If a slab of insulting material $4 \times 10^{- 3}$ m thick is introduced between the plates of a parallel plate capacitor,the separation between plates has to be increases by $3.5 \times 10^{- 3}$ m to restore the capacity to original value.The dielectric constant of the material will be

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Solution

The correct option is B 8
Thickness $\Rightarrow 4 \times 10^{- 3} m$
$d = 3.5 \times 10^{- 3} m$
$k = ?$
we know that
$\frac{A E o}{d} = \frac{A E o}{d + 3.5 - t + \frac{t}{k}}$ (all length are in $10^{- 3} m$ )
$0 = 3.5 - 4 + 4 / k$
$\frac{k}{2} = 4$ $k = 8$

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If a slab of insulting material 4×10−3m thick is introduced between the plates of a parallel plate capacitor,the separation between plates has to be increases by 3.5×10−3 m to restore the capacity to original value.The dielectric constant of the material will be

The correct option is B 8Thickness ⇒4×10−3md=3.5×10−3mk=? we know that AEod=AEod+3.5−t+tk (all length are in 10−3m)0=3.5−4+4/kk2=4 k=8

If a slab of insulting material $4 \times 10^{- 3}$ m thick is introduced between the plates of a parallel plate capacitor,the separation between plates has to be increases by $3.5 \times 10^{- 3}$ m to restore the capacity to original value.The dielectric constant of the material will be

The correct option is B 8
Thickness $\Rightarrow 4 \times 10^{- 3} m$
$d = 3.5 \times 10^{- 3} m$
$k = ?$
we know that
$\frac{A E o}{d} = \frac{A E o}{d + 3.5 - t + \frac{t}{k}}$ (all length are in $10^{- 3} m$ )
$0 = 3.5 - 4 + 4 / k$
$\frac{k}{2} = 4$ $k = 8$