Two metal spheres each 1.0 cm in radius are far apart. The firstsphere carries 4

Question

Two metal spheres each 1.0 cm in radius are far apart. The firstsphere carries 42 nC of charge, and the second sphere carries -10nC. a)What are the potential on each? V1=________kV V2=________kV b) If the spheres are connected by a thin wire, what will bethe potential on each once equilibrium is reached? V1=_______kV V2=_______kV c)How much charge must move between the spheres in order toachieve equilibrium? ______nC Please show all work-thanks a)What are the potential on each? V1=________kV V2=________kV b) If the spheres are connected by a thin wire, what will bethe potential on each once equilibrium is reached? V1=_______kV V2=_______kV c)How much charge must move between the spheres in order toachieve equilibrium? ______nC Please show all work-thanks

Explanation / Answer

   Potential on a sphere of radius R due tocharge on itself,    Vs = Qs/4R    Potential on an external sphere due to charge onanother sphere at a distance r,    Vext = Qext/4r    Vtotal = Vs + Vext = [Qs/R +Qext/r]*1/4     Now, in this case since the spheres are farapart, r->     So, Vext ˜ 0     => V = Q/4R (a)    V1 = Q1/4R1     V2 = Q2/4R2 (b) Since equilibrium can be reached only when there is nocurrent between the spheres,    both the spheres must be at the samepotential.    Thus, Q1+Q2 = Q (constant)    And V1=V2 => Q1'/R1 = (Q-Q1')/R2    => Q1' = (Q/R2) /[1/R1 + 1/R2] =QR1/[R1+R2] = (Q1+Q2)*R1/[R1+R2]    => Q2' = Q - Q1' = Q1+Q2-Q1' (c)   So, charge transferred = |Q1-Q1'| = |Q2-Q2'| =|[Q1R2 - Q2R1]/[R1+R2]| = ...........   

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