Question 194: Electrostats & Capacitance

Question

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?

Accepted Answer

$${C_{eq}} = {{{\varepsilon _0}A} \over {d - {d \over 2} + {d \over {2k}}}} = {{{\varepsilon _0}A} \over {{d \over 2}}} = {{2{\varepsilon _0}A} \over d}$$
If $$C = {{{\varepsilon _0}A} \over d}$$
$$ \Rightarrow {C_{eq}} = 2C$$ or $${{{C_{new}}} \over {{C_{old}}}} = {2 \over 1}$$

Answer

2 : 1

Answer

1 : 2

Answer

1 : 4

Answer

4 : 1

Electrostats & Capacitance

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?

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