Parallel plate capacitor by an electric field between the plates

AI Thread Summary
To prevent the 10 gram pith ball with a charge of 1μC from sliding down the vertical plate of a parallel plate capacitor, the electric force must counteract the gravitational force acting on the ball. The electric field between the plates generates a force that attracts the ball towards the plate, while gravity pulls it downward. The minimum voltage required can be calculated by considering the balance of these forces, factoring in the coefficient of static friction. Understanding the direction and magnitude of the electric force is crucial for solving the problem. A diagram illustrating all forces acting on the pith ball will aid in visualizing the scenario.
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Homework Statement



A 10 gram pith ball having a charge of 1μC is electrostatically attracted to one of the inside surfaces of a large parallel plate capacitor by an electric field between the plates. The plates are vertically oriented so that the electric field between the plates is horizontal. The distance between the plates of the capacitor is 10 cm, and the coefficient of static friction between the pith ball and the plate is 0.5.
What is the minimum voltage needed to keep the ball from sliding down the plate from gravity? Draw a picture to show all forces on the pith ball.

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That pith ball is "stuck" to a plate by electric force. How big is that force, and what is its direction? What other forces act on the pith ball?

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