Tutorial Solution To Parallel Plate Capacitor With Two Dielectrics PDF

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HW10...

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Solution for Open-Response Homework 10 Capacitance Solution to Open-Response Homework Problem 10.1(Parallel Plate Capacitor with Two Dielectrics) Problem: A square parallel plate capacitor with plate area A = 100cm2 has plate separation ℓ = 3mm. The airspace between the plates is partially filled by two d = 1mm thick dielectric slabs. One slab has κ1 = 2.5 and the other slab has κ2 = 4. Compute the capacitance, in the infinite parallel plate approximation of the system. Report both a symbolic and numeric value. Draw the electric field map with any bound charge. κ 1 κ 2

Solution Definitions Q ≡ Arbitrary Charge σ ≡ Charge Density A = 100cm2 ≡ Plate Area E0 ≡ Magnitude of Electric Field in Air κi ≡ Dielectric Constant i

κ1

Ei ≡ Magnitude of electric field in dielectric i d = 1mm ≡ Thickness of regions

κ2

∆V ≡ Potential Difference Across Plates ∆Vi ≡ Potential Difference Across Region

Strategy: Introduce charge on each plate. Compute the free space electric field, then correct for the dielectric. Compute the potential difference using the potential of a uniform field. Apply definition of capacitance. (a) Introduce Charge: Add an arbitrary charge Q = σA to the top plate and −Q = σA to the bottom plate. The σ must cancel out of the capacitance of we blew it. (b) Draw a Good Diagram: The equal and opposite charges added to the plates of the capacitor will produce a uniform field with magnitude E0 between the plates and zero outside. Draw 8 lines leaving the positive plate for this field. κ1 = 2.5 thins these lines by a factor of 2.5 giving 8/2.5 ≈ 3 lines crossing dielectric 1. Dielectric 2 thins the Free Space field by a factor of κ2 = 4, giving 2 lines crossing dielectric 2. Draw − charges where lines end. (c) Compute the Electric Field in Air: The electric field of an infinite plane of charge is ~E = 2ǫσ0 outward. If no dielectric were present the positive top conductor would contribute a field of σ2ǫ0 downward and the negative bottom conductor would contribute a field of σ2ǫ0 downward for a total field between the plates of ~0 = σ E ǫ0

Downward.

All charge is on the inner plates of the capacitor to cancel this field in the conductor. (d) Compute the Dielectric Fields: With the dielectrics present the region of the capacitor containing air is unchanged and has an electric field with magnitude E0 = σǫ0 . The dielectrics decrease this field by a factor of κ 1

so the magnitude of the fields in the dielectrics are E κ1 =

σ κ1 ǫ0

E κ2 =

σ κ2 ǫ0

(e) Compute the Potential Difference Across the Conductors: The potential difference across the conductors is the sum of the potential difference across the air space, dielectric 1, and dielectric 2, ∆V = ∆V0 + ∆Vκ1 + ∆Vκ2 Since the field in all regions points in the same direction, |∆V | = |∆V0 | + |∆Vκ1 | + |∆Vκ2 | For a uniform electric field the potential difference for a single region is |∆V | = Ed where d is the distance in the direction of the field. All regions in the problem have a thickness d = 1mm. The absolve value of the potential difference is then ¶ µ 1 dσ 1 + |∆V | = E0 d + Eκ1 d + Eκ2 d = 1+ κ1 κ2 ǫ0 |∆V | = 1.65

dσ ǫ0

(f) Apply Definition of Capacitance: By definition of capacitance, C=

Q = ∆V

Aσ µ dσ 1 + 1 + κ1 ǫ0 C=

1 κ2

¶ = µ d 1+

Aǫ0 1 κ1

+

1 κ2

¶

Aσ Aǫ0 = 0.61 dσ d 1.65 ǫ 0

(100 × 10 Aǫ0 C = 0.61 = 0.61 d

2

−4

C m2 )(8.85 × 10−12 Nm 2) = 54pF 0.001m

Total Points for Problem: 15 Points

Solution to Open-Response Homework Problem 10.2(Spherical Capacitor with Dielectric) Problem: The spherical capacitor at the right is formed from an inner conducting sphere with radius a = 4cm and an outer conducting shell of inner radius c = 12cm. Part of the airspace between the two conductor is filled with a dielectric shell with constant κ = 3. The dielectric has inner radius b = 8cm and outer radius c = 12cm. Compute the capacitance between the two conductors. Report both the symbolic and numeric value of the capacitance.

