Two identical conducting spheres are separated by afixed center to center distan
ID: 1764360 • Letter: T
Question
Two identical conducting spheres are separated by afixed center to center distance of 46.8 cm and have differentcharges. Initially, the spheres attract each other with a force of0.0965 N. The spheres are now connected by a thin, conducting wire.After the wire is removed, the spheres are positively charged andrepel one another with a force of 0.0318 N. Calculate the finalcharge on the spheres. Calculate the initial charge on the spherewhich carried a positive charge. Calculate the initial charge on the sphere whichcarried the negative charge.I think I know how to calculate the final charge, but I can'tfigure out how to solve for the intial charges. I tried vectoraddition and subtraction, but I'm not sure if I'm going about thisthe right way. Any help is appreciated, and I rate everyanswer!
relevant equations: F = kq/r2 k = 8.99 x 109 Nm2/C2 E = kq1q2/r2 Two identical conducting spheres are separated by afixed center to center distance of 46.8 cm and have differentcharges. Initially, the spheres attract each other with a force of0.0965 N. The spheres are now connected by a thin, conducting wire.After the wire is removed, the spheres are positively charged andrepel one another with a force of 0.0318 N. Calculate the finalcharge on the spheres. Calculate the initial charge on the spherewhich carried a positive charge. Calculate the initial charge on the sphere whichcarried the negative charge.
I think I know how to calculate the final charge, but I can'tfigure out how to solve for the intial charges. I tried vectoraddition and subtraction, but I'm not sure if I'm going about thisthe right way. Any help is appreciated, and I rate everyanswer!
relevant equations: F = kq/r2 k = 8.99 x 109 Nm2/C2 E = kq1q2/r2 E = kq1q2/r2
Explanation / Answer
Giventhat the distance between two centers is d = 46.8 cm = 0.468 m Theforce of attraction is F1 = 0.0965 N The force of repulsion is F2 = 0.0318 N -------------------------------------------------------------------------------------------- Let q1 and q2 areinitial charges ,then F1 = k*q1*q2 / d2 q1*q2 = F1*d2 / k =0.0965N*(0.468m)2 / (9*109N.m2/C2) = 5.01*10-12 C2 When two charges connected by conducting wire then the each chargeis q = (q1 +q2)/2 Then the repulsiveforce is F2 = k*q2 / d2 q = ( F2*d2 / k )1/2 = ( 0.0318N*(0.468m)2 / (9*109N.m2/C2) )1/2 =0.88*10-6 C Then q1 + q2 = 2q =1.75*10-6 C --------- (1) (q1 - q2 )2 = (q1 + q2 )2 +4q1*q2 q1-q2 = 4.80 *10-6 C ------------ (2) From equations (1) and (2) we getq1 = ------ C and q2 = --------- C
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