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A slab of material of dielectric cons...

A slab of material of dielectric constant K has the same area as the plates of a parallel plate capacitor but has a thickless `(1/2)d`. Where d is the separation of the plates . How is the capacitance changed when the slab is inserted between the plates

Text Solution

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`V=V_0 [(k+1)/( 2K)] and C= (2k)/(K+1) C_0`
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A slab of material of dielectric constant K has the same area as the plates of a parallel-plate capacitor but has a thickness (3/4)d, where d is the separation of the plates. How is the capacitance changed when the slab is inserted between the plates?

A slab of material of dielectric constant K has the same area as the plates of a parallel-plate capacitor but has a thickness ((3)/(4)) d where d is the separation of the plates. How Is the capacitance changed when the slab ls inserted between the plates ?

Knowledge Check

  • Which of the following one, the capacitance of parallel plate capacitor does not depend ?

    A
    On the area of the plate
    B
    On the distance between two plates
    C
    On the charge of the plate
    D
    On the shape of the plate

  • By increasing the charge on the plate of capacitor .......... .

    A
    its capacitance increases
    B
    p.d. between two plates increases
    C
    p.d. between two plates decreases
    D
    both option (A) and (B) increases

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