

Calculate the net electric field at the center of the square

Note that the electric field vectors directed along positive x-axis and negative y-axis due to the positive charges (for example) will cancel out with those directed along negative x-axis and positive y-axis due to the negative charges. The net electric field is therefore the sum of the remaining electric field vectors. By symmetry, the net electric field will be directed along the line joining the negative charges or opposite to the line joining the positive charges.
04

Compute the magnitude of the net electric field

While we found the direction of the net electric field in the previous step, the magnitude can be calculated by adding up the magnitudes of the vectors resulting from Step 2. Using the formula for the magnitude of the electric field given by \(E=k\frac{q}{r^2}\), we substitute for \(r = a/ \sqrt{2}\), and sum the x or y components of the electric field vectors, leading to the magnitude of the net electric field E_net = 2E_x = 2E_y = \(2k\frac{q}{a^2}\).

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Key Concepts

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

Coulomb's Law
Coulomb's law is a fundamental principle that describes how electrically charged objects interact with each other. Named after French physicist Charles-Augustin de Coulomb, who formulated the law in the 18th century, it states that the force between two point charges is directly proportional to the product of the magnitudes of the charges and inversely proportional to the square of the distance between them. The formula is given by:
\[ F = k \frac{q_1 q_2}{r^2} \]
where \(F\) is the magnitude of the force, \(k\) is Coulomb's constant (approximately \(8.9875 \times 10^9 N m^2 / C^2\)), \(q_1\) and \(q_2\) are the charges, and \(r\) is the distance between the charges.
  • Understanding Coulomb's law is essential for solving problems related to the electric force and field.
  • It allows us to calculate the electric field \(E\) created by any point charge at a certain distance.
  • This law is the electrostatic counterpart to Newton's law of universal gravitation.

In the context of the given exercise, Coulomb's Law is used to compute the electric field due to each charge at the center of the square, forming the first step in finding the net electric field.
Net Electric Field
The net electric field at a particular point in space is the vector sum of all the electric fields at that point due to various charges present in the vicinity. For point charges, the electric field produced by a single charge can be calculated using Coulomb's law, and then the principle of superposition is applied to find the net electric field.

The steps in finding the net electric field often involve calculating individual charge contributions, breaking them down into components, and then summing these components vectorially. The net electric field magnitude and direction are influenced by the arrangement of the charges and their amounts of charge.
  • The concept of 'net electric field' is critical when dealing with multiple sources of electric fields.
  • It simplifies complex arrangements of charges to understand their collective effect on a point in space.
  • It takes into account the cancellation effects when fields from opposite charges or in opposite directions are involved.

In our textbook solution, the net electric field is determined by adding up the electric field components from each of the four charges in the square's corners.
Electric Field Components
The concept of electric field components is useful in analyzing situations where the electric fields have directions other than along one of the axis in a coordinate system. By decomposing the field into its x- and y-axis components, or into radial and tangential components for polar coordinates, we can perform vector addition to find the net electric field.

To find the components in a Cartesian coordinate system, we use trigonometric functions:
  • The x-component is \(E_x = E \cos(\theta)\), where \(\theta\) is the angle the field makes with the x-axis.
  • The y-component is \(E_y = E \sin(\theta)\).

These components become especially important when electric fields have to be summed or subtracted due to multiple charges, as they allow us to apply the superposition principle in an organized manner. During the exercise, we calculated these components from the individual fields at 45 degrees, which simplified the subsequent vector addition for the net field.
Superposition Principle in Electrostatics
The superposition principle is a key concept in electrostatics that states the total electric field created by multiple charges is the vector sum of the individual fields produced by each charge independently. This principle applies due to the linear nature of electric field equations. It simplifies the calculation of electric fields created by a system of charges by allowing them to be treated separately and then combined.
  • The superposition principle is essential for analyzing complex electric field interactions.
  • It highlights that the fields due to each charge do not affect each other; they simply add up.
  • This principle also implies that the order in which we add the fields does not matter.

For the exercise we discussed, applying the superposition principle involves calculating the electric field at the square's center due to each charge and then summing their vector components to determine the net electric field at that point.

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Most popular questions from this chapter

What is the best explanation for the observation that the electric charge on the stem became positive as the charged bee approached (before it landed)? (a) Because air is a good conductor, the positive charge on the bee's surface flowed through the air from bee to plant. (b) Because the earth is a reservoir of large amounts of charge, positive ions were drawn up the stem from the ground toward the charged bee. (c) The plant became electrically polarized as the charged bee approached. (d) Bees that had visited the plant earlier deposited a positive charge on the stem.

Two very large parallel sheets are \(5.00 \mathrm{~cm}\) apart. Sheet \(A\) carries a uniform surface charge density of \(-8.80 \mu \mathrm{C} / \mathrm{m}^{2},\) and sheet \(B,\) which is to the right of \(A,\) carries a uniform charge density of \(-11.6 \mu \mathrm{C} / \mathrm{m}^{2} .\) Assume that the sheets are large enough to be treated as infinite. Find the magnitude and direction of the net electric field these sheets produce at a point (a) \(4.00 \mathrm{~cm}\) to the right of sheet \(A ;\) (b) \(4.00 \mathrm{~cm}\) to the left of sheet \(A ;\) (c) \(4.00 \mathrm{~cm}\) to the right of sheet \(B\).

A straight, nonconducting plastic wire \(8.50 \mathrm{~cm}\) long carries a charge density of \(+175 \mathrm{nC} / \mathrm{m}\) distributed uniformly along its length. It is lying on a horizontal tabletop. (a) Find the magnitude and direction of the electric field this wire produces at a point \(6.00 \mathrm{~cm}\) directly above its midpoint. (b) If the wire is now bent into a circle lying flat on the table, find the magnitude and direction of the electric field it produces at a point \(6.00 \mathrm{~cm}\) directly above its center.

An American penny is \(97.5 \%\) zinc and \(2.5 \%\) copper and has a mass of \(2.5 \mathrm{~g}\). (a) Use the approximation that a penny is pure zinc, which has an atomic mass of \(65.38 \mathrm{~g} / \mathrm{mol},\) to estimate the number of electrons in a penny. (Each zinc atom has 30 electrons.) (b) Estimate the net charge on all of the electrons in one penny. (c) The net positive charge on all of the protons in a penny has the same magnitude as the charge on the electrons. Estimate the force on either of two objects with this net magnitude of charge if the objects are separated by \(2 \mathrm{~cm}\). (d) Estimate the number of leaves on an oak tree that is 60 feet tall. (e) Imagine a forest filled with such trees, arranged in a square lattice, each \(10 \mathrm{~m}\) distant from its neighbors. Estimate how large such a forest would need to be to include as many leaves as there are electrons in one penny. (f) How does that area compare to the surface area of the earth?

Positive charge \(Q\) is distributed uniformly along the \(x\) axis from \(x=0\) to \(x=a\). A positive point charge \(q\) is located on the positive \(x\) -axis at \(x=a+r,\) a distance \(r\) to the right of the end of \(Q\) (Fig. P21.79). (a) Calculate the \(x\) - and \(y\) -components of the electric field produced by the charge distribution \(Q\) at points on the positive \(x\) -axis where \(x>a\). (b) Calculate the force (magnitude and direction) that the charge distribution \(Q\) exerts on \(q\). (c) Show that if \(r \gg a\), the magnitude of the force in part (b) is approximately \(Q q / 4 \pi \epsilon_{0} r^{2}\). Explain why this result is obtained.

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