A uniformly charged circular are ABof radis 30f3chargeoare 100 points (r v)plaze
ID: 1574178 • Letter: A
Question
A uniformly charged circular are ABof radis 30f3chargeoare 100 points (r v)plaze it s located is the second quadrast. The total charge on the are is Q > 0 and .1 R. dblocated in the moodqrtrait Thr total 2 charges are arranged in the (r. y) plane reynolds ür65763) HW#IC21 Electric Charge and Fildstom . (37m6) 3 What is the direction of the electric field distribation 1. Along the positive y-axis 2. In quadrant 3. Along the positive r-axis 4. Along the negative y-axis . In quadrast I 6. Along the negative a-axis vector £ at the origin, due to the charge 9, F-ec 10, F = 015 10.0 points Two charges are located in the (z) plane a shown. The Selds produced by these charges are obeerved at a point p with coordinates (0.0 -8.70 In quadrant IV 8. In quadrant Find the a-component of the electric field at p. The value of the Conlomb constant is 013 (part 2 of 3) 10.0 points the r-component of the electric field 898755 x 10N- t the origin due to thgtAnswer in nits of N/C foe a charge of1.7 Cand a raduso 0.64 m. The value of the Coulomb coustant is 8.98755 × 109 N·/ Answer in units of N/C 014 (part 3 of 3) 10.0 points If E-C at O, find the magnitude of the fall force F (not just the r-component) on an electron at this point 2, F =Explanation / Answer
the electric field for positive charge is directed away from it . in this case the electric field filed generated by each del q is going to be in quadrant , so the total electric field will be in the same quadrant.
option (7) is correct answer
dle E = k del q/ R^2
the charge del q per unit arc length del s is
lamda = Q/ ( pi/2) R
= 2Q/ pi R
del q= lamda del s
= lamda R del theta
= 2Q/ pi R * R del theta)
= 2Q/ pi R * del theta
E_x = kQ/ R^2 * 2 del theta/ pi * cos theta
integrate above equation between the limit 0 to pi/2
E = integral 0 to pi/2 kQ/ R^2 * 2 del theta/ pi * cos theta
= 2kQ/ pi R^2
= 2 (8.98755 * 10^9) ( 1.7 * 10^-6)/ pi ( 0.64)^2
=23.74 * 10^3 N/C
(14)
F = sqrt 2 eC = sqrt 2 e Ex
since Ex = C
so option (1) is correct answer
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