Question

Two point charges are placed on the x-axis as follows: charge q1 = 4.05 nC is located at x= 0.195 m , and charge q2 = 5.04 nC is at x= -0.300 m .

Part A

What is the magnitude of the total force exerted by these two charges on a negative point charge q3 = -6.00 nC that is placed at the origin?

F= ? N

Part B

What is the direction of the total force exerted by these two charges on a negative point charge q3 = -6.00 nC that is placed at the origin?

What is the direction of the total force exerted by these two charges on a negative point charge  = -6.00  that is placed at the origin?

A. to the +x direction

B. to the -x direction

C. perpendicular to the x-axis

D. the force is zero

Answer 1

part A:

force on q3 due to q1:

as two charges are of opposite sign, force is attractive in nature and towards positive x axis

force magnitude=k*q1*q3/distance^2

=9*10^9*4.05*10^(-9)*6*10^(-9)/0.195^2

=5.7515*10^(-6) N

force on q3 due to q2:

as two charges of opposite sign, force is attractive in nature.

hence force is towards -ve x axis

force magnitude=k*q2*q3/distance^2

=9*10^9*5.04*10^(-9)*6*10^(-9)/0.3^2

=3.024*10^(-6) N

so total force along +ve x axis

total force magnitude=5.7515*10^(-6)-3.024*10^(-6)

=2.7275*10^(-6) N

part B:

force is along +ve x direction

option A is correct.