d)
W = final energy - initial energy
W = 0.5* QV - 0.5* qv
W = 0.5* 1.8*10^-10* 34.2 - 0.5* 1.8* 10^-10* 12
W = 1.998*10^-9 J
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Problem 3 A parallel-plate capacitor with plate area 2.20 cm2 and air-gap separation 0.13 mm is...
Problem 3 A parallel-plate capacitor with plate area 5.00 cm2 and air-gap separation 0.29 mm is connected to a 12.00 V battery, and fully charged. The battery is then disconnected (a) What is the charge on the capacitor? 1.83x10-10 (c You are correct. Computer's answer now shown above Your receipt is 160-9143 Previous Tries (b) The plates are now pulled to a separation of 0.48 mm. What is the charge on the capacitor now? 183x1010 (c You are correct. Computer's...
A parallel-plate capacitor
with plate area 4.60 cm2 and air-gap separation 0.78 mm is
connected to a 12.00 V battery, and fully charged. The battery is
then disconnected. (a) What is the charge on the capacitor? (b) The
plates are now pulled to a separation of 0.98 mm. What is the
charge on the capacitor now? (c) What is the potential difference
across the plates now? (d) How much work was required to pull the
plates to their new separation?...
HELP PLS!!
Problem 3 A parallel-plate capacitor with plate area 8.60 cm2 and air-gap separation 0.43 mm is connected to a 12.00 V battery, and fully charged. The battery is then disconnected. (a) What is the charge on the capacitor? Submit Answer Tries 0/6 (b) The plates are now pulled to a separation of 0.68 mm. What is the charge on the capacitor now? Submit Answer Tries 0/6 c) What is the potential difference across the plates now? Submit Answer...
Problem3 A parallel-plate capacitor with plate area 8.30 cm2 and air-gap separation 0.57 mm is connected to a 12.00 V battery, and fully charged. The battery is then disconnected. a) What is the charge on the capacitor? Submit AnswerTries 0/6 atery and fully charged. The (b) The plates are now pulled to a separation of 0.81 mm. What is the charge on the capacitor now? Subnmit AnswerTries 0/6 (c) What is the potential difference across the plates now? Submit Answer...
85. A parallel-plate capacitor with plate area 3.0 cm2 and air- gap separation 0.50 mm is connected to a 12-V battery, and fully charged. The battery is then disconnected. (a) What is the charge on the capacitor? (b) The plates are now pulled to a separation of 0.75 mm. What is the charge on the capacitor now? (c) What is the potential difference between the plates now? (d) How much work was required to pull the plates to their new...
Problem 3 A parallel-plate capacitor with plate area 5.00 cm and air-gap separation 0.29 mm is connected to a 12.00 V battery, and fully charged. The battery is then disconnected. (a) What is the charge on the capacitor? Submit Answer Tries 0/6 (b) The plates are now pulled to a separation of 0.48 mm. What is the charge on the capacitor now? Submit Answer Tries 0/6 (c) What is the potential difference across the plates now? Submit Answer Tries 0/6...
A parallel-plate capacitor with plate area 2.5 cm2 and air-gap separation 0.50 mm is connected to a 12-V battery, and fully charged. The battery is then disconnected. The plates are now pulled to a separation of 0.75 mm. What is the potential difference between the plates now? Express your answer to two significant figures and include the appropriate units. It is asking for Vfinal. I got that the potential difference is 18 V but do not know how to get...
A parallel-plate capacitor with plate area 2.5 cm2 and air-gap separation 0.50 mm is connected to a 12-V battery, and fully charged. The battery is then disconnected. The plates are now pulled to a separation of 0.75 mm. What is the potential difference between the plates now? Express your answer to two significant figures and include the appropriate units. It is asking for Vfinal. I got that the potential difference is 18 V but do not know how to get...
A parallel-plate capacitor with plate area 4.0cm^2 and air-gap separation 0.50❝mm is connected to a 9.0-V battery, and fully charged. The battery is then disconnected. What is the charge on the capacitor? The plates are now pulled to a separation of 0.75?mm. What is the charge on the capacitor now? What is the potential difference between the plates now? How much work was required to pull the plates to their new separation?
A parallel-plate air capacitor of area A = 12.1 cm2 and plate separation of d = 3.90 mm is charged by a battery to a voltage of 53.0 V. a)What is the charge on the capacitor? b)If a dielectric material with κ = 5.00 is inserted so that it fills the volume between the plates (with the capacitor still connected to the battery), how much additional charge will flow from the battery onto the positive plate?