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Chapter 16: Problem 27

What are the magnitude and direction of the acceleration of a proton at a point where the electric field has magnitude \(33 \mathrm{kN} / \mathrm{C}\) and is directed straight up?

Short Answer

Expert verified
Answer: The magnitude of the acceleration is \(3.16 \times 10^{11} \, m/s^2\), and the direction of the acceleration is straight up.

Step by step solution

01

Identify the given information

We are given: - Electric field magnitude: \(E = 33 \, kN/C\) - Electric field direction: straight up - Charge of a proton: \(q_p = 1.602 \times 10^{-19} C\) - Mass of a proton: \(m_p = 1.673 \times 10^{-27} kg\)
02

Calculate the force on the proton

We can find the force exerted by the electric field on the proton using the formula: \(F = q_pE\) Substitute the given values: \(F = (1.602 \times 10^{-19} \, C)(33 \times 10^{3} \, N/C)\) Calculate the force: \(F = 5.286 \times 10^{-16} \, N\)
03

Calculate the acceleration of the proton

Now we can find the acceleration of the proton using the formula: \(a = \frac{F}{m_p}\) Substitute the given values: \(a = \frac{5.286 \times 10^{-16} \, N}{1.673 \times 10^{-27} \, kg}\) Calculate the acceleration: \(a = 3.16 \times 10^{11} \, m/s^2\)
04

Determine the direction of the acceleration

Since the direction of the electric field is straight up and the proton has a positive charge, the acceleration will be in the same direction as the electric field. Thus, the direction of the acceleration is also straight up.
05

State the final answer

The magnitude of the acceleration is \(3.16 \times 10^{11} \, m/s^2\), and the direction of the acceleration is straight up.

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Most popular questions from this chapter

Positive point charges \(q\) and \(2 q\) are located at \(x=0\) and \(x=3 d,\) respectively. Are there any points not on the \(x\) -axis where $\overrightarrow{\mathbf{E}}=0 ?$ Explain.
In fair weather, over flat ground, there is a downward electric field of about \(150 \mathrm{N} / \mathrm{C} .\) (a) Assume that the Earth is a conducting sphere with charge on its surface. If the electric field just outside is $150 \mathrm{N} / \mathrm{C}$ pointing radially inward, calculate the total charge on the Earth and the charge per unit area. (b) At an altitude of $250 \mathrm{m}\( above Earth's surface, the field is only \)120 \mathrm{N} / \mathrm{C}$ Calculate the charge density (charge per unit volume) of the air (assumed constant). [Hint: See Conceptual Example \(16.8 .]\)
A thin wire with positive charge evenly spread along its length is shaped into a semicircle. What is the direction of the electric field at the center of curvature of the semicircle? Explain.
Positive point charges \(q\) and \(2 q\) are located at \(x=0\) and \(x=3 d,\) respectively. What is the electric field at \(x=2 d\) (point \(S\) )?
Two metal spheres of radius \(5.0 \mathrm{cm}\) carry net charges of $+1.0 \mu \mathrm{C}\( and \)+0.2 \mu \mathrm{C} .$ (a) What (approximately) is the magnitude of the electrical repulsion on either sphere when their centers are \(1.00 \mathrm{m}\) apart? (b) Why cannot Coulomb's law be used to find the force of repulsion when their centers are \(12 \mathrm{cm}\) apart? (c) Would the actual force be larger or smaller than the result of using Coulomb's law with \(r=12 \mathrm{cm} ?\) Explain.
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