Solution

2

Definitions Q ≡ Charge on conductor κ ≡ Dielectric Constant ~ i ≡ Electric Field in Region i E ∆Vi ≡ Potential Difference Across Region i C ≡ Capacitance

Path of integration

Strategy: Introduce charge Q on inner conductor and −Q on outer conductor. Compute the field in the capacitor from Gauss’ Law. Integrate to compute the potential difference between the conductors, then apply the definition of capacitance. (a) Introduce Arbitrary Charge: Place a charge Q on the inner conductor and a charge −Q on the outer conductor. (b) Draw a Good Diagram: Select six field lines going out for the charge Q, so region II has six field lines. The dielectric in region III reduces the field by κ = 3, so two field lines cross region III. (c) Compute the Electric Field: The charge enclosed in region II is Q, so using Gauss’ Law in spherical coordinates, ~ II = kQ rˆ. E r2 In region III, the field without the dielectric is reduced by a factor of κ, so ~ III = kQrˆ. E κr2 (d) Compute the Potential Difference ∆VII : By definition of potential, ¯ Z b ¯ ¯ Z b ¯ ¯ ¯ ¯ kQ ¯¯ ~ II · rˆdr¯ = ¯ − |∆VII | = ¯¯ − E dr ¯ ¯ r2 ¯ a

a

¯µ ¯ ¶¯ ¯ µ ¶¯ ¯ kQ ¯ b ¯ ¯ 1 ¯ ¯ ¯ ¯ = ¯kQ 1 − ¯ |∆VII | = ¯ b r ¯a ¯ ¯ a ¯

Remove the absolute value and make sure the quantity is positive, ¶ µ 1 1 − |∆VII | = kQ a b (e) Compute ∆VIII :

¯ ¯ Z c ¯ ¯ Z c ¯ ¯ ¯ kQ ¯¯ ~EIII · rˆdr¯ = ¯ − dr |∆VIII | = ¯¯ − ¯ ¯ ¯ 2 b κr b ¯c ¶¯ ¯ ¶¯ ¯µ µ ¯ kQ ¯ ¯ ¯ kQ 1 1 ¯¯ ¯ ¯=¯ − |∆VIII | = ¯¯ ¯ ¯ ¯ κr b b ¯ κ c

Remove the absolute value so that the quantity is positive, |∆VIII | =

kQ κ

µ

¶ 1 1 − b c

(f) Compute Total Potential Difference: Since the fields point in the same direction in both region II and III, the magnitude of the potential differences add, so |∆V | = |∆VII | + |∆VIII |. Now, compute the total potential using κ = 3, ¶ µ ¶ µ kQ 1 1 1 1 − − . + |∆V | = |∆VII | + |∆VIII | = kQ b c a b κ 3

µ

1

¶ 1 1 1 − − + . b κb κc ¶ µa Q 1 1 1 1 |∆V | = − − + . κc 4πε0 a b κb |∆V | = kQ

(g) Apply Definition of Capacitance: The capacitance is defined as C=

Q = ∆V

Q Q 4πε0

µ

1 a

1 b

1 a

1 b

− +

1 κb

−

1 κc

¶

The symbolic form of the capacitance is then C=µ

4πε0 −

+

1 κb

−

1 κc

Substitute the radii,

¶

2

C=µ

C 4π(8.85 × 10−12 Nm 2) 1 4cm

−

1 8cm

+

1 (3)(8cm)

−

1 (3)(12cm)

C = 8pF. 4 point(s) : Electric Field in Airspace and Dielectric 4 point(s) : Potential Difference Across Dielectric 3 point(s) : Potential Difference Across Airspace 3 point(s) : Symbolic Expression for Capacitance 1 point(s) : Numeric Value of Capacitance Total Points for Problem: 15 Points

4

¶...