That's a good relevant equation, since at closest approach, the proton's velocity will be zero.AnthroMecha said:Homework Statement
A proton is moving with an initial velocity of 6.28x10^5 m/s straight towards a large conducting
sphere of radius 85.2 cm. The sphere has a charge of 4.83 μC on it. Protons have a mass of
1.69x10^-27 kg.
a) Determine how close the proton can come to the surface of the sphere.
b) What is the voltage on the surface of the sphere?
Homework Equations
possibly 1/2mv^2=-qΔV
The Attempt at a Solution
Don't even know where to begin on this one
SammyS said:That's a good relevant equation, since at closest approach, the proton's velocity will be zero.
What is the electric potential at the location of the proton, due to the sphere?
That's good for part (b).AnthroMecha said:I'm coming up with 50964.4V electric potential (I may be using the wrong equation).
I used V=(1/(4πε0))(q/r)
q=4.83μC, r=85.2cm
SammyS said:That's good for part (b).
To find the potential due to the sphere at the location of the proton at the time its velocity is 6.28×105 m/s, you have to know how far the proton is from the center of the sphere at that time.
Did they give that quantity ?
The Physics 2 exam covers a variety of topics, including electromagnetism, thermodynamics, optics, and quantum mechanics. It is important to review all of these topics, but pay extra attention to areas where you may have struggled in your coursework.
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