Problem 2: Two point charges of +1x10-4 and -1x10-4 coulombs are placed at a distance of 40cm from each other. A charge +6x10-5coulombs is placed mid way between them. What is the magnitude and direction of force on it?
Solution:
Step: 1 Overview
According to coulomb's law like charges (charges having similiar
signs either +ve or -ve) repel each other and unlike charges attract
each other. And force of that attraction or repulsion will be:
Force on Q3 due to Q1 (F1):
Note! +ve and -ve signs are only used to determine the direction of
force, that is why they are not used in the above calculation.
F=Kq1q2/r2 where value of constant K is 9x109Nm2/C2.
Step: 2 Calculation
Now naming the charges, suppose first charge (+1x10-4) is Q1, second (-1x10-4) is Q2 and third (+6x10-5) is Q3.
Since distance between Q1 and Q2 is 40cm, and Q3 is placed in a midway between Q1 and Q2, so distance between Q1 and Q3 will be 20cm (or 0.2m).Force on Q3 due to Q2 (F2):
F=Kq1q2/r2
F1=9x109(1x10-4)(6x10-5)/(0.2)2
F1=1350N.
Since both charges have +ve sign, it means both repel each other, so direction of force on Q3 due to Q1 will be along Q2.
Similiarly distance between Q2 and Q3 will also be 20cm (or 0.2m).Net Force on Q3:
F2=9x109(1x10-4)/(0.2)2
F2=1350N.
Since both Q3 and Q2 have inverse charges, it means they attract each other, so in this case force on Q3 will also be towards Q2.
F=F1+F2
F=1350N+1350N
F=2700N towards Q2(-ve charge) (Ans).
