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C www.webassign.net/web/Student/Assignment-Responses/submit?dep 15711383 C+ Ask Your Teacher -12 points KatzPSE11 24. P031.MI My Notes What is the electric field at point A in the figure if d 1.50 m, R 0.500 m, q1 18.0 nC, and a2 28.0 nC? Assume the positive x axis points to the right, through the center of the rings Express your answer in vector form.) N/C d Need Help? Read Master 5. O -/2 points KatzPSEf1 24.P021 My Notes Ask Your Teacher often we have distributions of charge for which integrating to find the electric field may not be possible in practice. In such cases, we may be able to get a good approximate solution by dividing the distribution into small but finite particles and taking the vector sum of the contributions of each. To see how this might work, consider a very thin rod of length L 26 cm with uniform linear charge density 62.0 nC/m. Estimate the magnitude of the electric field at a point Pa distance d cm from the end of the rod by dividing it into n segments of equal length as illustrated in the figure below for n 4. Treat each segment as a particle whose distance from pcint P is measured from its center Find estimates of Epfor n 1, 2, 4, and 8 segments. N/C n 2, E N/C N/C 4, E n 8, ED N/C 5:26 PM Ask rne /30/2017

Explanation / Answer

here,

d = 1.5 m , R = 0.5 m

q1 = 18 * 10^-9 C

q2 = - 28 * 10^-9 C

electric feild due to charged ring , E = K * Q * z /( z^2 + R^2)^1.5

then the electric feild at A , E = K * q1 * (d/2) /( (d/2)^2 + R^2)^1.5 + K * q2 * (d/2) /( (d/2)^2 + R^2)^1.5 i

E = (9 * 10^9 * 10^-9 * (1.5/2) /( (1.5/2)^2 + 0.5^2)^1.5 )* (18 + 28 ) i

E = 424 N/C i

the electric feild at point P due to the charged rings is 424 N/C i

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