Three identical point charges, each of mass 40 g and charge + q , hang from thre

Question

 

Three identical point charges, each of mass 40 g and charge +q, hang from three strings,

as in the figure.

If the lengths of the left and right strings are each 10.3 cm, and each forms an an

with the vertical, determine thegle of 45 degrees value ofq. The acceleration of gravity is 9.8 m/s^2, and the Coulomb constant is 8.98755 × 10^9 N · m^2/C^2.

Answer in units of

C

Three identical point charges, each of mass 40 g and charge +q, hang from three strings, as in the figure. If the lengths of the left and right strings are each 10.3 cm, and each forms an an with the vertical, determine thegle of 45 degrees value ofq. The acceleration of gravity is 9.8 m/s^2, and the Coulomb constant is 8.98755 10^9 N m^2/C^2. Answer in units of mu

Explanation / Answer

Find the distance between the *Further* left charge and the middle charge, because the further charge is actually the charge that matters. So the only variable you should have in these equations should be r1=(36.0cm)sin(45) = ... Fmid = ke [(q^2)/(r1^2)] r2 = 2(36.0cm*sin(45)) = ... Ffar= ke [(q^2)/(r2^2)] T(as in tension in the string, which is the only force acting upon the mass) is used in this equation. Tsin(45) - mg = 0 Tsin(45) - (.400kg)(9.80m/s^2) = 0 T= mg/sin(45) = ...N It should be small. like ranging 3-8N. my strings were 22.0cm and I got 4.606N. So since you have all of the forces on the string, you add them up to get a total net force. -Fmid - Ffar + Tcos(45) = 0 substitute these Forces with the equations and distances: -ke* [(q^2)/(r1^2)] - ke* [(q^2)/(r2^2)] + Tcos(45) = 0 through simplifying you then get: ke*q^2[(1/r1^2)+(1/r2^2)] = Tcos(45) (1/r1^2)+(1/r2^2) = ... [make sure you add these two values first, they should be a bit big] You should get: q^2 = [(Tcos(45))/(ke*(1/r1^2)+(1/r2^2))] through calculations you'll end up with a value with 10^-12 , but you have to square root both sides of the equations and you'll have your answer! it should be with units: .... 10^-6C

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