Academic Integrity: tutoring, explanations, and feedback — we don’t complete graded work or submit on a student’s behalf.

Constants A homemade capacitor is assembled by placing two 10-in. pie pans 4 cm

ID: 1879832 • Letter: C

Question

Constants

A homemade capacitor is assembled by placing two 10-in. pie pans 4 cm apart and connecting them to the opposite terminals of a 9-V battery.

Part A

Part complete

Estimate the capacitance.

Express your answer using one significant figure.

1×1011

Part B

Estimate the charge on each plate.

Express your answer using one significant figure.

nothing

Part C

Estimate the electric field halfway between the plates.

Express your answer using one significant figure.

nothing

*Note that Part A is correct*

SubmitRequest Answer

Part D

Estimate the work done by the battery to charge the plates.

Express your answer using one significant figure.

nothing

SubmitRequest Answer

Part E

Part complete

Which of the above values change if a dielectric is inserted?

SubmitPrevious Answers

Constants

A homemade capacitor is assembled by placing two 10-in. pie pans 4 cm apart and connecting them to the opposite terminals of a 9-V battery.

Part A

Part complete

Estimate the capacitance.

Express your answer using one significant figure.

C =

1×1011

  F

Part B

Estimate the charge on each plate.

Express your answer using one significant figure.

Q =

nothing

  C  

Part C

Estimate the electric field halfway between the plates.

Express your answer using one significant figure.

E =

nothing

  V/m  

*Note that Part A is correct*

SubmitRequest Answer

Part D

Estimate the work done by the battery to charge the plates.

Express your answer using one significant figure.

W =

nothing

  J  

SubmitRequest Answer

Part E

Part complete

Which of the above values change if a dielectric is inserted?

Capacitance. Charge. Electric field. Work done by the battery.

SubmitPrevious Answers

Explanation / Answer

A)

Capacitance is

C = A*eo / d

A = pi*r^2

diameter = 10 in = 0.254 m

radius = 0.254 m / 2 = 0.127 m

C = (pi x 0.127^2 x 8.854 x 10^-12) / (4 x 10^-2)

C = 1.12 x 10^-11 F

b)

Q = C x V

Q = 1.12 x 10^-11 F x 9 V

Q = 1 x 10^-10 C

c)

E = V/d

E = 9V / (4 x 10^-2 m)

E = 225 V/m

d)

W = 1/2*(Q*V)

W = 0.5 x 1 x 10^-10 x 9

W = 4.5 x 10^-10 J

e) Capacitance.