A) \[\frac{d}{4}\]
B) \[\frac{d}{2}\]
C) \[\frac{5d}{8}\]
D) \[d\]
Correct Answer: B
Solution :
The capacity in air \[C=\frac{{{\varepsilon }_{0}}A}{d}\] The capacity when dielectric slab of dielectric constant 5 is introduced between the plates \[C=\frac{{{\varepsilon }_{0}}A}{d-t+\frac{t}{5}}\] \[\therefore \] \[\frac{C}{C}=\frac{d-t+\frac{t}{5}}{d}\] \[C=\frac{166}{100}C\] \[\frac{100}{166}=\frac{d-t+\frac{t}{5}}{d}=\frac{d-\frac{4t}{5}}{d}\] or \[100d=166d-166\left( \frac{4t}{5} \right)\] \[166\left( \frac{4t}{5} \right)=66d\] \[t=\frac{66d\times 5}{166\times 4}=\frac{d}{2}\]
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