The size of the electric field force in A = 150 N/C and in B = 100 N/C. What's the size of the charges Q1 and Q2? Thank you!

We know that Electric field #vecE# is defined as the electric force produced at a distance #R#, per unit charge. The direction of the electric field is the direction of the force charge #Q# would exert on a positive test charge.

#vecE=(kQ)/R^2#
where for air #k~~9.0 xx 10^9 Nm^2C^-2#

Let direction #Q_1->Q_2# be positive #x#-axis.

Both Charges are shown as positive. Setting up the equation for electric field at #A# using SI units

#vecE_A=[(kQ_1)/(0.1)^2-(kQ_2)/(0.2)^2]hati=150hat iNC^-1#

Multiplying both sides with #(0.2)^2# to remove fraction from denominators and equating magnitudes we get

#(0.2)^2[(kQ_1)/(0.1)^2-(kQ_2)/(0.2)^2=150]#
#=>k(4Q_1-Q_2)=6#
#=>4Q_1-Q_2=6/k# .......(1)

Similarly for point #B#

#vecE_B=[(kQ_1)/(0.2)^2-(kQ_2)/(0.1)^2]hati=-100hat iNC^-1#

Multiplying both sides with #(0.2)^2# to remove fraction from denominators and equating magnitudes we get

#(0.2)^2[(kQ_1)/(0.2)^2-(kQ_2)/(0.1)^2=-100]#
#=>k(Q_1-4Q_2)=-4#
#=>Q_1-4Q_2=-4/k# .......(2)

Multiplying (2) with #4# and subtracting from (1) we get

#4Q_1-Q_2-4xx(Q_1-4Q_2)=6/k-(4xx(-4/k))#
#=>15Q_2=22/k#
#=>Q_2=22/(15k)#

Inserting value of #k#

#=>Q_2=22/(15xx9xx10^9)=1.63xx10^-10C#, rounded to two decimal places.

from (2)

#Q_1=4Q_2-4/k#

Inserting value of #Q_2# we get

#Q_1=4xx(22/(15k))-4/k#
#=>Q_1=88/(15k)-4/k#
#=>Q_1=28/(15k)#

Inserting value of #k#

#Q_1=28/(15xx9xx10^9)#
#=>Q_1=2.07xx10^-10C#, rounded to two decimal places.