A) \[14.4\pi \]
B) \[28.8\pi \]
C) \[144\pi \]
D) \[288\pi \]
Correct Answer: A
Solution :
Given: Diameter of sphere\[d=4\,cm\] Radius, r =2 cm Charge density of sphere \[\sigma ={{10}^{-4}}\]coulomb/c \[{{m}^{2}}\] \[\sigma ={{10}^{-4}}\times {{10}^{4}}\frac{\text{coulomb}}{\text{m}}\] \[\text{=1 coluomb/}{{\text{m}}^{\text{2}}}\] Charge \[{{q}_{0}}=40\]nanocoulomb \[=40\times {{10}^{-9}}=4\times {{10}^{-8}}\]coulomb Form the relation, Charge density is given \[\sigma =\frac{q}{A}\Rightarrow q=\sigma A=1\times 4\pi {{r}^{2}}\] \[q=4\times \pi (4\times {{10}^{-4}})=16\pi \times {{10}^{-4}}C\] The potential \[V=\frac{1}{4\pi {{\varepsilon }_{0}}}\times \frac{a}{(r+x)}\] (\[\because \,x=2\,cm\]given) \[V=\frac{9+\times {{10}^{9}}\times 16\pi \times {{10}^{-4}}}{(2+2)\times {{10}^{-2}}}\] \[=36\pi \times {{10}^{7}}\,\]volt Work done is give by \[W=Vq=36\pi \times {{10}^{7}}\times 4\times {{10}^{-8}}\] \[=14.4\,\pi \,J\]
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