Calculating Potential Difference and Work in a Parallel Plate Capacitor

In summary, the parallel plate capacitor has an area of 1 m2 and a spacing of 10 cm, with a potential difference of 1500 V. After being charged and disconnected from the surroundings, the plates are pulled apart to a spacing of 11 cm. The potential difference between the plates is now unknown, and the work required to pull the plates apart is also unknown. Assuming the pulling apart is done with a constant force of 10 N more than the force required to keep the plates apart, the speed at which the plates are moving away from each other when separated by 11 cm is also unknown. To answer part c, the mass of the plates is needed for energy conservation or kinematics equations.
  • #1
lacunae
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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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  • #2
You're right. You need the mass of the plates to answer the question.
 

Related to Calculating Potential Difference and Work in a Parallel Plate Capacitor

1. What is a parallel plate capacitor?

A parallel plate capacitor is a type of capacitor that consists of two parallel conducting plates separated by an insulating material, or dielectric. It is used to store electrical energy by accumulating opposite charges on the two plates.

2. How does a parallel plate capacitor work?

A parallel plate capacitor works by creating an electric field between the two plates. When a voltage is applied across the plates, one plate becomes positively charged and the other becomes negatively charged. This creates an electric potential between the plates, allowing the capacitor to store energy.

3. What are the factors that affect the capacitance of a parallel plate capacitor?

The capacitance of a parallel plate capacitor is affected by three main factors: the area of the plates, the distance between the plates, and the type of dielectric material used between the plates. The larger the area and smaller the distance, the higher the capacitance will be. Different dielectric materials have different abilities to store electric charge, which also affects the capacitance.

4. How is the capacitance of a parallel plate capacitor calculated?

The capacitance of a parallel plate capacitor can be calculated using the formula C = ε0A/d, where C is the capacitance in farads, ε0 is the permittivity of free space (8.85 x 10^-12 F/m), A is the area of the plates in square meters, and d is the distance between the plates in meters.

5. What are the practical applications of parallel plate capacitors?

Parallel plate capacitors have many practical applications, including in electronic circuits, power grids, and energy storage devices. They are also used in sensors, filters, and tuning circuits. In addition, parallel plate capacitors are essential components in many types of electronic devices such as computers, smartphones, and televisions.

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