Question

Use Coulomb's law to determine the magnitude o f the electric field at points A and B in (Figure 1) due to the two positive charges (Q = 6.9 μC ) shown. Suppose that a = 5.3 cm . Determine the magnitude of the electric field at A. Determine the angle between the direction of the electric field at point A and the positive x-direction. Determine the magnitude of the electric field at point B. Determine the angle between the direction of the electric field at point B and the positive x-direction.

Answer 1

magnitude of electric field due to a charge Q at a distance d is given by

E=k*Q/d^2

where k=9*10^9 m/F

field due to positive charge will be along the line connecting the charge with the field point and

the direction will be away from the charge.

part 1:

electric field at A:

distance between left charge Q and A=sqrt((2a)^2+a^2)=2.236*a=11.851 cm

magnitude of electric field=4.422*10^6 N/C

direction will be away from Q and towards A.

angle with +ve x axis=arctan(a/(2a))=26.56 degrees

hence x component of the field=4.422*10^6*cos(26.56)=3.955*10^6 N/C
y component =4.422*10^6*sin(26.56)=1.977*10^6 N/C

field due to right side Q:

distance between right charge Q and A=sqrt((2a)^2+a^2)=2.236*a=11.851 cm

magnitude of electric field=4.422*10^6 N/C

direction will be away from Q and towards A.

angle with -ve x axis=arctan(a/(2a))=26.56 degrees

hence x component of the field=-4.422*10^6*cos(26.56)=-3.955*10^6 N/C
y component =4.422*10^6*sin(26.56)=1.977*10^6 N/C

so the x component of both the fields will cancel each other out.

net electric field will be along +ve y axis.

direction with positive x direction=90 degrees

magntitude=3.954*10^6 N/C (ans)

part 2:

electric field at point B:

distance between left charge Q and B=sqrt((a)^2+a^2)=1.414*a=7.494 cm

magnitude of electric field=11.057*10^6 N/C

direction will be away from Q and towards B.

angle with +ve x axis=arctan(a/(a))=45 degrees

hence x component of the field=11.057*10^6*cos(45)=7.818*10^6 N/C
y component =11.057*10^6*sin(45)=7.818*10^6 N/C

electric field due to right side charge Q:

distance between right charge Q and B=sqrt((3a)^2+a^2)=3.162*a=16.7586 cm

magnitude of electric field=2.211*10^6 N/C

direction will be away from Q and towards B.

angle with -ve x axis=arctan(a/(3a))=18.435 degrees

hence x component of the field=-2.211*10^6*cos(18.435)=-2.0975*10^6 N/C
y component =2.211*10^6*sin(18.435)=0.7*10^6 N/C

net x component of electric field=7.818*10^6-2.0975*10^6=5.7205*10^6 N/C

net y component of electric field=7.818*10^6 + 0.7*10^6=8.518*10^6 N/C

hence net magnitude=sqrt(5.7205^2+8.518^2)*10^6 N/C=10.26*10^6 N/C

angle with +ve x axis=arctan(y component/x component)=56.115 degree