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A proton with a speed of 3.000×105 m/s has a circular orbit just outside a uniformly charged sphere of radius 7.00 cm. What is the charge on the sphere?

User TJD
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1 Answer

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To solve this problem we will apply the principles of energy conservation. The kinetic energy in the object must be maintained and transformed into the potential electrostatic energy. Therefore mathematically


KE = PE


(1)/(2) mv^2 = (kq_1q_2)/(r)

Here,

m = mass (At this case of the proton)

v = Velocity

k = Coulomb's constant


q_(1,2) = Charge of each object

r= Distance between them

Rearranging to find the second charge we have that


q_2 = ((1)/(2) mv^2 r)/(kq_1)

Replacing,


q_2 = ((1)/(2)(1.67*10^(-27))(3*10^5)^2(7*10^(-2)))/((9*10^9)(1.6*10^(-19)))


q_2 = 3.6531nC

Therefore the charge on the sphere is 3.6531nC

User Colm Bhandal
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