If a slab of insulating material 4 × {10^{ - 3}},m thick is introduced between plates of a parallel

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2. Electric Potential and Capacitance

hard

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

A

$6$

B

$8$

C

$10$

D

$12$

Solution

(b) $K = \frac{t}{{t – d'}} = \frac{{4 \times {{10}^{ – 3}}}}{{4 \times {{10}^{ – 3}} – 3.5 \times {{10}^{ – 3}}}} = 8$

Standard 12

Physics

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