Two point particles with charges q1 30 nC and q2 30 nC are

Two point particles (with charges q_1 = +30 nC and q_2 = -30 nC) are placed on the x axis 20 cm apart, with the negative particle to the left of the positive particle. A third point particle (with charge q_3 = +20 nC) is placed on the x axis halfway between the first two particles. What is the force (magnitude and direction) on the third particle? Two point particles (with charges q_1 = +120 nC and q_2 = +30 nC) are placed on the x axis 20 cm apart, with the latter to the left of the first. A third particle (with q3 = -60 nC) is placed on the x axis halfway between the first two particles. What is the force (magnitude and direction) on the third particle? Compute the ratio between the magnitudes of the gravitational force and the electrostatic force between two electrons separated by 1.0 m. How does this ratio change if we bring the electrons closer?

Solution

        -x axis <-------       (q1)    -30nC ------------------- (q2) +20nC    ----------------- (q3) +30nC q2     -----> +x axis

                                                          10cm                                        10cm

for simplicity let us consider the charges q1 , q2 and q3 labelled and seperated as shown above

Let Fr be the Force of repulsion on q2 by q3 , Fr = k q2 * q3 / r^2

Fr = + ( 8.98*10^ 9 * 20* 10^-9C * 30*10^-9 C) / 0.1^2

                                                                      = 5.3688*10^-4 N towards negative x axis

Let Fa be the Force of attraction on q2 by q1 , Fa = k q1 * q2 / r^2

Fa = + ( 8.98*10^ 9 * -30* 10^-9C * 20*10^-9 C) / 0.1^2

                                                                      = 5.3688*10^-4 N towards negative x direction

the net force on q2 due to q3 and q1 Fnet = Fr +Fa

                                                                    = 10.776 * 10^-4 N along negative x direction

(E1B.8)

-x axis <-------       (q1) +30nC <-------------------> (q2) -60nC <-----------------> (q3) +120nC q2     -----> +x axis

                                                    10cm                                          10cm

Let F23 be the Force of attraction on q2 due to q3 , F23 = k q2 * q3 / r^2

F23 = + ( 8.98*10^ 9 * -60* 10^-9C * 120*10^-9 C) / 0.1^2

                                                                     = 6.46 *10^-3 N along positive x axis

Let F21 be the Force of attraction on q2 by q1 , Fa = k q1 * q2 / r^2

Fa = ( 8.98*10^ 9 * -60* 10^-9C * +30*10^-9 C) / 0.1^2

                                                                      = 1.616*10^-3 N along negative x direction

the net force on Fnet on q2 due to q3 and q1 Fnet = F23 +F21

                                                                    = -1.616*10^-3 N+ 6.46 *10^-3 N

                                                          Fnet on q2 = 4.844 * 10^-3 N along postive x direction.