A metallic sphere ' A ' isolated from ground is charged to $+50 \mu \mathrm{C}$. This sphere is brought in…
A metallic sphere ' A ' isolated from ground is charged to $+50 \mu \mathrm{C}$. This sphere is brought in contact with other isolated metallic sphere ' $B$ ' of half the radius of sphere ' A '. Then the charge on the two isolated spheres $\mathrm{A} \& \mathrm{~B}$ are in the ratio
$1: 2$
$2: 1$
$4: 1$
$1: 1$
Solution
Charge will flow from sphere A to B till the potential becomes same
$\begin{aligned}
& \therefore \quad \mathrm{V}_{\mathrm{A}}=\mathrm{V}_{\mathrm{B}} \text {. } \\
& \therefore \quad \frac{1}{4 \pi \varepsilon_0} \frac{\mathrm{q}_{\mathrm{A}}}{\mathrm{r}_{\mathrm{A}}}=\frac{1}{4 \pi \varepsilon_0} \frac{\mathrm{q}_{\mathrm{B}}}{\mathrm{r}_{\mathrm{B}}} \\
& \therefore \quad \frac{q_A}{q_B}=\frac{r_A}{r_B} \\
& \therefore \quad \frac{q_A}{q_B}=\frac{2 r_B}{r_B} \\
& \ldots .\left(\because r_B=\frac{r_A}{2}\right) \\
& \therefore \quad \frac{q_A}{q_B}=\frac{2}{1}
\end{aligned}$