Question

Two conducting spheres are far apart.
smaller sphere carries a total charge Q. The
larger sphere has a radius that is twice that of the
smaller and is neutral. After the two spheres are
connected by a conducting wire, the charges on
the smaller and larger spheres, respectively, are:
(A)Q/2 and Q/2
(B) 2Q/3 and Q /3
(C) Q/3 and 2Q/3
(D) zero and Q​

Answer 1

Answer:

The charge on both spheres will get distributed until their potential becomes the same when connected by the wire. But the total charge on the system will remain the same. Therefore, the charge on the smaller and bigger spheres is $\frac{Q}{3}$ and $\frac{2Q}{3}$, respectively.

Explanation:

Charging by Conduction:

The process in which an uncharged body is charged by making it in touch with a charged body is known as charging by conduction.

• When the uncharged sphere is connected with the charged sphere by a conducting wire, some charge of the charged sphere gets transferred to the uncharged one.
• The transfer to the charge will take place until the potential on both the spheres become equal.

The potential of a conducting sphere can be given as,

$V=\frac{kq}{R}$

Here, k is the Columb's law constant, q is the charge on the sphere, and R is the radius of the sphere.

When the potential on both the given spheres is the same,

$V_{1} = V_{2}\\\frac{k q_{1} }{R_{1} } = \frac{k q_{2} }{R_{2} } \\\frac{k q_{1} }{R_{1} } = \frac{k q_{2} }{(2R_{1}) } \\q_{1} = \frac{q_{2} }{2}$

Here, the variables have their usual meaning. The variables with subscript 1 are for the smaller sphere and the ones with subscript 2 are for the bigger sphere.

Principle of conservation of charge:

The principle of conservation of charge states that the total charge on a system of charges remains the same.

The total charge on the system can be given as,

$q_{T}= q_{1} +q_{2}\\(Q)= (\frac{q_{2} }{2}) +q_{2}\\q_{2} = \frac{2Q}{3}$

Thus, the charge on the smaller sphere is,

$q_{1} = \frac{Q}{3}$

Therefore, the correct choice is (C).