Pictured on the right are three point charges Q1 = 15.4 ?C, Q2 = -30.6 ?C, and Q

Question

Pictured on the right are three point charges Q1 = 15.4 ?C, Q2 = -30.6 ?C, and Q3 = 67.3 ?C arranged according to the figure on the right. A fourth point charge is located at point A with a charge of QA = 13.5 ?C. Calculate the magnitude of the net force on the charge at point A.

Pictured on the right are three point charges Q1 15.4 AC, Q2 -30.6 UC, and Q3 67.3 AMC arranged according to the figure on the right. A fourth point charge is located at point A with a charge of QAE 13.5 PC. Calculate the magnitude of the net force on the charge at point A. Number 12 1.54 x 10 15.4 13.5 UC AT 40.1 cm 30.6 AMC Q2 Q, 67.3 pc 40.1 cm l

Explanation / Answer

let line joining Q2 and Q3 be x axis and Q2 and Q1 be y axis.

Q2 be at origin.

force due to Q1 on Qa:

as both charges are positive in nature, force will be repulsive.

direction of force is along +ve x axis.

force magnitude=9*10^9*15.4*10^(-6)*13.5*10^(-6)/0.401^2=11.636 N

in vector notation, force =F1=(11.636 N) i

force due to Q2 on Qa:

as the charges are of opposite sign, force is attractive in nature and hence directed from Qa towards Q2.

i.e. from (0.401, 0.401) to (0,0)

vector along this direction=(0,0)-(0.401,0.401)=(-0.401,-0.401)

distance=sqrt(2)*0.401=0.5671 m

unit vector along this direction=(-0.401,-0.401)/0.5671=(-0.707,-0.707)

force magnitude=9*10^9*13.5*10^(-6)*30.6*10^(-6)/0.5671^2=11.561 N

hence in vector notation, force =F2=-8.1749 i - 8.1749 j   

force due to Q3 on Qa:

both charges are of positive sign and force is repulsive.

force direction is along +ve y axis.

force magnitude=9*10^9*67.3*10^(-6)*13.5*10^(-6)/0.401^2=50.851 N

in vector notation, force=F3=50.851 j N

total force =F1+F2+F3

=3.4611 i + 42.676 j

force magnitude=sqrt(3.4611^2+42.676^2)=42.816 N

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