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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 given capacitor is equivalent to two capacitors joined in series,
where `C_(1)=(K_(1) epsilon_(0)A)/(d_(1)),C_(2)=(K_(2) epsilon_(0)A)/(d_(2))`
`:. (1)/(C_(eq))=(1)/(C_(1))+(1)/(C_(2))=(d_(1))/(K_(1)epsilon_(0)A)+(d_(2))/(K_(2) epsilon_(0)A)`
`=(d_(1)K_(2)+K_(1)d_(2))/(epsilon_(0)A K_(1)K_(2))`
or `C_(eq)=(epsilon_(0)A K_(1)K_(2))/(d_(1)K_(2)+K_(1)d_(1))`but `C_(eq)=(epsilon_(0)AK)/(d_(1)+d_(2))`
`:. K=(K_(1)K_(2)(d_(1)+d_(2)))/(d_(1)K_(2)+K_(1)d_(2))`.
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