Two conducting spheres, one of radius $6 \mathrm{~cm}$ and the other of radius $12 \mathrm{~cm}$, each carrying $3 \times 10^{-8} \mathrm{C}$, are placed very far apart. If the spheres are connected by a conducting wire, the magnitude of the charge transferred is $\mathrm{K}$. Find…

Question

Two conducting spheres, one of radius $6 \mathrm{~cm}$ and the other of radius $12 \mathrm{~cm}$, each carrying $3 \times 10^{-8} \mathrm{C}$, are placed very far apart. If the spheres are connected by a conducting wire, the magnitude of the charge transferred is $\mathrm{K}$. Find $\mathrm{K} \times 10^{8}$.

Aicharya · Accepted Answer

Since, the spheres are far apart, charges on them are uniformly distributed, so they may be supposed to be concentrated at the centre for calculating field and potential on and outside the surface.
$\begin{array}{l}
\therefore \quad V_{1}=9 \times 10^{9} \frac{3 \times 10^{-8}}{0.06}=4500 \mathrm{~V} \\
V_{2}=9 \times 10^{9} \frac{3 \times 10^{-8}}{0.12}=2250 \mathrm{~V}
\end{array}$
Let $V$ be the common potential after connection and $q_{1}$ and $q_{2}$ be the changes on the spheres.
Then $V=9 \times 10^{9} \frac{q_{1}}{0.06}=9 \times 10^{9} \frac{q_{3}}{0.12}$
Since total charge remains the same
$2 \times 3 \times 10^{-8}=q_{1}+q_{2}=\frac{V(0.06+0.12)}{9 \times 10^{9}}$
or $V=3000 \mathrm{~V}$
$\begin{array}{l}
\therefore \quad q_{1}=\frac{3000 \times 0.06}{9 \times 10^{9}}=2 \times 10^{-8} C \\
\qquad q_{2}=\frac{3000 \times 0.12}{9 \times 10^{9}}=4 \times 10^{-8} C
\end{array}$
Charge transferred $=3 \times 10^{-8}-2 \times 10^{-8}$
$\mathrm{K}=0.1 \mathrm{nC}=10^{-8}$ from the small to the large sphere.

Two conducting spheres, one of radius \(6 \mathrm{~cm}\) and the other of radius \(12 \mathrm{~cm}\), each…

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