Question 458: Electrostats & Capacitance

Question

A parallel plate air capacitor has capacity ' $C$ ' and distance of separation between plates is ' $d$ '. If a conducting sheet of thickness $\frac{2 d}{3}$ is inserted in between the plates, the capacitance becomes $C_1$. The ratio of $\frac{C_1}{C}$ is

Accepted Answer

Before inserting conductor,
$$ \mathrm{C}=\frac{\varepsilon_0 \mathrm{~A}}{\mathrm{~d}} $$
After inserting conductor of thickness $\frac{2 \mathrm{~d}}{3}$,
$$ \begin{aligned} & d^{\prime}=d-\frac{2 d}{3}=\frac{d}{3} \ & C_1=\frac{\varepsilon_0 A}{d^{\prime}}=\frac{3 \varepsilon_0 A}{d} \end{aligned} $$
The required ratio,
$$ 	herefore \quad \frac{C_1}{C}=\frac{\frac{3 \varepsilon_0 A}{d}}{\frac{\varepsilon_0 A}{d}}=\frac{3}{1} $$

Answer

$2: 1$

Answer

$4: 1$

Answer

$3: 1$

Answer

$5: 1$

Electrostats & Capacitance

A parallel plate air capacitor has capacity ' C ' and distance of separation between plates is ' d '. If a conducting sheet of thickness \frac{2 d}{3} is inserted in between the plates, the capacitance becomes C_1. The ratio of \frac{C_1}{C} is

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