# Consider a parallel plate capacitor of 8μF (micro-farad) with air filled in the gap between the plates. Now one half of the space between the plates is filled with a dielectric of dielectric constant 4, as shown in the figure. The capacity of the capacitor changes to Option 1) 8μF Option 2) 16μF Option 3) 20μF Option 4) 24μF

As we learnt in

Parallel Grouping -

$\dpi{100} C_{eq}=C_{1}+C_{2}+\cdots$

- wherein

$C = \frac{\epsilon _{0}A}{d} = 8\mu F$

After inserting dielectric, we will have Capacitance, $C = \frac{\epsilon _{0}\left ( \frac{A}{2} \right )}{d}+ \frac{K\epsilon _{0}\left ( \frac{A}{2} \right )}{d}$

$=\frac{\epsilon _{0}A}{2d}\left ( K+1 \right ) = \frac{8\mu F}{2}\times 5\ \mu F$

$C= 20\ \mu F$

Option 1)

8μF

This option is incorrect

Option 2)

16μF

This option is incorrect

Option 3)

20μF

This option is correct

Option 4)

24μF

This option is incorrect

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