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  • Solve it, If the distance between the plates of parallel plate capacitor is halved and the dielectric constant of dielectric is doubled, then its capacity will

  If the distance between the plates of parallel plate capacitor is halved and the dielectric constant of dielectric is doubled, then its capacity will

  • Option 1)

    Increase by 16 times

  • Option 2)

    Increase by 4 times

  • Option 3)

    Increase by 2 times

  • Option 4)

    Remain the same

 

Answers (2)

best_answer

As we have learned

If metallic slab insert between the plates -

\dpi{100} {C}'=\frac{\epsilon _{0}A}{d-t}

- wherein

 

 

    C= \frac{K\varepsilon _0A}{d}\propto \frac{K}{d} Hence, \frac{C_1}{C_2}=\frac{K_1}{K_2}\times \frac{d_2}{d_1}= \frac{K}{2K}\times \frac{d/2}{d}=1/4 Therefore, C_2= 4C_1


Option 1)

Increase by 16 times

Option 2)

Increase by 4 times

Option 3)

Increase by 2 times

Option 4)

Remain the same

Posted by

Plabita

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JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

it decreases by 2times

Posted by

swesue

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