Question

WILL THUMBS UP CORRECT SOLUTIONS!

Two objects cany initial charges that are q and 42, respectively, where lzll. They ae lceted 0.340 m apart and bellave ike pcint cherges, They attract each other with a force that hes a magnitude of 1.30 N. The objects are then brought nto contact, so the net charge is shared equaly. and then they are retumed to their initial position. Now it ifound that the objects repel ona another with a force whosc magnitude i equal to the magritude of the initial attrective forca. What are the magnitudes of the iritial chargas nn tha chjacta? Addiional Matarials ф Section 18.6 -1 point@ CJ10 13P049 My Notes An clectron is released fom rcst at the negative plate of ง paralla platc capacitor. The charge par unit area on cach plate is ơ-1.61 x 10-7 C/m2, and the Platc sc aration 1s 1.57 10 2 m. How fast is thc electron mcvng Just bcforc it reachcs te Postic plate? m/s Addirional Materials ■ Section 18.8 -3 points CJ10 18P056 My Notes A surface completely surrounds a 12.7 x 10 C chrge, Find the electric nux through this surface when the surlace is a sphere with a radius of 0.90 m, a sphere with a radus of 0.45 m, and a cube with edges that are 0.45 m lng (a) a sphare with a radius of 0.90 m N- mic sphere with radius ot 0.45 m N mic a cbe with edges that are 0.45 m long N m2,c

Answer 1

given , initial charge on object 1 = q1
initial charge on object 2 = q2
r = distance between both charges = 0.340 m
Force between both charges = 1.30 N
Now electrostatic force is given by:

F = k*q1*q2/r^2
q1*q2 = F*r^2/k
Using given values
q1*q2 = 1.30*0.340^2/(9*10^9)
q1*q2 = -1.67*10^-11 C
[See that q1*q2 will be negative as force is attractive which means one of
them is negative and other one is positive].
Now when both spheres are brought into contact, after that charge will be
equally distributed. Now charge on each sphere will be Q, where

Q = (q1 + q2)/2
Now when returned to distance r = 46.9 cm, force will be repulsive because
both charge will have same sign either positive or negative, So
F1 = k*Q*Q/r^2
Q^2 = F1*r^2/k
Q = sqrt (1.30*0.340^2/(9*10^9))
Q = 4.09*10^-6 C
So,
(q1 + q2)/2 = 4.09*10^-6 C
q1 + q2 = 8.18*10^-6 C
We know that
q1*q2= -1.67*10^-11 C
q1*(8.18*10^-6 - q1) = -1.67*10^-11
Solving above equation
q1 = -1.69*10^-6 C or q1 = 9.87*10^-6 C

given, |q2| > |q1|

So, q1 = -1.69*10^-6 C

and q2 = 8.18*10^-6 - (-1.69*10^-6)

q2 = 9.87*10^-6 C

So, magnitudes is given by,

|q1| =1.69*10^-6 C = 1.69 $\mu$C

|q2| = 9.87*10^-6 C = 9.87 $\mu$C
Please Upvote. Let me know if you have any doubt.