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If a capacitor having capacittance 2F and plate separation of 0.5 cm will have areaA. `1130cm^(2)`B. `1130m^(2)`C. `1130 km^(2)`D. none of these |
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Answer» Correct Answer - C `therefore` Capacitance of a parallel plate capacitor, `C=(epsilon_(0)A)/dA=(Cd)/epsilon_(0)=(2 xx 0.5 xx 10^(-2))/(8.85 xx 10^(-12))` `=1.130 xx 10^(9)m^(2)` = 1130 `km^(2)` |
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