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A parallel plate capacitor is made of tw...

A parallel plate capacitor is made of two dielectric blocks in series. One of the blocks has thickness `d_(1)` and dielectric constant `K_(1)` and the other has thickness `d_(2)` and dielectric constant `K_(2)` as shown in figure. This arrangement can be through as a dielectric slab of thickness `d (= d_(1) + d_(2))` and effective dielectric constant `K`. The `K` is.
.

A

`(K_(1)d_(1)+K_(2)d_(2))/(d_(1)+d_(2))`

B

`(K_(1)d_(1)+K_(2)d_(2))/(K_(1)+K_(2))`

C

`(K_(1)K_(2)(d_(1)+d_(2)))/(K_(2)d_(1)+K_(1)d_(2))`

D

`(2K_(1)K_(2))/(K_(1)+K_(2))`

Text Solution

Verified by Experts

The correct Answer is:
C
The capacities of two individual condensers are `C_(1) = (k_(1) epsi_(0)A)/(d_(1))` and `C_(2)=(K_(2)epsi_(0)A)/(d_(2))`
The arrangement is equaivalent ot tow capacitor joined in series.
`therefore ` Equivalent capacitance `(1)/(C_(eq)) = (1)/(C_(1)) +(1)/(C_(2)) = (d_(1))/(k_(1)epsi_(0)A)+(d_(2))/(k_(2)epsi_(0)A)`
` =(1)/(epsi_(0)A) [(d_(1))/(k_(1))+(d_(2))/(k_(2))] = (1)/(epsi_(0)A)[(k_(2)d_(1)+k_(1)d_(2))/(k_(1)k_(2))]`
`or C_(eq) = epsi_(0)A ((k_(1)k_(2))/(k_(2)d_(1) +k_(1)d_(2))) " ".......(i)`
Also `C_(eq) = (kepsi_(0)A)/(d_(1)+d_(2))`
From(i)and (ii)
`epsi_(0)A((k_(1)k_(2))/(d_(2)k_(1) + d_(1)+k_(2))) = epsi_(0)((k)/(d_(1)+d_(2)))`
`therefore K = (k_(1) K_(2)(d_(1)+d_(2)))/(d_(1)k_(1) + d_(1)k_(2))`
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