JEE MAIN - Physics (2019 - 10th January Morning Slot - No. 16)
A parallel plate capacitor is of area 6 cm2
and a separation 3 mm. The gap is filled with three dielectric
materials of equal thickness (see figure) with dielectric constants K1 = 10, K2 = 12 and K3 = 14. The dielectric constant of a material which when fully inserted in above capacitor, gives same capacitance would be -
_10th_January_Morning_Slot_en_16_1.png)
12
36
14
4
Explanation
Let dielectric constant of material used be K.
$$ \therefore $$ $${{10{ \in _0}\,A/3} \over d} + {{12{ \in _0}\,A/3} \over d} + {{14{ \in _0}\,A/3} \over d} = {{K{ \in _0}A} \over d}$$
$$ \Rightarrow $$ K $$=$$ 12
$$ \therefore $$ $${{10{ \in _0}\,A/3} \over d} + {{12{ \in _0}\,A/3} \over d} + {{14{ \in _0}\,A/3} \over d} = {{K{ \in _0}A} \over d}$$
$$ \Rightarrow $$ K $$=$$ 12
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