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Capacitance of a capacitor made by a thin metal foil is 2 muF. If the foil is foilded with paper of thickness 0.15 mm, dielectric constant of paper is 2.5 and width of paper is 400 mm, the length of foil will be

Answer»

0.34 m
1.33 m
13.4 m
33.9 m

Solution :If LENGTH of the foil is l, then `C = (Kepsilon_(0)(lxxb))/(d)`
`rArr 2XX10^(-6) = (2.5xx8.85xx10^(-12)(lxx400xx10^(-3)))/(0.01xx10^(-3))`
`l = 33.9 m`


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