Explanation Formula used: A body thrown with velocity u in horizontal direction is influenced by gravity. Therefore, Acceleration of the proton perpendicular to the motion is given by; a = F m Here, F = force m = mass of proton. In the given question the electric force is present. So, F = E × q Here, E = σ ε 0 The sideway deflection is given by; s = a t 2 2 We know that t=d/v with d being distance and v being velocity. The acceleration a is given by; a = σ q m ε 0 Here, ï³ = surface charge density. q = charge of proton. m = mass of proton. Calculation: The sideway deflection is given by; s = a t 2 2 s = a ( 2 × 10 − 8 ) 2 2 …(1) Here, a = σ q m ε 0 so, a = σ q m ε 0 a = 1 × 10 − 6 C/m 2 × 1

Physics for Scientists and Engineers: A Strategic Approach, Vol. 1 (Chs 1-21) (4th Edition)

4th Edition

ISBN: 9780134110684

Author: Randall D. Knight (Professor Emeritus)

Publisher: PEARSON

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Chapter 23, Problem 51EAP

To determine

The distance through which the proton gets deflected sideways.

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Chapter 23, Problem 50EAP

Chapter 23, Problem 52EAP

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Chapter 23 Solutions

Physics for Scientists and Engineers: A Strategic Approach, Vol. 1 (Chs 1-21) (4th Edition)

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