Two charges of (-6x10^-9)C and (-2.2x10^-9)C are separated by a distance of 43 c

Question

Two charges of (-6x10^-9)C and (-2.2x10^-9)C are separated by a distance of 43 cm. Find the equilibrium position for a third charge of (+1.4x10^-8)C by identifying its distance from the ?rst charge q1.Answer in units of cm.

I would like for you guys to rework it and see if you got the same answer. I got 16.2 cm. I assumed that q3 must be in the middle. Thefore the force from q1 acting on q3 must equal the force of q2 acting on q3. I think my logic is right but I don't want to lose points due to crappy algebra. Please show work.

it actually may be 26.7 because I used (-2.2x10^-9) as my first charge and the question uses the other as q1. None the less do as the question asks and tell me what you guys get

Explanation / Answer

Charges q = -6 x 10 -9 C             q ' = -2.2 x 10 -9 C Distance r = 43 cm                 = 0.43 m Third charge Q = 1.4 x 10 -8 C Let the equilibrium position point from charge q be x. If the force on Q due to q = force on Q due to q' then Q be in equilibrium position                       KQq /x 2 = KQq' /(r-x) 2                               q /x 2 = q' /(r-x) 2                            (r-x) / x = [q' /q]                                         = 0.6055                                   r-x = 0.6055 x                                       r = 1.6055 x                                       x = r/1.6055                                          = 26.78 cm                            (r-x) / x = [q' /q]                                         = 0.6055                                   r-x = 0.6055 x                                       r = 1.6055 x                                       x = r/1.6055                                          = 26.78 cm

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