Help for easy electric problems

[x-men]

A proton with speed v=3 . 10*5m/s , (.=coss) orbits just outside the charged sphere of radius = 1cm What is the charge of sphere?

 

[Tide]

What, exactly, have you done so far?

 

[quicknote]

Based on the given information, we can use the formula for centripetal force to determine the charge of the sphere. The centripetal force, F, is equal to the product of the mass of the proton, m, and the square of its velocity, v, divided by the radius of the orbit, r:

F = mv^2/r

Since the proton is in a circular orbit, the force of attraction between the proton and the charged sphere is equal to the centripetal force. We can express this as:

F = k(q1q2)/r^2

Where k is the Coulomb's constant and q1 and q2 are the charges of the proton and the charged sphere, respectively.

Equating the two expressions for F, we get:

mv^2/r = k(q1q2)/r^2

Rearranging this equation, we can solve for q2, the charge of the sphere:

q2 = (mv^2r)/kq1

Substituting the given values, we get:

q2 = [(1.67 x 10^-27 kg)(3 x 10^5 m/s)^2(0.01 m)]/(8.99 x 10^9 Nm^2/C^2)(1.6 x 10^-19 C)

Simplifying, we get:

q2 = 3.75 x 10^-14 C

Therefore, the charge of the sphere is 3.75 x 10^-14 Coulombs.