Question 35: Electrostats & Capacitance

Question

When a dielectric slab of dielectric constant $\mathrm{k}=2$ is used to fill the space between the plates of a parallel plate capacitor, the capacitance of the capacitor is found to be $20 \mu \mathrm{~F}$. What will be the capacitance when the slab is replaced with air?

Accepted Answer

The capacitance with a dielectric is related to the air (vacuum) capacitance by
$$C = k\,C_0$$
where
Ã¢ÂÂ¢ $C=20\ \mu	ext F$ is the measured value with dielectric constant $k=2$,
Ã¢ÂÂ¢ $C_0$ is the capacitance with air.
So
$$C_0 = \frac{C}{k} = \frac{20\ \mu	ext F}{2} = 10\ \mu	ext F\,. $$
Answer: 10 ÃÂ¼F (Option A).

Answer

$10 \mu F$

Answer

$40 \mu F$

Answer

$20 \mu F$

Answer

$5 \mu F$

Electrostats & Capacitance

When a dielectric slab of dielectric constant \mathrm{k}=2 is used to fill the space between the plates of a parallel plate capacitor, the capacitance of the capacitor is found to be 20 \mu \mathrm{~F}. What will be the capacitance when the slab is replaced with air?

Choose the correct answer: