A parallel-plate capacitor with plate area 2.0 cm2and air-gap separation 0.50mm

Question

A parallel-plate capacitor with plate area 2.0 cm2and air-gap separation 0.50mm is connected to a 9.0-V battery, and fully charged. The battery is then disconnected.

Part A

What is the charge on the capacitor?

Express your answer to two significant figures and include the appropriate units.

nothingnothing

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Part B

The plates are now pulled to a separation of 0.75mm. What is the charge on the capacitor now?

Express your answer to two significant figures and include the appropriate units.

19.527 times 10 Superscript negative 1219.527•1012Upper CC

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Part C

What is the potential difference between the plates now?

Express your answer to two significant figures and include the appropriate units.

nothingnothing

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Part D

How much work was required to pull the plates to their new separation?

Express your answer to two significant figures and include the appropriate units.

A parallel-plate capacitor with plate area 2.0 cm2and air-gap separation 0.50mm is connected to a 9.0-V battery, and fully charged. The battery is then disconnected.

Part A

What is the charge on the capacitor?

Express your answer to two significant figures and include the appropriate units.

Q =

nothingnothing

SubmitPrevious AnswersRequest Answer

Incorrect; Try Again; 3 attempts remaining

Part B

The plates are now pulled to a separation of 0.75mm. What is the charge on the capacitor now?

Express your answer to two significant figures and include the appropriate units.

Q =

19.527 times 10 Superscript negative 1219.527•1012Upper CC

SubmitPrevious AnswersRequest Answer

Incorrect; Try Again; 3 attempts remaining

Part C

What is the potential difference between the plates now?

Express your answer to two significant figures and include the appropriate units.

Vfinal =

nothingnothing

SubmitRequest Answer

Part D

How much work was required to pull the plates to their new separation?

Express your answer to two significant figures and include the appropriate units.

W =

Explanation / Answer

(a)

Capacitance of capacitor,

C = e0*A / d

C = 8.85*10^(-12)*2*10^(-4) / 0.5*10^(-3)

C = 3.54*10^(-12) F

Charge on capacitor,

Q = CV =  3.54*10^(-12) * 9

Q = 31.86*10^(-12) C

(b)

since battery is disconnected, charge remains same.

so, Q' = 31.86*10^(-12) C

(c)

New capacitance, C' = 8.85*10^(-12)*2*10^(-4) / 0.75*10^(-3)

C' = 2.36*10^(-12) C

New voltage, V' = Q' / C'

V' = 31.86*10^(-12) / 2.36*10^(-12)

V' = 13.5 V

(d)

Work required to pull the plates to new separation,

W = V'*Q

W = 13.5* 31.86*10^(-12)

W = 4.3*10^(-10) J

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