Please answer all parts to the question and show all work ! THANK YOU !!!!! Thre

Question

Please answer all parts to the question and show all work ! THANK YOU !!!!!

Three point charges, A = 1.75 mu C, B = 7.25 mu C, and C = -4.60 mu C, are located at the corners of an equilateral triangle as in the figure above. Find the magnitude and direction of the electric field at the position of the 1.75 mu C charge. (b) How would the electric field at that point be affected if the charge there were doubled? The magnitude of the field would be halved. The field would be unchanged. The magnitude of the field would double. The magnitude of the field would quadruple. Would the magnitude of the electric force be affected? Yes No

Explanation / Answer

part a:

field due to 7.25 uC charge:

as it is a positive charge, direction will be away from 7 uC and towards origin.

unit vector along this direction=(-cos(60),-sin(60))=(-0.5,-0.866)

distance=L=0.5 m

then electric field magnitude=9*10^9*7.25*10^(-6)/0.5^2=261000 N/C

electric field in vector notation=E1=261000*(-0.5,-0.866) N/C

field due to -4.6 uC charge:

as charge is negative, field is directed towards the -4.6 uC charge.

unit vector along this direction=(1,0)

distance=L=0.5 m

field magnitude=9*10^9*4.6*10^(-6)/0.5^2=165600 N/C

then in vector notation,field due to -4.6 uC=E2=165600*(1,0) N/C

so total field at origin=E1+E2

=(35100, -226032.63) N/C

magnitude=sqrt(35100^2+226032.63^2)=228741.688 N/C

angle with +ve x axis=arctan(-226032.63/35100)=-81.173 degrees

part b:

as field does not depend upon charge at that point, it will remain unchanged.

as magnitude of electric force=agnitude of electric field*magntiude of charge

force will be affected.

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