GATE 2015 EE - SET 1 - Question 36

Solution :

E_{r} = 4 (glass) , E_{r}=1 (air)

C_{1} =  \frac{\epsilon_{r}\epsilon_{0}A} {d}= \frac{1\epsilon_{0} A}{d} = \frac{\epsilon_{0} A}{d}

C_{2} =  \frac{\epsilon_{o}\epsilon_{r}A} {d}= \frac{4\epsilon_{0} A}{d}

C_eq = \frac{C_{1}C_{2}}{C_{1}+C_{2}}

C_eq = \frac{4\epsilon_{0}A}{5d}

Electric field  E = \frac{C_{eq}V}{A\epsilon_{0}}

V = \frac{EA\epsilon _{0} }{\frac{4\epsilon_{0}A }{5d}} = \frac{5Ed}{4}

V =  \frac{5*30*10^{3}*10^{2}*10^{-3}*5}{4} = 18750 v = 18.75kv