HW i Begin Date: 3/27/2017 12:00:00 AM Due Date: 46/2017 11:59:00 PM End Date: 6
ID: 1603246 • Letter: H
Question
HW i Begin Date: 3/27/2017 12:00:00 AM Due Date: 46/2017 11:59:00 PM End Date: 610/2011 39 00 PM 33%) Problem 2: A positiv e charge of magnitude Qi S5 nc is located at the origin A negative charge Q2--8.5 nc is located on the positive x-axis atx 125 an from the origin. The point P is located y ssam above charge Q2. ent Status here for ed view otheexpertta.com a Status 17% Part (a) Calculate the x-component off the electric field at point P due to charge ei Write your answer in units of NIC. Completed Grade Sumamary Completed Potential 100% Attempts remaining 415 6 atan0 acotan0 sinh0 detailed view cosh0 tanh0 e Degrees Radians Feedback deduction per feedback. Hines: 35 deduction per hint. Hints rmaaiaisg 17% Part b) Calculate the y omponent of the electmc field at poantPdue to charge Write your answer in units ofNc lii 17% Part (e) calculate the c write your answer in units ofNc y-component of the electric field at pomt P due to the charge Q2 a 17% Part (do component of the electric field at point P due to both charges. Write your answer in units of Nc lil 17% Part (e) Calculate the write your answer in units ofNc Calculate the magnitude of the electric field at point P due to both charges A 17% Part calculate the angle in degrees of the electric field at point Prelative to the positve x-axis.Explanation / Answer
a)Ex=kQ/r2=[k*5.5*10^-9/(12.52+5.52)](12.5)
Ex=3.318N/C
b)
Ey=kQ/r2=[k*5.5*10^-9/(12.52+5.52)](5.5)
EY=1.46N/C
C)
Ey=kQ/r2=[k*8.5*10^-9/(5.5)2]
Ey=2.53N/C
d)
Ey=kQ/r2=[k*5.5*10^-9/(12.52+5.52)](5.5)
EY=1.46N/C
C)Ey due to Q1
Ey=kQ/r2=[k*8.5*10^-9/(5.5)2]
Ey=2.53N/C
Ey due to Q2
Ey=kQ/r2=[k*5.5*10^-9/(12.52+5.52)](5.5)
EY=1.46N/C
EY,net=2.53+1.46=3.99 N/C
e)E=[(3.318)2+(3.99)2]1/2=26.93N/C
f)angle=tan-1(Ey/Ex)=tan-1(3.99/3.318)=50.25 degree
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