When a dielectric slab of thickness 6 cm is introduced between the plates of parallel plate condenser it is found that the distance between the plates has to he increased by 4 cm to restore to capacity to original value. The dielectric constant of the slab is

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Let initial capacitance be CC-x3B5-oAd-xA0- -xA0- -xA0- -xA0-A-xA0-and-xA0-d are in cmFinal capacitance is same as initial capacitance-d-x3B5-oA-d-4-x2212-6-x3B5-oA-6K-x3B5-oASolving- we get-K-3

When a dielectric slab of thickness 6 cm is introduced between the plates of parallel plate condenser it is found that the distance between the plates has to he increased by 4 cm to restore to capacity to original value. The dielectric constant of the slab is

$1.5$
$2 / 3$
$3$
$4$

Let initial capacitance be $C$
$C = \frac{ε_{o} A}{d}$ $A a n d d$ are in $c m$

Final capacitance is same as initial capacitance.
$\frac{d}{ε_{o} A} = \frac{d + 4 - 6}{ε_{o} A} + \frac{6}{K ε_{o} A}$
Solving, we get:
$K = 3$

When a dielectric slab of thickness 6 cm is introduced between the plates of parallel plate condenser it is found that the distance between the plates has to he increased by 4 cm to restore to capacity to original value. The dielectric constant of the slab is1.52/334

Let initial capacitance be CC=εoAd A and d are in cmFinal capacitance is same as initial capacitance.dεoA=d+4−6εoA+6KεoASolving, we get:K=3

When a dielectric slab of thickness 6 cm is introduced between the plates of parallel plate condenser it is found that the distance between the plates has to he increased by 4 cm to restore to capacity to original value. The dielectric constant of the slab is

$1.5$
$2 / 3$
$3$
$4$

Let initial capacitance be $C$
$C = \frac{ε_{o} A}{d}$ $A a n d d$ are in $c m$

Final capacitance is same as initial capacitance.
$\frac{d}{ε_{o} A} = \frac{d + 4 - 6}{ε_{o} A} + \frac{6}{K ε_{o} A}$
Solving, we get:
$K = 3$