Question

Three identical metallic conducting spheres carry the following charges: q 1 1 = +8.20 μC, q 2 2 = +8.20 μC, and q 3 3 = −3.60 μC. The spheres that carry the charges q 1 1 and q 2 2 are brought into contact. Then they are separated. After that, one of those two spheres is brought into contact with the third sphere that carries the charge q 3 3; those two are then separated as well. What is the final charge on the third sphere and how many excess (or deficiency) electrons make up the final charge on the third sphere?

Answer 1

from the equilibrium condition when two charges brough together , the total charge of system equially divided in to two bodies .

from the data, initially q1 and q2 are brough into contact

total charge on system Q= q1 + q2= 8.2μC+8.2μC = 16.4 μC

16.4 μC get dived into sphere 1 and 2

q1 = 16.4 μC/2 = 8.2μC

q2 = 16.4 μC/2 = 8.2μC

now q1 or q2 brough contant with thrid sphere

total charge = q1 + q3 = 8.2μC+( −3.60 μC) = 4.6 μC

the final charge on the third sphere is

q3 = total charge /2 = 4.6 μC/2 = 2.3 μC

the number of electrons is

n = q3/e = 2.3* 10^-6/-1.6 * 10^-19 = 1.4375 * 10^13 electrons