3 points My Notes Ask Your Teacher radius R and surface charge density +4 o HC/m
ID: 1594513 • Letter: 3
Question
3 points My Notes Ask Your Teacher radius R and surface charge density +4 o HC/m2) lies in the xy plane with its center at the origin. th Sh On Det ly charged disk The marked point P is exactly a distance R above the center of the disk. An electron is to be released from rest at several points along the z axis. (a) What is the electron's initial acceleration if it is released at point P? m/s2 (b) if it is released 00 th of the way from the disk to point P? m/s2 k (c) if it is released 000 th of the way from the disk to point p? m/s2 k (d) Did your answer change more from part a to part b or from part b to part c? Explain why this makes sense conceptually.Explanation / Answer
field at central axis due to charged disc,
E = 2 pi sigma k [ 1 - x/sqrt(x^2 + R^2) ]
a) x = R
E = 2pi ( 4.70 x 10^-6 x 9 x 10^9 ) [ 1 - R / sqrt(R^2 + R^2) ]
E = 2pi(42300) (0.293)
E = 77844.79 N/C
F = ma = qE
a = qE/m = (1.6 x 10^-19 x 77844.79 ) / (9.109 x 10^-31)
a = 1.367 x 10^16 m/s^2
b) E = 2pi ( 4.70 x 10^-6 x 9 x 10^9 ) [ 1 - 0.01R / sqrt(0.01R^2 + R^2) ]
E = 2pi(42300) (0.99)
E = 263121.08 N/C
F = ma = qE
a = qE/m = (1.6 x 10^-19 x 263121.8 ) / (9.109 x 10^-31)
a = 4.631 x 10^16 m/s^2
c) E = 2pi ( 4.70 x 10^-6 x 9 x 10^9 ) [ 1 - 0.001R / sqrt(0.001R^2 + R^2) ]
E = 2pi(42300) (0.99)
E = 263120.95 N/C
F = ma = qE
a = qE/m = (1.6 x 10^-19 x 263120.95 ) / (9.109 x 10^-31)
a = 4.622 x 10^16 m/s^2
d) it changes more from b to C.
as it comes closer, field doesn't . (very very small changes)
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