IV Two small spheres, each carrying a net positive charge, are separated by 0.40

Question

IV Two small spheres, each carrying a net positive charge, are separated by 0.400m You have been asked to perform measurements that will allow you to determine the charge on each sphere. You set up a coordinate system with one sphere (charge ql) at the origin and the other sphere (charge g2) at x 0.400 m Available to you are a third sphere with net charge q3-4.00 10-6 C and an apparatus that can accurately on it. First you place the third sphere on the x-axis at x- net force on it to be 4.50 N in the tx-direction. Then you move the third sphere to x = +0.600 mn and measure the net force on it now to be 3.50 N in the +x- direction measure the location of this sphere and the net force 0.200 m; you measure the A. Calculate ql and q2. B. What is the net force (magnitude and direction) on q3 if it is placed on the x- axis at x -0.200 m? C. At what value ofx (other than x = {q) could q3 be placed so that the net force on it is zero?

Explanation / Answer

let,

charges,

q1 at x1=0

q2 at x2=0.4 m

distance between q1 and q2 is r12=0.4 m

q3=4*10^-6 C

if charge q3 is at x3=0.2

Fnet=F13-F23

  

4.5=K*q1*q3/r13^2 - K*q2*q3/r23^2

4.5=k*q3*(q1/r13^2 -q2/r23^2)

4.5=9*10^9*4*10^-6*(q1/0.2^2 - q2/0.2^2)

4.5=9*10^9*4*10^-6/0.2^2*(q1 - q2)

===> q1-q2=5*10^-6 ----(1)

if charge q3 is at x3=0.6

Fnet=F13+F23

  

3.5=K*q1*q3/r13^2 + K*q2*q3/r23^2

3.5=k*q3*(q1/r13^2 +q2/r23^2)

3.5=9*10^9*4*10^-6*(q1/0.6^2 + q2/0.2^2)

3.5=9*10^9*4*10^-6*(q1/0.36 + q2/0.04)

3.5=9*10^9*4*10^-6*(100*q1/36 + 100*q2/4)

3.5=900*10^9*4*10^-6*(q1/36 + q2/4)

3.5=900*10^9*4*10^-6/(36)*(q1 + 9*q2)

===> q1+9*q2=3.5*10^-5 ---------(2)

from equation (1) and (2)

q1=8*10^-6 C or 8 uC

q2=3.03*10^-6 C or 3.03 uC

b)

if q3 is at x3=-0.2 m

Fnet=F13+F23

=k*q1*q3/r13^2 + k*q2*q3/r23^2

=9*10^9*8*10^-6*4*10^-6/0.2^2 + 9*10^9*3.03*10^-6*4*10^-6/0.6^2

=7.5 N

c)

let x at which net force is zero

Fnet=F13+F23

0=k*q1*q3/x^2 +k*q2*q3/(0.4-x)^2

==>

0=q1/x^2 + q2/(0.4-x)^2

0=8*10^-6/x^2 + 3.03*10^-6/(0.4-x)^2

0=8/x^2 + 3.03/(0.4-x)^2

-8/x^2 = 3.03/(0.4-x)^2

-((0.4-x)/x)^2=3.03/8

(0.4-x)/x=0.615

===> x=0.25 m

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