A) \[32\mu C\] from A to B
B) \[24\text{ }\mu C\] from A to B
C) \[32\text{ }\mu C\] from B to A
D) None of the above
Correct Answer: A
Solution :
Given, \[{{r}_{1}}=4cm,\] \[{{r}_{2}}=6cm\] \[{{q}_{1}}=80\mu C,\] \[{{q}_{2}}=40\mu C\] Capacitances of spheres, \[{{C}_{1}}=4\pi {{\varepsilon }_{0}}{{r}_{1}},\] \[{{C}_{2}}=4\pi {{\varepsilon }_{0}}{{r}_{2}}\] On connecting the spheres, charge is redistributed in the ratio of their capacitances. \[\therefore \] \[\frac{{{q}_{1}}'}{{{q}_{2}}'}=\frac{{{C}_{1}}}{{{C}_{2}}}=\frac{4\pi {{\varepsilon }_{0}}{{r}_{1}}}{4\pi {{\varepsilon }_{0}}{{r}_{2}}}\] \[\therefore \] \[\frac{{{q}_{1}}'}{{{q}_{2}}'}=\frac{4}{6}=\frac{2}{3}\] \[\therefore \] \[{{q}_{1}}'=\frac{2\times (80+40)}{(2+3)}=\frac{2\times 120}{5}\] \[=48\mu C\] \[\therefore \] Charge flown from first sphere to second sphere \[=80-48-32\mu C\]
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