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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 thought as a dielectric slab of thickness `d(=d_(1)+d_(2))` and effective dielectric constant k. The k is :

A

`(k_1d_1+k_2d_2)/(d_1+d_2)`

B

`(k_1+d_1+k_2d_2)/(k_1+k_2)`

C

`(k_1k_2(d_1+d_2))/((k_1d_1+k_2d_2))`

D

`(2k_1k_2)/(k_1+k_2)`

Text Solution

Verified by Experts

The correct Answer is:
C
The capacitance of parallel plate capacitor filled with dielectric block of thickness `d_1` and dielectric constant `k_1` is given by
`C_1 = (k_1epsilon_0A)/(d_1)`
Similarly, capacitance of parallel plate capacitor filled with dielectric block of thickness `d_2` and dielectric constant `k_2` is given by
`C_2 = (k_2epsilon_0A)/(d_2)`
Since the two capacitors are in series combination, the equivalent capacitance is given by
`1/C = 1/(C_1) + 1/(C_2)`
or `C=(C_1C_2)/(C_1+C_2)= ((k_1epsilon_0A)/(d_1).(k_2epsilon_0A)/(d_2))/((k_1epsilon_0A)/(d_1)+(k_2epsilon_0A)/(d_2))=(k_1k_2 epsilon_0A)/(k_1d_2+k_2d_1)`
But the equivalent capacitances is given by
`C= (kepsilon_0A)/(d_1+d_2)`
On comparing, we have
`k= (k_1k_2(d_1+d_2))/(k_1d_2+k_2d_1)`
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