QUESTION 4 Find the acceleration in Pm/s2 (1P = 1015) for the electron in problem 22.48.a. using a charge density of a = 6.94 µC/m²
QUESTION 4 Find the acceleration in Pm/s2 (1P = 1015) for the electron in problem 22.48.a. using a charge density of a = 6.94 µC/m²
Related questions
Question
Find the acceleration in Pm/s2 (1P = 1015, for the
electron in problem #48.a. using a charge
density of o = 6.94 uC/m2. Answer in 5 significant digi

Transcribed Image Text: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.

Transcribed Image Text:QUESTION 4
Find the acceleration in Pm/s² (1P = 1015) for the electron in problem 22.48.a. using a charge density of a = 6.94 µC/m²
Expert Solution

GIVEN:
The surface charge density of the uniform charge disk is given as, .
The charge of electron is given as, .
DIAGRAM:
TO DETERMINE:
(a) The initial acceleration of electron if it is released at a distance-R, .
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