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

 

Answers (1)

As we learnt in 

Parallel Grouping -

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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