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:
[tex]V_1 = V_2[/tex] (1)
Keeping in mind the relationship between charge, voltage and capacitance:
[tex]C=\frac{Q}{V}[/tex]
we can re-write (1) as
[tex]\frac{Q_1}{C_1}=\frac{Q_2}{C_2}[/tex] (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
[tex]C=4 \pi \epsilon_0 R[/tex]
where R is the radius of the sphere. Substituting this into (2), we find
[tex]\frac{Q_1}{4 \pi \epsilon_0 R_1}=\frac{Q_2}{4 \pi \epsilon_0 R_2}[/tex] (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:
[tex]R_2 = 2R_1[/tex]
So the eq.(3) becomes
[tex]\frac{Q_1}{4 \pi \epsilon_0 R_1}=\frac{Q_2}{4 \pi \epsilon_0 2R_1}[/tex]
And re-arranging it we find:
[tex]Q_2 = 2Q_1[/tex]
And since we know that the total charge is
[tex]Q_1 + Q_2 = 7.00 nC[/tex]
we find
[tex]Q_1 = 2.33 nC\\Q_2 = 4.67 nC[/tex]
The final charges on the first and second spheres are 2.33 nC and 4.67 nC respectively.
The capacitance of a capacitor at a constant voltage is given by the following formula;
[tex]V = \frac{Q}{C} \\\\\frac{Q_1}{C_1} = \frac{Q_2}{C_2} \\\\[/tex]
For a metal sphere, C = 4πεR
[tex]\frac{Q_1}{4\pi \varepsilon _0 R_1} = \frac{Q_2}{4\pi \varepsilon _0 R_2}[/tex]
The diameter of the second sphere = 2 times the first sphere
R₂ = 2R₁
[tex]\frac{Q_1}{4\pi \varepsilon R_1} = \frac{Q_2}{4\pi \varepsilon 2R_1}\\\\Q_2 = 2Q_1[/tex]
Q₁ + Q₂ = 7
Q₁ + 2Q₁ = 7
3Q₁ = 7
Q₁ = 7/3
Q₁ = 2.33 nC
Q₂ = 2Q₁ = 2(2.33) = 4.67 nC
Thus, the final charges on the first and second spheres are 2.33 nC and 4.67 nC respectively.
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