Calculating Potential Difference and Work in a Parallel Plate Capacitor

AI Thread Summary
The discussion focuses on calculating the potential difference and work involved in a parallel plate capacitor after the plates are pulled apart. With an initial area of 1 m², a spacing of 10 cm, and a potential difference of 1500 V, the new spacing is 11 cm after disconnection from the surroundings. The first two parts of the problem are manageable under the assumption of constant charge, but the third part raises challenges due to the lack of mass information needed for calculations. Participants emphasize the necessity of knowing the mass to determine the velocity of the plates when separated by 11 cm. Overall, the conversation highlights the complexities of applying energy conservation and kinematics in this scenario.
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Homework Statement



The area of the plates of a parallel plate capacitor is 1 m2, the spacing between the plates is 10 cm, the potential difference between the plates is 1500 V. After beign charged, the plates are disonnected electrically from the surroundings. The plates are then pulled farther apart so that the new spacing is 11 cm.


a. What is now the potential difference between the plates?
b. How much work is required to pull the plates apart?
c. Assume that the pulling apart of the plates is accomplished by a constant force of 10 N more than the force required to keep the plates apart. How fast are the plates moving away from one another when the plates are separated by 11 cm?


Homework Equations




W=F*dr

The Attempt at a Solution



The first parts are doable, if, like I'm assuming, the charge is going to be constant. However, without a mass given, I can't see how part c is going to work out. I figured it'd have something to do with energy conservation or kinematics, but I don't know how to solve for v without an m to put into the energy equation, or to figure the acceleration. Any help is appreciated.
 
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You're right. You need the mass of the plates to answer the question.
 

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