Question

ch 18 # 2

Four identical metal spheres have charges of q_A = -8.0 μC, q_B=-2.0 μC, q_C=+5.0 μC, and q_D=+12.0 μC.

(a) Two of the spheres are brought together so they touch, and then they are separated. Which spheres are they, if the final charge on each one is +5.0 μC?
Choose the answer for part (a) from the menu in accordance to the question statement                                                          C and DB and DB and CA and BA and DA and C

(b) In a similar manner, which three spheres are brought together and then separated, if the final charge on each of the three is +3.0 μC?
Choose the answer for part (b) from the menu in accordance to the question statement                                                          A+C+DB+C+DA+B+CA+B+D

(c) The final charge on each of the three separated spheres in part (b) is +3.0 μC. How many electrons would have to be added to one of these spheres to make it electrically neutral?

N = Number

Answer 1

When two or more identical spheres are brought together, the charges on the Spheres redistributed among the spheres and final charge of the Spheres will be same.

If two sphere of charges q1 and q2 are added then final charge on each sphere will be (q1 + q2)/2. If three spheres, then (q1 + q2 + q3)/3. And so on.

a) if we brought together sphere C and D the final charge on each = (+5+12)/2 = 8.5 C

If B and D, then charge on each = (–2 + 12)/2 = +5 C.(Ans)

B and C gives (–2+5)/2 = 1.5C. A and B gives (–8–2)/2 = –5 C, A and D gives (–8+12)/2 = 2C, Aand C gives (–8+5)/2 = –1.5 C.

b) A+C+D, final charge on each = (–8+5+12)/3 =+3C (ans)

B+C+D, final charge on each = (–2+5+12)/3 = 5C

A+B+C, final charge on each = (–8–2+5)/3 = –1.67C

A+B+D, final charge on each = (–8–2+12)/3 = .67 C

c) each charge is equal to the integral multiple of the charge of an electron.

To make neutral we need to add Q=–3.0C = –3× 10^–6 C charge. This charge should be integral multiple of the charge of an electron, e = –1.602 × 10^-19 C.(ans)

So, the required electon number N = Q/e =(–3 × 10^–6)/(–1.602×10^–19) = 1.8727 × 10^13