This is a context-rich question where I don’t have access to the problem with it’s quantities, but rather I need to come up with a strategy to solve it (what equations to use, diagrams to draw, describe what’s happening etc). This question falls under the Electric, Gravitational and Magnetic Fields Unit.
Question:
You and a friend have been given the task of designing a display for the science department that will demonstrate the strength of the electric force. Your friend comes up with an idea that sounds neat theoretically, but you are not sure it is practical. They suggest you use an electric force to hold a glass marble in place on a sloped plywood ramp. They would get the electric force by attaching uniformly charged spheres in a semicircle near the bottom of the ramp (see diagram below). Your friend claims that if the charges on the marble and spheres and the slope of the ramp are chosen properly, the marble would be balanced at the centre of the semicircle. To test this idea, you examine a system with a ramp at some angle and five spheres with some charge arranged uniformly in a semicircle of some radius. The marble would sit in a slot of some length cut lengthwise into the center of the ramp. You measure the mass of the lightest marble you can find. What charge would you need to put on the marble?
What would I do to solve this? Thank you!
Thanks for asking. The key principle in solving problems with equilibrium is... equilibrium. So, if you have a sloped plywood and a marble, the marble can move along this incline. Thus, the force of gravity acts on it. On the other hand, we have electric charges that prevent the marble from rolling down, which means that the component of the electric force along the incline is equal to the component of the force of gravity that is along the slope.
Make a drawing and use arrows to indicate forces: the slope goes from right (up_ to left, the force of gravity is applied to the centre of the marble, you can project this force on the incline. The electric force (coulomb's force) acts along the incline in opposite direction. This will help you figure out how to start cracking the problem.
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