Answer:
Q'sphere=2.7*10^-9 C
Q'rod=-4.7*10^-9 C
Explanation:
given data:
charge on metallic sphere Qsphere=3.1*10^-9 C ∴1n=10^-9
charge on rod Qrod =-4*10^-9 C
no of electron n= 9.2×10^9 electrons
To find:
we are asked to find the charges Q'sphere on the sphere and Q'rod on the rod after the rod touches the sphere.
solution:
the total charge transferred when the rod touches the sphere equal to the no of electrons transferred multiplied by the charge of each electron:
Q(transferred)= nq_(e)
=(9.2×10^9)(1.6×10^-19)
=-1.312×10^-9 C
because electron are negative they move from the negatively charged rod to the positively charged rod so that new charged of the sphere is:
Q'sphere =Qsphere+Q(transferred)
=(3.1*10^-9 )-(1.312×10^-9)
=2.7*10^-9 C
similarly the new charge of the rod is:
Q'rod = Qrod-Q(transferred)
= (-6*10^-9 C)-(1.312*10^-9 C)
= -4.7*10^-9 C
∴note: there maybe error in calculation but the method is correct.
