question_answer

If a dielectric plate of thickness t is placed between the plates of a parallel plate capacitor of plate distance d, the capacitance becomes half of the original value. The dielectric constant of the plate will be:

A) \[\frac{2t}{2d+t}\]          

B)                        \[\frac{2t}{2d-t}\]           

C) \[\frac{t}{d+t}\]               

D)                        \[\frac{t}{d-t}\]

Correct Answer: C

Solution :

The capacitor of the capacitance  without placing the dielectric plate is given by \[C=\frac{{{\varepsilon }_{0}}A}{d}\]                      ?(1) When the dielectric plate of dielectric constant K and thickness t is inserted between the parallel plates of a capacitor, then it becomes the series combination of two capacitors as shown in figure, so the equivalent capacitance \[{{C}_{eq}}\]is \[{{C}_{eq}}=\frac{C}{2}=\frac{{{C}_{1}}{{C}_{2}}}{{{C}_{1}}+{{C}_{2}}}\] Here,                     \[{{C}_{1}}=\frac{{{\varepsilon }_{0}}A}{d-t}\] and                        \[{{C}_{2}}=\frac{K{{\varepsilon }_{0}}A}{t}\] or            \[\frac{C}{2}=\frac{\left( \frac{{{\varepsilon }_{0}}A}{d-t} \right)\times \left( \frac{K{{\varepsilon }_{0}}A}{t} \right)}{{{\varepsilon }_{0}}A\left\{ \frac{1}{d-t}+\frac{K}{t} \right\}}=\frac{K{{\varepsilon }_{0}}A}{t+(d-t)K}\] or            \[\frac{C}{2}=\frac{K{{\varepsilon }_{0}}A}{t+(d-t)K}\] As \[C=\frac{{{\varepsilon }_{0}}A}{d}\]or \[{{\varepsilon }_{0}}A=Cd\] or            \[\frac{C}{2}=\frac{KCd}{t+(d-t)K}\] or            \[t+(d-t)K=2Kd\] or            \[K\{-(d-t)+2d\}=t\] or            \[K=\frac{t}{d+t}\]