Two spheres A & B of radii $4 \mathrm{~cm} \& 6 \mathrm{~cm}$ are given charges of $80 \mu \mathrm{C} \& 40…
- $32 \mu \mathrm{C}$ from B to A
- $32 \mu \mathrm{C}$ from A to B
- $20 \mu \mathrm{C}$ from A to B
- $16 \mu \mathrm{C}$ from B to A
Solution

$\mathrm{Q}_1=80 \mu \mathrm{C}, \mathrm{Q}_2=40 \mu \mathrm{C}$ $\mathrm{r}_1=4 \mathrm{~cm}, \mathrm{r}_2=6 \mathrm{~cm}$ $\therefore \quad$ After connecting wire, $\mathrm{Q}_1^{\prime}=\left(\frac{\mathrm{r}_1}{\mathrm{r}_1+\mathrm{r}_2}\right)\left(\mathrm{Q}_1+\mathrm{Q}_2\right)=\left(\frac{4}{4+6}\right)(80+40)=48 \mu \mathrm{C}$ $\therefore \quad$ Charge flow from A to B is $\Delta \mathrm{Q}=\mathrm{Q}_1-\mathrm{Q}_1^{\prime}=80-48=32 \mu \mathrm{C}$
Asked in: AP EAMCET 2024 (20 May Shift 1)