When a dielectric slab of thickness 6cm is introduced between the plates of parallel plate condenser, it is found that the distance between the plates has to be increased by 4 cm to restore the capacity to the original value. The dielectric constant of the slab is

Answers (1)
S safeer

Capacitance

C=\frac{\epsilon_0A}{d}

Capacitance when a dielectric of thickness t is placed between the plates

{C}'=\frac{\epsilon _{0}A}{d-t+\frac{t}{k}}

Given when the separation between plates is d+4,  C'=C

Therefore

\\\\\frac{\epsilon _{0}A}{d+4-t+\frac{t}{k}}=\frac{\epsilon _{0}A}{d}\\\\\text{implies}\\\\d+4-6+\frac{6}{k}=d\\\text{implies}\\\\k=3

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