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  • Can someone help me with this, 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 t

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)

best_answer

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

Posted by

Aadil

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