Assignment I Begin Date: 120 2018 12:00:00 AM- Due Date: 2/3/2018 11:59:00 PM En

Question

Assignment I Begin Date: 120 2018 12:00:00 AM- Due Date: 2/3/2018 11:59:00 PM End Date: 2/5/2018 11:59:00 PM (6%) Proble|n 8: A positive charge of magnitude Q,-9.5 nC is located at the origin. A negative charge 02-4.5 nC is located on the positive x-axis at x 4.5 cm from the origin. The point P is located y 18 cm above charge Qz. ©theexpert ta.com 17% Part (a) Calculate the x-component of the electric field at point P due to charge Q1 Write your answer in units of Nic. Grade Summary 100% sin) cotan sinO atanacotan sinb cosh cOs) Attempts remaining (2% per templ) delailed view acos) tanhcotanh O Degrees O Radians T gihe up! Hinls: 3% deluclion per hinl. Hinls remaining:4 Feedback: 0% deducionper leedback. 17% Part (b) Calculate the y component of the electric field at point P due to charge )1 Write your answer in units of N C. 1796 Part (c) Calculate the y component of the electric field at point P due to the Charge write your answer in units of NK. 17% Part (d) Calculate the y-component of the electric field at point P due to both charges. Write your answer n units of No. 17% Part (e) Calculate the magnitude ofthe electric field at point P due to both charges. Write your answer in units ofNC 17% Part (f) Calculate the angle in degrees of the electric field nt point P relative to the positive x axis

Explanation / Answer

part(a)

E1x = k*Q1*x/(x^2+y^2)^(3/2)

E1x = 9*10^9*9.5*10^-9*0.045/(0.045^2+0.18^2)^(3/2)

E1x = 602.4 N/C

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part(b)

E1y = k*Q1*y/(x^2+y^2)^(3/2)

E1y = 9*10^9*9.5*10^-9*0.18/(0.045^2+0.18^2)^(3/2)

E1y = 2409.5 N/C

====================

part(c)

E2y = -k*Q2/y^2

E2y = -9*10^9*4.5*10^-9/0.18^2

E2y = -1250 N/C

======================

part(d)

Ey = E1y + E2y = 1159.5 N/C

==============================

part(e)

E =sqrt(Ex1^2 + Ey^2)

E = sqrt(602.4^2+1159.5^2)

E = 1307 N/C

======================

part(f)

angle = tan^-1(Ey/Ex)

angle = 62.5

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