Question

A capacitor consists of two parallel plates, each with an area of 15.0 cm2 , separated by a distance of 0.200 cm . The material that fills the volume between the plates has a dielectric constant of 4.00. The plates of the capacitor are connected to a 200 V battery.Part AWhat is the capacitance of the capacitor?C = ______ FPart BWhat is the charge on either plate?Q = ____ CPart CHow much energy is stored in the charged capacitor?U = _____ J

Answer 1

Answer:

Part A.   $2.66 \times 10^{-11}\text{ F}$

Part B. $5.32\times 10^{-9}\text{ C}$

Part C. $5.32\times 10^{-7}\text{ J}$

Explanation:

Part A

The capacitance of a parallel-plate capacitor is given by

$C = \dfrac{\epsilon A}{d}$

A is the area of the plates, d is the separation distance and ∈ is the permittivity of the material between the plates.

∈ is given as $\epsilon_r\epsilon_0$

$\epsilon_r$ is the dielectric constant and $\epsilon_0$ is the permittivity of free space with a value of $8.854 \times 10^{-12} \text{ F/m}$.

Hence,

$C = \dfrac{4.00\times(8.854\times10^{-12}\text{ F/m})\times(15.0\times10^{-4} \text{ m}^2)}{0.2\times 10^{-2} \text{ m}} = 2.66 \times 10^{-11}\text{ F}$

Part B

The charge, Q, on a capacitor is given by Q = CV

where V is its voltage and C its capacitance.

$Q = (2.66 \times 10^{-11}\text{ F})\times(200\text{ V}) = 5.32\times 10^{-9}\text{ C}$

Part C

The energy of a capacitor is given by

$E = \frac{1}{2}QV$

$E = \frac{1}{2} \times (5.32\times10^{-9}\text{ C})\times (200\text{ V}) = 5.32\times10^{-7}\text{ J}$