When a capacitor of capacitance C is filled with a dielectric (i.e. thickness of the dielectric is equal to the spacing between the plates) of dielectric constant k, what is the new capacitance?

C k

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\frac{C}{k}

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C k^{2}

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\frac{C}{k^{2}}

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Solution

The correct option is A $C k$
When t=d for the dielectric introduced,
The formula for new capacitance,
$C^{'} = \frac{ε_{0} A}{d - t + \frac{t}{k}}$
It changes to,
$C^{'} = \frac{k ε_{0} A}{t}$
Since t=d,
C'=kC

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Q. When a capacitor of capacitance C is filled with a dielectric (i.e. thickness of the dielectric is equal to the spacing between the plates) of dielectric constant k, what is the new capacitance?

Q. A parallel plate capacitor with air between the plates has a capacitance of 9

p F

. The separation between its plates is

d

. The space between the plates is now filled with two dielectrics. One of the dielectric has dielectric constant

K_{1} = 3

and thickness

\frac{d}{3}

while the other one has dielectric constant

K_{2} = 6

and thickness

\frac{2 d}{3}

. Capacitance of the capacitor is now :

A parallel plate capacitor with air between the plates has a capacitance of $9 p F$ . The separation between its plates is $d$ . The space between the plate is now partially filled with two dielectrics. One of the dielectric has dielectric constant $K_{1} = 3$ and thickness $\frac{d}{3}$ while the other one has dielectric constant $K_{2} = 6$ and thickness $\frac{d}{3}$ . The remaining $\frac{d}{3}$ space is filled with air. The new capacitance is :