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The area of each plate of a parallel capacitor is 100 cm^(2) and the distance between them is 0.05 cm. It is filled with a dielectric and its capacitance becomes 3.54 xx 10^(-4) muF. Find the dielectric constant of the substance. |
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Answer» Solution :Data supplied, `A=100 cm^(2)=100 xx 10^(-4) m^(2)` `C=3.54 xx 10^(-4) muF= 3.54 xx 10^(-4) xx 10^(-6) F =3.54 xx 10^(-10) F` `d=0.05 cm =0.05 xx 10^(-2)m ""epsi_(0) =8.854 xx C^(-12) C^(2) N^(-1) m^(-2)` `C=(epsi_(0) KA)/(d) ""K=(Cd)/(epsi_(0) A)=(3.54 xx 10^(-10) xx 0.05 xx 10^(-2))/(8.854 xx 10^(-12) xx 100 xx 10^(-4)) K=2` |
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