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Help me please, - Electrostatics - JEE Main-6

 Plate separation of a 15\mu F  capacitor is 2 mm. A dielectric slab (K = 2) of thickness 1 mm is inserted between the plates. Then new capacitance is given by 

  • Option 1)

    15\mu F

  • Option 2)

    20\mu F

  • Option 3)

    30\mu F

  • Option 4)

    25\mu F

 
Answers (1)
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As we have learned

If a number of dielectric slab inserted between the Plate -

{C}'=\frac{\epsilon _{0}A}{d-\left ( t_{1}+t_{2} +\cdots \right )+\left ( \frac{R_{1}}{k_{1}}+\frac{R_{2}}{k_{2}}+\cdots \right )}

- wherein

 

 given C = \frac{\varepsilon _0A}{d}..........(1)

   Then by using

   C' = \frac{\varepsilon _0A}{d-t+\frac{t}{k}}= \frac{\varepsilon _0A}{2*10^{-3}-10^{-3}+\frac{10^{-3}}{2}}= 2/3*\varepsilon _0A*10^{3}                            

 ; From equation (i) C' =20\mu F

 


Option 1)

15\mu F

Option 2)

20\mu F

Option 3)

30\mu F

Option 4)

25\mu F

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