Two point charges are placed on the x-axis as follows: charge q1= 3.96 is located at x=0.202 , and charge q2= 4.99 is at x=-0.302 .What is the magnitude of the total force exerted by these two charges on a negative point charge q3= -5.95 that is placed at the origin?

Question

Two point charges are placed on the x-axis as follows: charge  q1= 3.96  is located at   x=0.202 , and charge  q2= 4.99  is at   x=-0.302 .What is the magnitude of the total force exerted by these two charges on a negative point charge  q3= -5.95  that is placed at the origin?

egenriether · Answer

The force at the origin will be the sum of each individual one.  One force will be given by Kc*(Q1*Q2)/R^2.  Where Kc = 1/(4*pi*eps_0)  Note the Q's carry the sign.  Since these are all on the x axis, the vector components aren't hard but the direction matters.  A negative force means they are attracted, as these will be.  Get each answer and add them up.

anonymous · Answer

the total force on the negative charge placed at the origin=F1+F2
where $$F1=-(3.96*5.95)/(4*3.14*8.854e-12*0.202*0.202)$$

F2=−(4.99∗5.95)/(4∗3.14∗8.854e−12∗0.302∗0.302)

Now F1 and F2 are opposite to each other and we have the three charges in the x-axis so it is straight away to find the difference between the two forces F1 and F2 ie.F1-F2

resultant force is F=5.193-2.928
                          =2.265*10^12 newton   (magnitude)
If we consider the given numericals in the SI units, charge in coulombs, distance in metres. The direction  of the resultant force will be towards q1 and as a result the charge q3 is attracted towards  q1.