Equipotential surface of a infinite plane?

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The equipotential surface of an infinite plane of charge is a series of parallel planes. This is derived from the symmetry of the electric field created by the infinite plane, which is uniform and perpendicular to the surface. The discussion emphasizes considering symmetries to understand the shape, suggesting that if one grasps the concept of symmetries, further calculations may be unnecessary. Additionally, confirming findings through external sources is recommended. Understanding these principles is crucial for grasping the behavior of electric fields and potentials in physics.
squirrelschaser
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Homework Statement



So the equipotential surface of a point charge is sphere with the charge in the center, and the equipotential surface of a infinite line is a cylinder with the line of charge as the axis. I was wondering what is the shape of the equipotential surface of a infinite plane?

Homework Equations

The Attempt at a Solution

 
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squirrelschaser said:

Homework Statement



So the equipotential surface of a point charge is sphere with the charge in the center, and the equipotential surface of a infinite line is a cylinder with the line of charge as the axis. I was wondering what is the shape of the equipotential surface of a infinite plane?

Homework Equations

The Attempt at a Solution


i) consider symmetries. ii) compute it iii) make an educated guess. iv) confirm guess by looking up what other sources say the potential of a infinite plane of charge looks like. Actually if you think really hard about i) you can skip the others. If you don't know what 'symmetries' is all about you and you can't compute it, skip the first options and just do the others. Variety of choices.
 
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The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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