Question

Metal sphere 1 has a positive charge of 7.00 nc . metal sphere 2, which is twice the diameter of sphere 1, is initially uncharged. the spheres are then connected together by a long, thin metal wire. what are the final charges on each sphere?

Answer 1

Answer:

2.33 nC, 4.67 nC

Explanation:

when the two spheres are connected through the wire, the total charge (Q=7.00 nC) re-distribute to the two sphere in such a way that the two spheres are at same potential:

$V_1 = V_2$ (1)

Keeping in mind the relationship between charge, voltage and capacitance:

$C=\frac{Q}{V}$

we can re-write (1) as

$\frac{Q_1}{C_1}=\frac{Q_2}{C_2}$ (2)

where:

Q1, Q2 are the charges on the two spheres

C1, C2 are the capacitances of the two spheres

The capacitance of a sphere is given by

$C=4 \pi \epsilon_0 R$

where R is the radius of the sphere. Substituting this into (2), we find

$\frac{Q_1}{4 \pi \epsilon_0 R_1}=\frac{Q_2}{4 \pi \epsilon_0 R_2}$ (3)

we also know that sphere 2 has twice the diameter of sphere 1, so the radius of sphere 2 is twice the radius of sphere 1:

$R_2 = 2R_1$

So the eq.(3) becomes

$\frac{Q_1}{4 \pi \epsilon_0 R_1}=\frac{Q_2}{4 \pi \epsilon_0 2R_1}$

And re-arranging it we find:

$Q_2 = 2Q_1$

And since we know that the total charge is

$Q_1 + Q_2 = 7.00 nC$

we find

$Q_1 = 2.33 nC\\Q_2 = 4.67 nC$