Find the acceleration in Pm/s2 (1P = 1015, for theelectron in problem #48.a. using a chargedensity of o = 3.54 uC/m2. QUESTION 4Find the acceleration in Pm/s² (1P = 1015) for the electron in problem 22.48.a. using a charge density of a = 3.54 µC/m². celerated the proton through a distanceof 1.00 cm?e48 In Fig. 22-59, an electron (e) is tobe released from rest on the central axisof a uniformly charged disk of radius R.The surface charge density on the disk is+4.00 μC/m². What is the magnitude ofthe electron's initial acceleration if it isreleased at a distance (a) R, (b) R/100,and (c) R/1000 from the center of the disk? (d) Why does the ac-celeration magnitude increase only slightly as the release point ismoved closer to the disk?Figure 22-59Problem 48.

4) particle = electron magnitude of charge on electron (q) = 1.6×10-19 C charge density (σ) = 3.54…

Answered: QUESTION 4 Find the acceleration in…

College Physics

11th Edition

ISBN:9781305952300

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter1: Units, Trigonometry. And Vectors

Section: Chapter Questions

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Find the acceleration in Pm/s2 (1P = 1015, for the
electron in problem #48.a. using a charge
density of o = 3.54 uC/m2.

celerated the proton through a distance
of 1.00 cm?
e
48 In Fig. 22-59, an electron (e) is to
be released from rest on the central axis
of a uniformly charged disk of radius R.
The surface charge density on the disk is
+4.00 μC/m². What is the magnitude of
the electron's initial acceleration if it is
released at a distance (a) R, (b) R/100,
and (c) R/1000 from the center of the disk? (d) Why does the ac-
celeration magnitude increase only slightly as the release point is
moved closer to the disk?
Figure 22-59
Problem 48.

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$4) p a r t i c l e = e l e c t r o n m a g n i t u d e o f c h a r g e o n e l e c t r o n (q) = 1.6 \times 10^{- 19} C c h a r g e d e n s i t y (σ) = 3.54 \frac{μ C}{m^{2}}$

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