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A substance has a dielectric constant 2.0 and its dielectric strength is 20 xx 10^(6) V/m . It is taken as a dielectric material in a parallel plate capacitor. The minimum area of its each plate such that its capacitance becomes 8.85xx10^(-2) muFand it can withstand a potential difference of 2000 V is ...... |
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Answer» Solution :The charge on capacitor `:. Q =CV = (8.85 xx10^(-8) ) xx(2000)` `:. Q = 17.7 xx10^(-5) C` The surface charge density on the PLATE `sigma = (Q)/(A) = (17.7xx10^(-5))/(A) C//m^(2)` The electric field when DIELECTRIC INSERTED `E = (E_(0))/(K) ` but `E_(0) = (sigma)/(epsilon_(0))` `:. E = (sigma)/(K epsilon_(0))= (Q)/(K epsilon_(0)A)` `:. A = (Q)/ (KEepsilon_(0))(17.7xx10^(-5))/(2xx20xx10^(6)xx8.85xx10^(-12))` `:. A = 0.5 m^(2)` |
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