question_answer

Separation between the plates of a parallel plate capacitor is \[d\]and the area of each plate is \[A\]. When a slab of material of dielectric constant\[k\]and thickness \[t(t<d)\] is introduced between the plates,  its capacitance becomes                                                                                          [MP PMT 1989]

A)                    \[\frac{{{\varepsilon }_{0}}A}{d+t\left( 1-\frac{1}{k} \right)}\] 

B)            \[\frac{{{\varepsilon }_{0}}A}{d+t\left( 1+\frac{1}{k} \right)}\]

C)                    \[\frac{{{\varepsilon }_{0}}A}{d-t\left( 1-\frac{1}{k} \right)}\]

D)            \[\frac{{{\varepsilon }_{0}}A}{d-t\left( 1+\frac{1}{k} \right)}\]

Correct Answer: C

Solution :

           Potential difference between the plates \[V={{V}_{air}}+{{V}_{medium}}\] \[=\frac{\sigma }{{{\varepsilon }_{0}}}\times (d-t)+\frac{\sigma }{K{{\varepsilon }_{0}}}\times t\] Þ \[V=\frac{\sigma }{{{\varepsilon }_{0}}}(d-t+\frac{t}{K})\] \[=\frac{Q}{A{{\varepsilon }_{0}}}(d-t+\frac{t}{K})\] Hence capacitance \[C=\frac{Q}{V}=\frac{Q}{\frac{Q}{A{{\varepsilon }_{0}}}(d-t+\frac{t}{K})}\] \[=\frac{{{\varepsilon }_{0}}A}{(d-t+\frac{t}{k})}=\frac{{{\varepsilon }_{0}}A}{d-t\,\left( 1-\frac{1}{k} \right)}\]