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A parallel-plate air capacitor of plate separation 2mm and capacitance 1muF is charged to 100V. A dielectric slab of relative permittivity 50 is now inserted so as to fill the space between the plates. (i)Find the polarisation charge on one of the boundaries of the dielectric slab. (ii) Find the magnitude of the polarisation of the dielectric slab. (epsilon_(0) = 8.85 xx 10^(-12) C^(2)//N.m^(2)) |
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Answer» Solution :Data: `C_(0) = 10^(-6) F, d=2 XX 10^(-3) m, k=50, V=100 V`, `epsilon_(0) = 8.85 xx 10^(-12) C^(2)//N.m^(2)` `therefore Q_(p) = Q(1-1/k)= C_(0)V(1-1/R)` `therefore Q_(P) = Q(1-1/k) = C_(0)V(1-1/k)(therefore Q=C_(0)V)` `=10^(-6) xx 100 xx 49/50 = 49 xx 2 xx 10^(-6) = 9.8 xx 10^(-5)C = 98 MU` C (ii) SINCE the magnitude of the polarisation, `P=sigma_(p) = Q_(p)/A` and `C_(0) = (Aepsilon_(0))/d`. `P=(Q_(p).epsilon_(0))/(C_(0)d) = ((9.8 xx 10^(-5))(8.85 xx 10^(-12)))/((10^(-6))(2 xx 10^(-3)))` `=0.49 xx 8.85 xx 10^(-7) = 4.3365 xx 10^(-7) C//m^(2)` |
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