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prohojiy [21]
3 years ago
8

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
Physics
2 answers:
Llana [10]3 years ago
3 0

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

liq [111]3 years ago
3 0

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

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