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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

A conductor in electrostatic equilibrium contains a cavity in which there are two point charges: \(q_{1}=+5 \mu \mathrm{C}\) and \(q_{2}=-12 \mu \mathrm{C} .\) The conductor itself carries a net charge \(-4 \mu C .\) How much charge is on (a) the inner surface of the conductor? (b) the outer surface of the conductor?
Point charges are arranged on the vertices of a square with sides of $2.50 \mathrm{cm} .$ Starting at the upper left corner and going clockwise, we have charge A with a charge of \(0.200 \mu \mathrm{C}, \mathrm{B}\) with a charge of \(-0.150 \mu \mathrm{C}, \mathrm{C}\) with a charge of \(0.300 \mu \mathrm{C},\) and \(\mathrm{D}\) with a mass of \(2.00 \mathrm{g},\) but with an unknown charge. Charges \(\mathrm{A}, \mathrm{B},\) and \(\mathrm{C}\) are fixed in place, and \(\mathrm{D}\) is free to move. Particle D's instantaneous acceleration at point \(D\) is \(248 \mathrm{m} / \mathrm{s}^{2}\) in a direction \(30.8^{\circ}\) below the negative \(x\) -axis. What is the charge on D?
Two otherwise identical conducting spheres carry charges of $+5.0 \mu \mathrm{C}\( and \)-1.0 \mu \mathrm{C} .$ They are initially a large distance \(L\) apart. The spheres are brought together, touched together, and then returned to their original separation \(L\). What is the ratio of the magnitude of the force on either sphere after they are touched to that before they were touched?
Suppose a 1.0 -g nugget of pure gold has zero net charge. What would be its net charge after it has \(1.0 \%\) of its electrons removed?
\(\mathrm{A}+2.0\) -nC point charge is \(3.0 \mathrm{cm}\) away from a $-3.0 \mathrm{-nC}$ point charge. (a) What are the magnitude and direction of the electric force acting on the +2.0 -nC charge? (b) What are the magnitude and direction of the electric force acting on the -3.0 -nC charge?
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