A) \[7.5\mu C\]
B) \[15\mu C\]
C) \[10\mu C\]
D) \[20\mu C\]
Correct Answer: C
Solution :
: Common potential of two spheres \[=\frac{{{Q}_{1}}+{{Q}_{2}}}{{{C}_{1}}+{{C}_{2}}}\] \[\therefore \]\[V=\frac{(15+15)\times {{10}^{-6}}}{4\pi {{\varepsilon }_{0}}(5+10)\times {{10}^{-2}}}\] or\[V=\frac{30\times {{10}^{-4}}}{4\pi {{\varepsilon }_{0}}\times 15}=\frac{2\times {{10}^{-4}}}{4\pi {{\varepsilon }_{0}}}\] \[\therefore \]Change on smaller sphere = capacity\[\times V\] \[q=(4\pi {{\varepsilon }_{0}}\times 5\times {{10}^{-2}})\times \frac{2\times {{10}^{-4}}}{4\pi {{\varepsilon }_{0}}}C\] \[=10\times {{10}^{-6}}C=10\,\mu C\]You need to login to perform this action.
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