Calculate the magnitude and direction (relative to the +x+x-axis) of the electric field in Example 21.621.6. Example 21.621.6: A point charge q=−8.0q = -8.0 nC is located at the origin. Find the electric-field vector at the field point x=1.2x = 1.2 m, y=−1.6y = -1.6 m.

Ch 21: Electric Charge and Electric Field

Young & Freedman Calc - University Physics 14th Edition

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Young & Freedman Calc 14th Edition

Ch 21: Electric Charge and Electric Field

Problem 30a

Chapter 21, Problem 30a

Calculate the magnitude and direction (relative to the $+ x$ -axis) of the electric field in Example $21.6$ . Example $21.6$ : A point charge $q = - 8.0$ nC is located at the origin. Find the electric-field vector at the field point $x equals 1.2$ m, $y = - 1.6$ m.

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Identify the position vector $ \mathbf{r} $ from the charge to the field point. The charge is at the origin, so $ \mathbf{r} = (1.2 \text{ m}) \hat{i} + (-1.6 \text{ m}) \hat{j} $.

Calculate the magnitude of the position vector $ r $ using the formula $ r = \sqrt{x^2 + y^2} $. Substitute $ x = 1.2 \text{ m} $ and $ y = -1.6 \text{ m} $ into the formula.

Use Coulomb's Law to find the magnitude of the electric field $ E $ at the point. The formula is $ E = \frac{k |q|}{r^2} $, where $ k $ is Coulomb's constant $ 8.99 \times 10^9 \text{ N m}^2/\text{C}^2 $ and $ q = -8.0 \text{ nC} $.

Determine the direction of the electric field vector. Since the charge is negative, the electric field points towards the charge. Calculate the angle $ \theta $ relative to the +x-axis using $ \theta = \tan^{-1}\left(\frac{y}{x}\right) $.

Express the electric field vector $ \mathbf{E} $ in component form using $ \mathbf{E} = E_x \hat{i} + E_y \hat{j} $, where $ E_x = E \cos(\theta) $ and $ E_y = E \sin(\theta) $.

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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 small positive test charge q. The direction of the electric field is the direction of the force on a positive charge, and its magnitude is measured in newtons per coulomb (N/C).

Coulomb's Law

Coulomb's Law describes the force between two point charges. It states that the magnitude of the force F between two charges q1 and q2 is proportional to the product of the magnitudes of the charges and inversely proportional to the square of the distance r between them: F = k * |q1 * q2| / r^2, where k is Coulomb's constant. This law is essential for calculating the electric field due to a point charge.

Vector Components

Vector components are the projections of a vector along the axes of a coordinate system. For a vector in two dimensions, such as the electric field, it can be broken down into x and y components using trigonometry: E_x = E * cos(θ) and E_y = E * sin(θ), where θ is the angle with respect to the +x-axis. Understanding vector components is crucial for determining the direction and magnitude of the electric field vector.

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Related Practice

Textbook Question

A uniform electric field exists in the region between two oppositely charged plane parallel plates. A proton is released from rest at the surface of the positively charged plate and strikes the surface of the opposite plate, $1.60$ cm distant from the first, in a time interval of $3.20 \times 10^{- 6}$ s. Find the magnitude of the electric field.

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Textbook Question

A uniform electric field exists in the region between two oppositely charged plane parallel plates. A proton is released from rest at the surface of the positively charged plate and strikes the surface of the opposite plate, $1.60$ cm distant from the first, in a time interval of $3.20$\times$10^{-6}$ s. Find the speed of the proton when it strikes the negatively charged plate.

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Textbook Question

The earth has a net electric charge that causes a field at points near its surface equal to $150$ $N/C$ and directed in toward the center of the earth. What would be the force of repulsion between two people each with the charge calculated in part (a) and separated by a distance of $100$ m? Is use of the earth's electric field a feasible means of flight? Why or why not? Note: Part (a) asked for what magnitude and sign of charge would a $60$ -kg human have to acquire to overcome his or her weight by the force exerted by the earth's electric field.

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Textbook Question

The earth has a net electric charge that causes a field at points near its surface equal to $150$ $N / C$ and directed in toward the center of the earth. What magnitude and sign of charge would a $60$ -kg human have to acquire to overcome his or her weight by the force exerted by the earth's electric field?

Textbook Question

A proton is traveling horizontally to the right at $4.50$\times$10^6$ m/s. What minimum field (magnitude and direction) would be needed to stop an electron under the conditions of part (a)? Note: Part (a) asks for how much time does it take the proton to stop after entering the field.

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Textbook Question

A $positive 8.75$ -mC point charge is glued down on a horizontal frictionless table. It is tied to a $negative 6.50$ -mC point charge by a light, nonconducting $2.50$ -cm wire. A uniform electric field of magnitude $1.85 \times 10^{8}$ $N / C$ is directed parallel to the wire, as shown in Fig. E $21.34$ . Find the tension in the wire.

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