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.
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Options

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_(1)d_(2)+k_(2)d_(1)))`
D. `(2k_(1)k_(2))/(k_(1)+k_(2))`

Answer

C

Explanation

The given combination is equivalent to series connection of two separate capacitors `C_(1) and C_(2)` with dielectric slabs `k_(1) and k_(2)`, respectively.
`C_(1)=(k_(1)epsilon_(0)A)/(d_(1)) and C_(2)=(k_(2)epsilon_(0)A)/(d_(2))`
`C_("net")=(C_(1)xxC_(2))/(C_(1)+C_(2))=((k_(1)epsilon_(0)A)/(d_(1))xx(k_(2)epsilon_(0)A)/(d_(2)))/((k_(1)epsilon_(0)A)/(d_(1))+(k_(2)epsilon_(0)A)/(d_(2)))=(epsilon_(0)A)/(d)=(k_(1)k_(2)d)/(k_(1)d_(2)+k_(2)d_(1))`
Also, `C_("net")=(k epsilon_(0)A)/(d)`
Thus, `k=(k_(1)k_(2)d)/(k_(1)d_(2)+k_(2)d_(1))=(k_(1)k_(2)(d_(1)+d_(2)))/(k_(1)d_(2)+k_(2)d_(1))`