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

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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 = 6.94 uC/m2. Answer in 5 significant digi

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: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.
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
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, σ=6.94 μC/m2=6.94×10-6 C/m2.

The charge of electron is given as, e=1.6×10-19 C.

DIAGRAM:

Advanced Physics homework question answer, step 1, image 1

TO DETERMINE:

(a) The initial acceleration of electron if it is released at a distance-R, a.

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