What is the area of the plates of a $3\,F$ parallel plate capacitor, if the separation between the plates is $5\,mm$
A$1.694 \times {10^9}\,{m^2}$
B$4.529 \times {10^9}\,{m^2}$
C$9.281 \times {10^9}\;{m^2}$
D$12.981 \times {10^9}\;{m^2}$
AIIMS 1998, Easy
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A$1.694 \times {10^9}\,{m^2}$
a (a) We have $C = \frac{{{\varepsilon _0}A}}{d}$ $==>$ $A = \frac{{Cd}}{{{\varepsilon _0}}} = \frac{{3 \times 5 \times {{10}^{ - 3}}}}{{8.85 \times {{10}^{ - 12}}}}$
$ = 1.7 \times {10^9}\,{m^2}$
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