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A parallel plate capacitor has a capacity $C$. The separation between the plates is doubled and a dielectric medium is inserted between the plates. If the capacity is $3 C$, then the dielectric constant of the medium will be
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Verified Answer
The correct answer is:
$6$
We know that, capacitance of a parallel plate capacitor,
$$
C=\frac{\varepsilon_{0} A}{d}
$$
where, $A=$ area of plate
$d=$ separation between the plates.
When $d^{\prime}=2 d$, then
$C^{\prime}=\frac{\varepsilon_{0} A K}{d^{\prime}}$
$\Rightarrow \quad 3 C=\frac{\varepsilon_{0} A K}{2 d}$
$\left[\because C^{\prime}=3 C\right]$
$\Rightarrow \quad 3 C=\frac{\varepsilon_{0} A}{d} \cdot \frac{K}{2}$
$\Rightarrow \quad 3 C=C \cdot \frac{K}{2} \quad$ [from Eq. (i)]
$\Rightarrow \quad 3=\frac{K}{2}$
$\Rightarrow \quad K=6$
$$
C=\frac{\varepsilon_{0} A}{d}
$$
where, $A=$ area of plate
$d=$ separation between the plates.
When $d^{\prime}=2 d$, then
$C^{\prime}=\frac{\varepsilon_{0} A K}{d^{\prime}}$
$\Rightarrow \quad 3 C=\frac{\varepsilon_{0} A K}{2 d}$
$\left[\because C^{\prime}=3 C\right]$
$\Rightarrow \quad 3 C=\frac{\varepsilon_{0} A}{d} \cdot \frac{K}{2}$
$\Rightarrow \quad 3 C=C \cdot \frac{K}{2} \quad$ [from Eq. (i)]
$\Rightarrow \quad 3=\frac{K}{2}$
$\Rightarrow \quad K=6$
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