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Ch 21: Electric Charge and Electric Field
Young & Freedman Calc - University Physics 15th Edition
Young & Freedman Calc15th EditionUniversity PhysicsISBN: 9780135159552Not the one you use?Change textbook
Chapter 21, Problem 38

A point charge is placed at each corner of a square with side length aa. All charges have magnitude qq. Two of the charges are positive and two are negative (Fig. E21.4221.42). 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 qq and aa?
Square with charges: +q at top corners, -q at bottom corners, side length a, dot at center representing net electric field.

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First, identify the configuration of the charges. The square has four corners, and each corner has a point charge. Two charges are positive (+q) and two are negative (-q). Arrange them such that opposite corners have the same type of charge.
Next, consider the symmetry of the problem. The center of the square is equidistant from all four charges. Due to symmetry, the electric field contributions from opposite charges will have components that cancel each other out.
Calculate the electric field due to a single charge at the center of the square. The distance from the center to any corner is \( \frac{a}{\sqrt{2}} \). Use the formula for the electric field due to a point charge: \( E = \frac{k \cdot q}{r^2} \), where \( k \) is Coulomb's constant and \( r \) is the distance from the charge to the point of interest.
Determine the direction of the electric field vectors. The electric field due to a positive charge points away from the charge, while the field due to a negative charge points towards the charge. Analyze the vector components along the x and y axes for each charge.
Finally, sum the vector components of the electric fields from all four charges to find the net electric field at the center. Due to symmetry, the net electric field will have a specific direction along one of the diagonals of the square. Calculate the magnitude using vector addition of the individual electric fields.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Electric Field

The electric field is a vector field around a charged particle that represents the force exerted per unit charge at any point in space. It is defined as E = F/q, where F is the force experienced by a test charge q. The direction of the electric field is the direction of the force on a positive test charge.
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Superposition Principle

The superposition principle states that the net electric field due to multiple charges is the vector sum of the electric fields produced by each charge individually. This principle allows us to calculate the resultant field at a point by adding the contributions from each charge, considering both magnitude and direction.
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Symmetry in Electric Fields

Symmetry plays a crucial role in simplifying electric field calculations. In this problem, the square configuration and equal magnitude of charges create symmetrical electric field contributions. This symmetry can be used to determine the direction and simplify the calculation of the net electric field at the center of the square.
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Related Practice
Textbook Question

A point charge q1=4.00q_1=-4.00 nC is at the point x=0.600x = 0.600 m, y=0.800y = 0.800 m, and a second point charge q2=+6.00q_2=+6.00 nC is at the point x=0.600x = 0.600 m, y=0y = 0. Calculate the magnitude and direction of the net electric field at the origin due to these two point charges.

Textbook Question

Point charges q1=4.5q_1=-4.5 nC and q2=+4.5q_2=+4.5 nC are separated by 3.13.1 mm, forming an electric dipole. The charges are in a uniform electric field whose direction makes an angle of 36.936.9° with the line connecting the charges. What is the magnitude of this field if the torque exerted on the dipole has magnitude 7.2×1097.2\(\times\)10^{-9} Nm?

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Textbook Question

A 4.00-4.00-nC point charge is at the origin, and a second 5.00-5.00-nC point charge is on the xx-axis at x=0.800x = 0.800 m. Find the electric field (magnitude and direction) at each of the following points on the xx-axis: (i) x=0.200x = 0.200 m; (ii) x=1.20x = 1.20 m; (iii) x=0.200x = -0.200 m.

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Textbook Question

A +8.75+8.75-mC point charge is glued down on a horizontal frictionless table. It is tied to a 6.50-6.50-mC point charge by a light, nonconducting 2.502.50-cm wire. A uniform electric field of magnitude 1.85×1081.85\(\times\)10^8 N/CN/C is directed parallel to the wire, as shown in Fig. E21.3421.34. What would the tension be if both charges were negative?

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Textbook Question

A very long, straight wire has charge per unit length 3.20×10103.20\(\times\)10^{-10} C/m. At what distance from the wire is the electric field magnitude equal to 2.502.50 N/C?

Textbook Question

A 4.00-4.00-nC point charge is at the origin, and a second 5.00-5.00-nC point charge is on the xx-axis at x=0.800x = 0.800 m. Find the net electric force that the two charges would exert on an electron placed at each point in part (a). Note: Part (a) asked to find the electric field (magnitude and direction) at each of the following points on the xx-axis: (i) x=0.200x = 0.200 m; (ii) x=1.20x = 1.20 m; (iii) x=0.200x = -0.200 m.

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