Electrostatics Question 804

Question: A parallel plate capacitor with air between the plates has a capacitance of 8 pF. Calculate the capacitance if the distance between the plates is reduced by half and the space between them is filled with a substance of dielectric constant. $ ( {\varepsilon _{r}}=6 ) $

Options:

A) 72 pF

B) 81 pF

C) 84 pF

D) 96 PF

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Answer:

Correct Answer: D

Solution:

[d] Capacity of parallel plate capacitor

$ C=\frac{{\varepsilon _{r}}{\varepsilon _{0}}A}{d}\text{ }( \text{For air }{\varepsilon _{r}}=i ) $

So, $ \frac{{\varepsilon _{0}}A}{d}=8\times {{10}^{-12}} $

If $ d\to \frac{d}{2} $

and $ {\varepsilon _{r}}\to 6 $

then new capacitance $ C’=6\times \frac{{\varepsilon _{0}}A}{d/2}=12\frac{{\varepsilon _{0}}A}{d}=12\times 8pF=96pF $



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