Phy213: General Physics III 5/20/2007 Chapter 25 Worksheet 1

Parallel Plate Capacitors:

1. Consider parallel plate capacitor (air filled) with a surface area of 225 cm2 and a charge of 1.5

µC (q) on each of its plates and a plate separation distance of 1.0x10-4 m.

a. What is the capacitance of the capacitor?

Ans. -9

o

AC = = 2.0x10 F

dε

b. What is the potential difference across the capacitor?

Ans. q

V = = 750VC

c. How much energy is stored in the capacitor?

Ans. 2 -41

U = CV = 5.6x10 J2

d. If the capacitor were filled with a dielectric material, κ=3.3 (while still maintaining the same

amount of charge on the plates) what is the new capacitance?

Ans. -9

airC = C = 6.6x10 Fκ

e. How much charge would be stored in this capacitor with this dielectric material, at the same potential difference as in part (b)?

Ans. -6q = CV = 5.0x10 C

Cylindrical Capacitors:

2. A cylindrical air-filled capacitor (R1 = 0.5x10-4m and R2 = 0.5x10

-2m) of length L=2.0 m has a potential difference of +120 V (outside – inside) between the inner and outer conductors.

a. What is the capacitance of this capacitor?

Ans. ( )2

1

-11o

R

R

2 LC = = 2.4x10 F

ln

πε

b. How much charge is stored on each face of the capacitor?

Ans. -9q = CV = 2.9x10 C

c. Determine the equation for the electric field vector inside the capacitor.

Ans. Apply Gauss’s Law for a Gaussian cylinder (noting that R1 & R2 << L):

ˆE cylinder shaft r

o o o

q q = EA = E(2 rL) = E or i

2 rL 2 r

λπ

ε π ε π ε

Φ ⇒ ≅

�

+

+

+

+

+

+

+

+

+

-

-

-

-

-

-

-

-

-

Phy213: General Physics III 5/20/2007 Chapter 25 Worksheet 2

Spherical Capacitors:

3. A spherical air-filled capacitor (R1 = 0.5x10-4m and R2 = 0.5x10

-2m) has a potential difference of +120 V (outside – inside) between the surfaces.

a. What is the capacitance of this spherical capacitor?

Ans. -151 2

o

2 1

R RC = 4 = 5.6x10 F

R -Rπε

b. How much charge is stored on each face of the capacitor?

Ans. -13q = CV = 6.7x10 C

c. Determine the equation for the electric field vector inside the capacitor.

Ans. Apply Gauss’s Law for a “Gaussian sphere”: ˆ2E sphere r2

o o

q q=EA =E(4 r )= E i

4 rπ

ε π εΦ ⇒ ≅

�

4. Consider a spherical biological cell of radius 5.0 x 10-6 m. The membrane of the cell has capacitive properties no different that the electrolytic capacitors used in the lab. For this cell, the potential difference across the cell membrane (Vin-Vout) is -90 mV and the capacitance per unit surface area is 0.01 F/m2.

a. What is the capacitance of the cell?

Ans. ( )2 2 -12Fcell cellmC = 0.01 4 R = 3.1x10 Fπ

b. How much charge is stored on each face of the membrane?

Ans. -13

cellq = C V = 2.8x10 C

c. How much energy is stored in the cell membrane?

Ans. 2 -14

cell

1U = C V = 1.3x10 J

2

d. If the dielectric constant for the cell membrane is κ = 9.0, what is the thickness of the cell

membrane?

Ans. since Rcell ≈ R1 ≈ R2, the cell capacitance can be approximated by:

2 2

-9cell cello 2 1 o

2 1

R RC 4 or d = R -R 4 = 8.0x10 m

R -R Cπκε πκε

≈ ≈

5. The measured capacitance for a blood cell is measured to be 1.6x10-2 pF. Assume that the membrane thickness and the dielectric constant is the same as for the previous problem.

a. What is the surface area of the cell?

Ans. 2

-12 2cell cello o cell

o

4 R A CdC = A = = 1.6x10 m

d d

πκε κε

κε

≈ ⇒

b. Calculate the specific capacitance (capacitance per unit area] for this cell.

Ans. 2

-14

F-12 2 m

cell

C 1.6x10 F = = = 0.010

A 1.6x10 m

R

R

Phy213: General Physics III 5/20/2007 Chapter 25 Worksheet 3

Capacitors in Series & Parallel

5. Two capacitors, C1=10µF and C2=50µF, are connected together in parallel. The potential

difference across the capacitors is 100V. a. What is the equivalent capacitance of the parallel capacitors.

Ans. Ceq = C1 + C2 = 60µF

b. How much total energy is stored in each capacitor separately and combined?

Ans. 21

tot eq2U = C V =0.3 J

2 21 1

1 1 2 22 2U = C V =0.05 J & U = C V =0.25 J

c. How much charge is stored in each capacitor separately and combined?

Ans. -3

tot eqq = C V = 6.0x10 C

-3 -3

1 1 2 2q = C V = 1.0x10 C & q = C V = 5.0x10 C

6. Two capacitors, C1=10µF and C2=50µF, are connected together in series. The potential

difference from the end of C1 to the opposite end of C2 is 100V.

a. What is the equivalent capacitance of the series capacitors.

Ans. -3

eq

1 2

1C = = 8.3x10 C

1 1+

C C

b. How much total energy is stored in each capacitor separately and combined?

Ans. 21

tot eq2U = C V =0.042 J , using the charge values obtained from part c (below):

2 2

1 2

1 2

q qU = =0.035 J & U = =0.0069 J

2C 2C

c. How much charge is stored in each capacitor separately and combined?

Ans. -4tot 1 2 eqq = q = q = C V =8.3x10 C