A metallic sphere has a charge of + 3.1 nC. A negatively charged rod has a charge of - 4.0 nC. When the rod touches the sphere, 9.2*109 electrons are transferred. What are the charges of the sphere and the rod now?

Q'sphere=2.7*10^-9 C

Q'rod=-4.7*10^-9 C

Explanation:

given dа ta:

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.