JEE MAIN - Physics (2022 - 25th June Evening Shift - No. 8)
Two metallic plates form a parallel plate capacitor. The distance between the plates is 'd'. A metal sheet of thickness $${d \over 2}$$ and of area equal to area of each plate is introduced between the plates. What will be the ratio of the new capacitance to the original capacitance of the capacitor?
2 : 1
1 : 2
1 : 4
4 : 1
Explanation
$${C_{eq}} = {{{\varepsilon _0}A} \over {d - {d \over 2} + {d \over {2k}}}} = {{{\varepsilon _0}A} \over {{d \over 2}}} = {{2{\varepsilon _0}A} \over d}$$
If $$C = {{{\varepsilon _0}A} \over d}$$
$$ \Rightarrow {C_{eq}} = 2C$$ or $${{{C_{new}}} \over {{C_{old}}}} = {2 \over 1}$$
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