A) \[6.6\times {{10}^{-6}}m,{{10}^{3}}{{m}^{2}}\]
B) \[6.6\times {{10}^{-5}}m,{{10}^{4}}{{m}^{2}}\]
C) \[6.6\times {{10}^{-4}}m,\text{ }{{10}^{5}}{{m}^{2}}\]
D) \[6.6\times {{10}^{-6}}m,\text{ }{{10}^{2}}{{m}^{2}}\]
Correct Answer: A
Solution :
We knows, electric field is given by \[E=\frac{V}{d}\] \[{{E}_{\max }}=\frac{V}{{{d}_{\min }}}\] \[{{d}_{\min }}=\frac{V}{{{E}_{\max }}}\] Substitute\[V=20\text{ }V\]and \[{{E}_{\max }}=3\times {{10}^{6}}V\] \[{{d}_{\min }}=\frac{20}{3\times {{10}^{6}}}=6.6\times {{10}^{-6}}m\] We knows, the capacitance is given by \[C=\frac{k{{\varepsilon }_{0}}A}{d}\] \[\frac{A}{d}=\frac{C}{k{{\varepsilon }_{0}}}\] \[A=\frac{C}{k{{\varepsilon }_{0}}}\] Putting \[C=1.77\text{ }\mu F=1.77\times {{10}^{-6}}C,\] \[d=6.6\times {{10}^{-6}},k=200,\] \[{{\varepsilon }_{0}}=8.85\times {{10}^{-12}}{{C}^{2}}{{N}^{-1}}{{m}^{2}}\] \[A=\frac{1.77\times {{10}^{-6}}\times 6.6\times {{10}^{-6}}}{200\times 8.85\times {{10}^{-12}}}={{10}^{3}}{{m}^{2}}\]
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