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Four point charges have the same magnitude of \(2.4 \times 10^{-12} \mathrm{C}\) and are fixed to the corners of a square that is \(4.0 \mathrm{cm}\) on a side. Three of the charges are positive and one is negative. Determine the magnitude of the net electric field that exists at the center of the square.

Short Answer

Expert verified
The net electric field magnitude at the center is approximately \(4.5 \times 10^3 \text{ N/C}\).

Step by step solution

01

Identify Charge Positions

Label the four corners of the square as A, B, C, and D. Let the charges at A, B, and C be positive and the charge at D be negative, each with a magnitude of \(2.4 \times 10^{-12} \text{ C}\).
02

Determine Electric Field Contribution by a Single Charge

The electric field due to a point charge is given by the formula \( E = \frac{k |q|}{r^2} \), where \( k = 8.99 \times 10^9 \text{ N m}^2/\text{C}^2 \) is Coulomb's constant, \( q = 2.4 \times 10^{-12} \text{ C} \), and \( r \) is the distance from the charge to the center of the square. The distance \( r \) from a corner of the square to the center where each charge is located is \( \frac{\sqrt{(0.04^2 + 0.04^2)}}{2} = \frac{\sqrt{0.0032}}{2} = 0.0283 \text{ m} \).
03

Calculate Electric Field Magnitude from One Charge

Substitute the values into the formula for a single charge: \[ E = \frac{8.99 \times 10^9 \times 2.4 \times 10^{-12}}{(0.0283)^2} \text{ N/C} \]The result is \( E = 2.7 \times 10^3 \text{ N/C} \) for a single charge.
04

Determine the Direction of Electric Fields

At the center, the electric field due to each charge is directed radially. Use vector addition to combine the fields. Positive charges contribute fields away from themselves, and the negative charge's field points toward itself.
05

Superpose Electric Fields from All Charges

Add the fields vectorially: the fields due to the positive charges at A, B, and C push outward while the field due to the negative charge at D pulls inward. This requires vector addition of E-fields directed opposite the diagonal of the square.
06

Calculate Net Electric Field Magnitude

Using symmetry and vector components, calculate the resultant electric field vector. Since the opposite diagonal charges cancel, only consider vector components effectively on the same axis. The magnitude of the net E-field due to non-cancelled components is \[ E_{net} = 4.5 \times 10^3 \text{ N/C} \].

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Key Concepts

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

Point Charges
In electrostatics, point charges are fundamental entities representing charged particles confined to a small region, essentially a point in space. Think of them as tiny charged spheres that generate an electric field around them. Though considered dimensionless, these point charges effectively model real-world charged particles, like electrons or protons.
The exercise deals with point charges placed at the corners of a square. Each has the same charge, magnitudes of which influence both the strength and direction of the electric fields they create. These fields interact with each other through the principles of electrical forces.
Coulomb's Law
Coulomb's law defines the force between two point charges. It states the force \( F \) between two charges, \( q_1 \) and \( q_2 \), separated by distance \( r \), is given by:
  • \( F = k \frac{|q_1 q_2|}{r^2} \)
Here, \( k \) is Coulomb's constant. This law shows that the force increases with greater charge magnitudes and decreases as the square of the distance between them.
In this exercise, Coulomb's law is used to calculate the electric field, \( E \), which is closely related to force. The electric field due to a single charge \( q \) at a distance \( r \) is given by:
  • \( E = \frac{k |q|}{r^2} \)
By applying Coulomb's law, the electric field for each charge can be individually calculated.
Vector Addition
Electric fields are vectors, meaning they have both magnitude and direction. To understand the net electric field due to multiple point charges, vector addition is crucial. When combining these fields:
  • The fields from positive charges point away from the charge.
  • The field from a negative charge points toward the charge.
Since they are vectors, these fields add according to vector rules. The direction and magnitude must both be considered. In our exercise, fields are combined using vector addition, especially due to the symmetry and positioning of the charges around the square.
Calculate the diagonal components, recognize symmetry, and apply vector addition rules to find the overall electric field at the center.
Superposition Principle
The superposition principle allows us to analyze complex electric fields from multiple point charges by considering each charge's effect independently. According to this principle, the total electric field from a group of charges is the vector sum of the fields due to individual charges.
In this problem, the superposition principle simplifies calculations by facilitating the addition of fields from four point charges at the corners of the square. Each field is calculated separately using Coulomb's law and then summed at the center of the square using vector addition.
This principle is a powerful tool in electrostatics, significantly simplifying calculations and predictions of electric fields in configurations like those in the exercise.

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Most popular questions from this chapter

A charge \(Q\) is located inside a rectangular box. The electric flux through each of the six surfaces of the box is: \(\Phi_{1}=+1500 \mathrm{N} \cdot \mathrm{m}^{2} / \mathrm{C},\) \(\Phi_{2}=+2200 \mathrm{N} \cdot \mathrm{m}^{2} / \mathrm{C}, \Phi_{3}=+4600 \mathrm{N} \cdot \mathrm{m}^{2} / \mathrm{C}, \Phi_{4}=-1800 \mathrm{N} \cdot \mathrm{m}^{2} / \mathrm{C}\) \(\Phi_{5}=-3500 \mathrm{N} \cdot \mathrm{m}^{2} / \mathrm{C},\) and \(\Phi_{6}=-5400 \mathrm{N} \cdot \mathrm{m}^{2} / \mathrm{C} .\) What is \(Q ?\)

A small drop of water is suspended motionless in air by a uniform electric field that is directed upward and has a magnitude of \(8480 \mathrm{N} / \mathrm{C}\). The mass of the water drop is \(3.50 \times 10^{-9} \mathrm{kg} .\) (a) Is the excess charge on the water drop positive or negative? Why? (b) How many excess electrons or protons reside on the drop?

mmh Two spherical shells have a common center. \(A-1.6 \times 10^{-6}\) C charge is spread uniformly over the inner shell, which has a radius of \(0.050 \mathrm{m} . \mathrm{A}+5.1 \times 10^{-6} \mathrm{C}\) charge is spread uniformly over the outer shell, which has a radius of \(0.15 \mathrm{m}\). Find the magnitude and direction of the electric field at a distance (measured from the common center) of (a) \(0.20 \mathrm{m}\) (b) \(0.10 \mathrm{m},\) and (c) \(0.025 \mathrm{m}\)

Four identical metallic objects carry the following charges: +1.6 \(+6.2,-4.8,\) and \(-9.4 \mu \mathrm{C} .\) The objects are brought simultaneously into contact, so that each touches the others. Then they are separated. (a) What is the final charge on each object? (b) How many electrons (or protons) make up the final charge on each object?

The membrane surrounding a living cell consists of an inner and an outer wall that are separated by a small space. Assume that the membrane acts like a parallel plate capacitor in which the effective charge density on the inner and outer walls has a magnitude of \(7.1 \times 10^{-6} \mathrm{C} / \mathrm{m}^{2} .\) (a) What is the magnitude of the electric field within the cell membrane? (b) Find the magnitude of the electric force that would be exerted on a potassium ion \(\left(\mathrm{K}^{+} ;\right.\) charge \(\left.=+e\right)\) placed inside the membrane.

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