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Ch 22: Electric Charges and Forces
Knight Calc - Physics for Scientists and Engineers 5th Edition
Knight Calc5th EditionPhysics for Scientists and EngineersISBN: 9780137344796Not the one you use?Change textbook
Chapter 22, Problem 70a

Starting from rest, how long does it take an electron to move 1.0 cm in a steady electric field of magnitude 100 N/C?

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Step 1: Identify the forces acting on the electron. The electric field exerts a force on the electron given by \( F = qE \), where \( q \) is the charge of the electron (\( -1.6 \times 10^{-19} \, \text{C} \)) and \( E \) is the magnitude of the electric field (\( 100 \; \text{N/C} \)).
Step 2: Use Newton's second law \( F = ma \) to find the acceleration of the electron. Rearrange the formula to \( a = \frac{F}{m} \), where \( m \) is the mass of the electron (\( 9.11 \times 10^{-31} \; \text{kg} \)). Substitute \( F \) from Step 1 into this equation.
Step 3: Use the kinematic equation \( x = v_0 t + \frac{1}{2} a t^2 \) to solve for the time \( t \). Since the electron starts from rest, \( v_0 = 0 \), and the equation simplifies to \( x = \frac{1}{2} a t^2 \). Substitute \( x = 1.0 \; \text{cm} \) (convert to meters: \( 0.01 \; \text{m} \)) and \( a \) from Step 2 into this equation.
Step 4: Rearrange the simplified kinematic equation \( x = \frac{1}{2} a t^2 \) to solve for \( t \). The formula becomes \( t = \sqrt{\frac{2x}{a}} \). Substitute the values of \( x \) and \( a \) to find \( t \).
Step 5: Perform the calculation to determine the time \( t \). Ensure all units are consistent (meters, seconds, etc.) and verify the result.

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Key Concepts

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

Electric Field

An electric field is a region around a charged particle where other charged particles experience a force. The strength of the electric field is measured in newtons per coulomb (N/C) and indicates how much force a unit charge would experience. In this question, the electric field of 100 N/C exerts a force on the electron, causing it to accelerate.
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Acceleration

Acceleration is the rate of change of velocity of an object with respect to time. In the context of an electron in an electric field, it can be calculated using Newton's second law, F = ma, where F is the force acting on the electron, m is its mass, and a is the acceleration. The force on the electron is determined by the electric field strength and the charge of the electron.
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Kinematic Equations

Kinematic equations describe the motion of objects under constant acceleration. They relate displacement, initial velocity, final velocity, acceleration, and time. For an electron starting from rest, the relevant equation is s = ut + (1/2)at², where s is the displacement, u is the initial velocity (zero in this case), a is the acceleration, and t is the time taken to travel the distance.
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Related Practice
Textbook Question

A small 1.0 g block charged to 75 nC is placed on a 30° inclined plane. The coefficients of static and kinetic friction are 0.20 and 0.10, respectively. What minimum strength horizontal electric field is needed to keep the block from sliding down the plane?

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Textbook Question

Space explorers discover an 8.7×1017 kg asteroid that happens to have a positive charge of 4400 C. They would like to place their 3.3×105 kg spaceship in orbit around the asteroid. Interestingly, the solar wind has given their spaceship a charge of −1.2C. What speed must their spaceship have to achieve a 7500-km-diameter circular orbit?

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Textbook Question

An electric field E=200,000i^\(\overrightarrow{E}\)=200,000\(\hat{i}\) N/C causes the point charge in FIGURE P22.68 to hang at an angle. What is θ?

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Textbook Question

Three 1.0 nC charges are placed as shown in FIGURE P22.66. Each of these charges creates an electric field E at a point 3.0 cm in front of the middle charge. What are the three fields E₁, E₂, and E₃ created by the three charges? Write your answer for each as a vector in component form.

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Textbook Question

A 5.0 g ball charged to 1.5 μC is tied to a 25-cm-long string. It swings at 250 rpm in a horizontal circle around a stationary ball charged to −2.5 μC. What is the tension in the string?

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Textbook Question

A 10.0 nC charge is located at position (x, y)=(1.0 cm, 2.0 cm). At what (x, y) position(s) is the electric field (21,600i^28,800j^)(21,600\(\hat{i}\)-28,800\(\hat{j}\)) N/C?

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