Answer to Question #183051 in General Chemistry for KyKy

Question #183051

Spheres A, B, and C have charges +8.0 C, +12.0 C, and -5.0 C, respectively. Spheres A and B touched each other and then separated. Sphere A then touched sphere C and then separated. (a) What is the total charge on the three spheres before and after touching each other? (b) What is the final charge on each sphere assuming they are identical? (c) What is the final charge on each sphere assuming that rA = rB = 2rC

1
Expert's answer
2021-04-22T03:03:09-0400

When spheres are of the same size, the charge redistribution occurs equally.

(a) The total charge of the spheres equals to their sum and remains the same no matter how many times three spheres were interacting:

"q_A+q_B+q_C = (+8.0C) + (+12.0C) + (-5.0C) = +15.0C"


(b) When identical spheres A and B touched their charge redistributed:

"q_A' = q_B'="

"=(q_A+q_B)\/2 = \\\\\n= ((+8.0C) + (+12.0C) )\/ 2 = +10.0C"

Next, sphere A with a new charge touched sphere C:

"q_A'' = q_C''=\\\\\n(q_A'+q_C)\/2=\\\\\n=((+10.0C)+(-5.0C))\/2 = +2.5 C"

Thus, the final charges on identical A, B and C spheres are +2.5 C, +10.0 C and +2.5 C respectively.

(c) When assuming that "r_A = r_B = 2r_C" , spheres A and B remain identical, so only charge for A and C spheres will change:


"q_A''=\\frac{q_A'+q_c}{r_A+r_C}r_A=\\\\\n=\\frac{q_A'+q_c}{r_A(1+2)}r_A= \\frac{q_A'+q_c}{3}=\\\\\n=\\frac{(+10.0C)+(-5.0 C)}{3}= +\\frac{5}{3}C"


"q_C'=\\frac{q_A'+q_c}{r_A+r_C}r_C=\\\\\n=\\frac{q_A'+q_c}{r_A+2r_A}2r_A=\\\\\n=\\frac{q_A'+q_c}{r_A(1+2)}2r_A= \\frac{q_A'+q_c}{3}2=\\\\\n=\\frac{2((+10.0C)+(-5.0 C))}{3}= +\\frac{10}{3}C"

When "r_A = r_B = 2r_C" , the final charges on A, B and C spheres are +5/3 C, +10.0 C and +10/3 C respectively.


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