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Calculate the initial (from rest) acceleration of a proton in a $5.00 \times 10^{6} \mathrm{N} / \mathrm{C}$ electric field (such as created by a research van de Graaff). Explicitly show you follow thesteps in the Problem-Solving Strategy for electrostatics.

$a=4.79 \cdot 10^{14} \mathrm{m} / \mathrm{s}^{2}$

Physics 102 Electricity and Magnetism

Chapter 18

Electric Charge and Electric Field

Cornell University

University of Michigan - Ann Arbor

Hope College

McMaster University

Lectures

11:53

In physics, a charge is a …

10:30

The electric force is a ph…

00:59

Calculate the initial (fr…

06:26

A proton is projected in t…

04:26

00:56

Determine the final veloci…

06:31

W A proton is projected in…

04:15

A proton is placed in a un…

03:00

A constant electric field …

05:55

A proton moves through a u…

04:29

Calculate the speed of a p…

0:00

04:38

A proton is released from …

03:08

A proton moves through a r…

03:47

06:45

A proton accelerates from …

02:42

(a) Calculate the speed of…

An electron and a proton a…

02:10

A Van de Graaff generator …

05:45

A proton moves at $4.50 \t…

03:06

A proton moves in the elec…

A proton and an electron a…

So here in this problem, we have to find the acceleration off the proton and an electric field off five times 10 to the power six Newton per column. Now we know that the charge on the proton as 1.60 times 10 to the power minus 19 column. Now we have a relationship between the electric field intensity on the electrostatic force, which is equals two if over. Q. So from here, we can say that if force acting on the charge would be eight times Q on from second loft, motion begins. Know that Net force acting on a particle is equals to mass times acceleration so they can incubate both this equation. And we can say that e que will be called to mass times acceleration. From here. The expression for acceleration will get this thing as eight times. Q. Divide by the Moss and the mosque off Proton. We have 1.67 times 10 to the power minus 27 K g. Now we have all the release within target This equation to calculate the acceleration. Acceleration is equals to e. We have five times 10 to the power six Newton Poke will, um, times the charge. On column. We have 1.60 times 10 to the power minus 19 column Divide by the mosque off Proton is 1.67 times 10 to the power minus 27 k g. Nor will simplify this on wicked acceleration as 4.79 times 10 to the power 14 media per second square. So is the final answer for the given problem. I hope you have understand the problem. Thank you.

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