Two small spheres, each carrying a net positive charge, are separated by 0.400m

Question

Two small spheres, each carrying a net positive charge, are separated by 0.400m You have been asked to the charge on each sphere. You set up a coordinate system with one sphere (charge q 1) 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 . 106 C and an apparatus that can accurately measure the location of this sphere and the net force on it. First you place the third sphere on the x-axis at 0.200 m; you measure the net force on it to be 4.50 N in the +x-direction. Then you move the third sphere to x +0.600 m and measure the net force on it now to be 350N direction. A. Calculate q1 and g2. in the +x- What is the net force (magnitude and direction) on q3 if it is placed on the x- axis at x--0.200 m? B. C. At what value ofr (other than x = {q) could q3 be placed so that the net force on it is zero?

Explanation / Answer

Let the charges be q1=x and q2 = y

for the first setup.

total force on q3 = force due to q2 + force due to q1

so,

Force = k*q1*q3/d1^2 + k*q2*q3/d2^2

or 4.5 = 9*10^9*4*10^-6*x/0.2^2 + 9*10^9*4*10^-6*y/0.2^2

also, for the second setup,

3.5 = 9*10^9*4*10^-6*x/0.6^2 + 9*10^9*4*10^-6*y/0.2^2

solving the above two equaitons,

we get q1 = 1.25 uC and q2 = 2.75 uC

b)force = k*q1*q3/d1^2 + k*q2*q3/d2^2

or F = 9*10^9*4*10^-6*1.25*10^-6/0.2^2 + 9*10^9*4*10^-6*3.75*10^-6/0.6^2

or F = 1.5 N(in the negative x direction)

C)let the distance on the x axis be x. The point will be in the positive x axis since both the charges are positive.so,

kq1q3/d1^2 = kq2q3/(0.4-d1)^2

or 1.25/(x^2) = 3.75/(0.4-x)^2

or x=0.146 m

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