Figure a shows charged particles 1 and 2 that are fixed in p

Figure (a) shows charged particles 1 and 2 that are fixed in place on an x axis. Particle 1 has a charge with a magnitude of |q_1| = 22e. Particle 3 of charge q_3 = +22e is initially on the x axis near particle 2.Then particle 3 is gradually moved in the positive direction of the x axis. As a result, the magnitude of the net electrostatic force on particle 2 due to particles 1 and 3 changes. Figure (b) gives the x component of that net force as a function of the position x of particle 3. The scale of the x axis is set by x_s = 2.60 m. The plot has an asymptote of F_2,net = 0.9613 times 10^-25 N as x - infinity. As a multiple of e and including the sign, what is the charge q_2 of particle 2?

Solution

The force of charge 1 on particle 2 is

                                                F12 = k q1 *q2 / r12 ^2     here r12 is the distance between charges q1 and q2

the force of charge 3 on charge 2 is

                                                 F32 = k q1 *q2 / r32 ^2 ; here r32 is the distance between charges q3 and q2

from the net force plot on q2 , we can clearly notice that at one point the net force on q2 is 0 for the position of q3 where r32 = 2 times 2.60m = 5.2 m

so when q3 is at 5.2 m from q2 the net force on q2 is 0 ,

i.e the force due to q1 and q3 on q2 are equal and opposite is 0 => F12 = F32   when q3 stays at 5.2 m on x axis

                             F12 = k * q1 *q2 / r12 ^2 = - k * q3 *q2 / r32 ^2 = 0

                                     = q1 / r12^2 = q3 / r32^2 = q3 / 5.2 m

                                         r12^2 = (q1* 5.2 ^2 / q3 ) =

                                             r12 = 5.2 m

when the position of q3 is infinity , then the only force on q2 is due to q1 ; from the plot we can see that the force is along postive direction and asymptote Fnet = 0.9613 * 10^-25N

when q3 is at infinet distance the only force on q2 is due to q1

                                k q1 * q2 / 5.2 ^2 = Fnet = 0.9613 * 10^-25N

                                              q2 = 0.9613 * 10^-25N *5.2^2 / k * q1) = 5133.151 e

and the charge on q1 is postive q2 is positive , this is due to because of the direction of asymtotic force which is in postive x direction which concludes that q1 and q2 arerepelling each other and carry same charge

when q3 brought near q2 the net force on q2 is along negative direction which indicates q3 is repelling q2 which mean q3 and q2 have same sign and is positive.