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