A) electric field in the dielectric-filled region is higher than that in the air-filled region
B) on the two halves of the bottom plate the charge densities are unequal
C) charge on the half of the top plate above the air-filled part is\[\frac{Q}{K+1}\]
D) capacitance of the capacitor shown above is\[(1+K)\frac{{{C}_{0}}}{2},\]where \[{{C}_{0}}\]is the capacitance of the same capacitor with the dielectric removed.
Correct Answer: B
Solution :
We know that \[{{C}_{1}}=\frac{K{{\varepsilon }_{0}}A}{2d},{{C}_{2}}\frac{{{\varepsilon }_{0}}A}{2d}\] and \[{{C}_{eq}}=\frac{\varepsilon .A}{2d}(K+1)\] \[{{C}_{eq}}=\frac{{{C}_{0}}}{2}(K+1)\] \[\frac{{{Q}_{1}}}{{{Q}_{2}}}=\frac{{{C}_{1}}}{{{C}_{2}}}=\frac{K}{1}\] \[\Rightarrow \] \[\frac{{{\sigma }_{1}}}{{{\sigma }_{2}}}=\frac{K}{1}\] \[{{Q}_{1}}=\frac{KQ}{K+1}\]and \[{{Q}_{2}}=\frac{Q}{K+1}\] So, \[E=\frac{\sigma }{{{\varepsilon }_{0}}K}\Rightarrow \frac{{{E}_{1}}}{{{E}_{2}}}=\frac{{{\sigma }_{1}}}{{{\sigma }_{2}}}\times \frac{{{K}_{1}}}{1}\] \[=\frac{{{\sigma }_{1}}}{{{\sigma }_{2}}}\times \frac{{{K}_{2}}}{{{K}_{1}}}=\frac{K}{1}\times \frac{1}{K}\] \[1:1\] So, [b], [c] and [d] are correct options.
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