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A paralel plate capacitor having cross s...

A paralel plate capacitor having cross sectional area A and seperation d has air in between plates. Now an insulating slab of same area but thickness d/2 is inserted between plates having dielectric constant k =4. Ratio of new capacitance to its original capacitance will be

A

`8:5`

B

`6:5`

C

`4:1`

D

`2:1`

Text Solution

Verified by Experts

The correct Answer is:
A
`C_(K)= (epsi_(0)A)/(d-t+ (t)/(K))= (epsi_(0)A)/(d-(d)/(2) + (d)/(8))= (8)/(5) (epsi_(0)A)/(d)`
`C_(a)= (epsi_(0)A)/(d)`
`therefore C_(k)= (8)/(5) C_(a) rArr (C_(k))/(C_(B))=(8)/(5)`
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