hint Determine the and components of the acceleration of the electron. This is a
ID: 1652307 • Letter: H
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hint
Determine the and components of the acceleration of the electron. This is a problem involving motion with constant acceleration (very similar to projectile motion). See the Examples in sec. 23.7; also see p. 81 - 84 in our textbook (on Webassign) to review motion in a plane with constant acceleration.
answer(a) 37.7 N/C, away from center. (b) 41.6 N/C, away from center. (c) 28.6 N/C, toward center. (d) 33.8 N/C, toward center.
8. A sphere is composed of two regions of charge (diagram at right): an inner spherical core, of radius R,-2 m and positive uniform charge density = 10-9 Cm'; and a spherical shell surrounding the core, of outer radius R2 3 m and negative uniform charge density --10-9 Chn. Find the magnitude and direction of the electric field at the following distances from the center: (a) r = 1 m; (b) r = 2.2 m; (c) r-2.8 m; (d) r-3.5 m. Note: to receive credit, you must show a clearly labeled diagram, and the Gaussian surface you are using for each part of this problem; label its radius.Explanation / Answer
A] For this case, the Gaussian surface would be a sphere of radius 1m i.e. half of the radius odf inner core.
Total flux 4 pi r ^2 * E = [4/3]*pi r^3 * rho / e0
E = rho r /3 e0
= 1*10^-9/[3*8.85e-12]
= 37.7 N/C away from the center
b] For this case, the Gaussian surface would be a sphere of radius 2.2m i.e. more than R1 but less than R2.
Total flux 4 pi r ^2 * E = charge enclosed / e0 = = [volume1*rho1 + volume2*rho2]/e0
4 pi *2.2 ^2 * E = [ 4/3*pi 2^3*10^-9 + 4/3*pi*(2.2^3-2^3)*-10^-9 ] / e0
E = [ 1/3*2^3*10^-9/2.2^2 + 1/3 *(2.2^3-2^3)*-10^-9/2.2^2]/8.85e-12
= 41.6 N/C away from the center
c] similar to last part, in place of 2.2m, have r = 2.8 m, so Gaussian surface will be larger,
E = [ 1/3*2^3*10^-9/2.8^2 + 1/3 *(2.8^3-2^3)*-10^-9/2.8^2]/8.85e-12
= -28.6 N/C away from the center
or 28.6 N/C towards the center.
d] In this case the gaussian surface will be sphere of radius 3.5m, which is more than R2 also.
E.4pi r = charge enclosed / e0 = [volume1*rho1 + volume2*rho2]/e0
4 pi *3.5 ^2 * E = [ 4/3*pi 2^3*10^-9 + 4/3*pi*(3^3-2^3)*-10^-9 ] / e0
E = [ 1/3*2^3*10^-9/3.5^2 + 1/3 *(3^3-2^3)*-10^-9/3.5^2]/8.85e-12
= -33.8 N/C away from the center
or 33.8 N/C towards center
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