(5%) Problem 17: A 3-D printer lays down a semicircular arc of positively charge
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(5%) Problem 17: A 3-D printer lays down a semicircular arc of positively charged plastic with a radius R = 3.3 cm, and a linear charge density of +1.7 C/m. After the printer has finished the arc, the stylus moves to the center of the arc as shown. The minute segment of the plastic arc highlighted in the diagram subtends an angle de. Note the measurement of the angle shown in the figure d Stylus tip Otheexpertta.com 14% Part (a) Input a symbolic expression for the charge dq on the segment of charge of size de in terms of given parameters. R de dq Correct! 14% Part (b) Input a symbolic expression for the electric force vector, exerted by the minute segment of plastic subtending the arc d on an electron (charge -e), within the stylus at the center of the arc. Express your answer in terms of given parameters, fundamental constants, and the unit vectors i and j in the specified coordinate system. dF = ( ( k de )R2 ) ( cos(e) 1-sin(8) j ) X Incorrect! 14% Part (c) Find the indefinite integral of the x-component Fx from part (b), but do not evaluate the limits Grade Summary Deductions Potential F-|( (-kA)/R ) sin()- 10% 90%Explanation / Answer
given
R = 3.3 cm = 0.033 m
lambda = 1.7 uC/m
a. dq = lambda*dl
dl = R*d(theta)
hence
dq = lambda*R*d(theta)
b. electric field at point P is
dE = k*dq(-cos(theta)i + sin(theta)j)/R^2
hence
dF = -e*dE
dF = ke*lambda*(cos(tehta)i - sin(theta)j)*d(tehta)/R
c. integrating
F = ke*lambda*(sin(theta)i + cos(theta)j)/R + C
d. upper limit, theta = pi/2
lower limit, theta = -pi/2
e. F = ke*lambda*(2i)/R = 2ke*lambd i/R
f. F = 8.98*10^9*1.7*10^-6*2 *1.6*10^-19/ 0.033 = 1.48*10^-13 N
g. the force is along +ve x axis
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