JEE MAIN - Physics (2021 - 27th July Morning Shift - No. 7)
In the reported figure, a capacitor is formed by placing a compound dielectric between the plates of parallel plate capacitor. The expression for the capacity of the said capacitor will be :
(Given area of plate = A)
_27th_July_Morning_Shift_en_7_1.png)
(Given area of plate = A)
$${{15} \over {34}}{{K{\varepsilon _0}A} \over d}$$
$${{15} \over 6}{{K{\varepsilon _0}A} \over d}$$
$${{25} \over 6}{{K{\varepsilon _0}A} \over d}$$
$${9 \over 6}{{K{\varepsilon _0}A} \over d}$$
Explanation
$${1 \over {{C_{eff}}}} = {d \over {K{ \in _0}A}} + {{2d} \over {3K{ \in _0}A}} + {{3d} \over {5K{ \in _0}A}}$$
$${C_{eff}} = {{15K{\varepsilon _0}A} \over {34d}}$$
$${C_{eff}} = {{15K{\varepsilon _0}A} \over {34d}}$$
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