Question

A capacitor consists of two parallel plates, each with an area of 17.0 cm2 , separated by a distance of 0.150 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 400 V battery.
Part A

What is the capacitance of the capacitor?

C = ______ F

Part B

What is the charge on either plate?

Q = ____ C

Part C

How much energy is stored in the charged capacitor?

U = _____ J

Answer 1

Answer:

a) $C = 40.138\,pF$, b) $q = 16.056\,nC$, c) $U = 3.212\,\mu J$

Explanation:

a) The capacitance of two parallel plates capacitor with dielectric is given by the following expression:

$C = K\cdot \epsilon_{o}\cdot \frac{A}{d}$

Where:

$K$ - Dielectric constant.

$\epsilon_{o}$ - Vaccum permitivity.

$A$ - Plate area.

$d$ - Distance between plates.

Hence, the capacitance of the system is:

$C = (4.00)\cdot (8.854\times 10^{-12}\,\frac{F}{m} )\cdot \left(\frac{17\times 10^{-4}\,m^{2}}{0.150\times 10^{-2}\,m}\right)$

$C = 4.014\cdot 10^{-11}\,F$

$C = 40.138\,pF$

b) The charge can be found by using the definition of capacitance:

$q = C\cdot V_{batt}$

$q = (4.014\times 10^{-11}\,F)\cdot (400\,V)$

$q = 1.606\times 10^{-8}\,C$

$q = 16.056\,nC$

c) The energy stored in the charged capacitor is:

$U=\frac{1}{2}\cdot Q\cdot V_{batt}$

$U=\frac{1}{2}\cdot (1.606\times 10^{-8}\,C)\cdot (400\,V)$

$U = 3.212\times 10^{-6}\,J$

$U = 3.212\,\mu J$

Answer 2

Answer:

Part A: 40.12×10⁻¹² F

Part B: 1.6048×10⁻⁸ C.

Part C: 3.2096×10⁻⁶ J

Explanation:

Part A

The formula of the capacitance of a capacitor is given as,

C = εrε₀A/d............... Equation 1

Where, C = capacitance of the capacitor,  A = Area of the plate, d = distance of separation of the plates, εr = dielectric constant, ε₀ = permitivity of free space.

Given: A = 17 cm² = 0.0017 m², d = 0.150 cm = 0.0015 m, εr = 4.00, ε₀ = 8.85×10⁻¹² F/m

Substitute into equation 1

C = 4(0.0017)(8.85×10⁻¹²)/(0.0015)

C = 40.12×10⁻¹² F.

Part B

Using

Q = CV........................ Equation 2

Where Q = charge on either plates, V = Voltage.

Given: V = 400 V, C  = 40.12×10⁻¹² F

Substitute into equation 3

Q = 400(40.12×10⁻¹²)

Q = 1.6048×10⁻⁸ C.

Part C

The formula for the energy stores in a charged capacitor is given as

U = 1/2CV²..................... Equation 3

Where U = Energy stored in the capacitor

Given: C = 40.12×10⁻¹² F,  V = 400 V

Substitute into equation 3

U = 1/2(40.12×10⁻¹²)(400²)

U = 3.2096×10⁻⁶ J