Question 676: Electrostatics & Capacitance

Question

Capacity of a parallel plate capacitor in the absence of dielectric medium is $C_0$. A sheet of dielectric constant $k$ and thickness of one-fourth of the plate separation is inserted between the plates. If the new capacity is $C$, then $\frac{C}{C_0}$ is

Accepted Answer

Without dielectric, $C_0=\frac{\varepsilon_0 A}{d}$ for partial insertion of dielectric,
$$ C=\frac{\epsilon_0 A}{d-t\left(1-\frac{1}{k}ight)} $$
On putting $t=\frac{d}{4}$, we get
$$ \begin{aligned} & C=\frac{\varepsilon_0 A}{d-\frac{d}{4}\left(1-\frac{1}{k}ight)} \ & C=\frac{4 k \varepsilon_0 A}{d(3 k+1)} \ & \frac{C}{C_0}=\frac{4 k}{3 k+1} \end{aligned} $$

Answer

$\frac{3 k+1}{2 k}$

Answer

$\frac{3 k+1}{4 k}$

Answer

$\frac{2 k}{3 k+1}$

Answer

$\frac{4 k}{3 k+1}$

Electrostatics & Capacitance

Capacity of a parallel plate capacitor in the absence of dielectric medium is C_0. A sheet of dielectric constant k and thickness of one-fourth of the plate separation is inserted between the plates. If the new capacity is C, then \frac{C}{C_0} is

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