Question

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?

Answer 1

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

$KE = PE$

$\frac{1}{2} mv^2 = \frac{kq_1q_2}{r}$

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

$q_2 = \frac{\frac{1}{2} mv^2 r}{kq_1}$

Replacing,

$q_2 = \frac{\frac{1}{2}(1.67*10^{-27})(3*10^5)^2(7*10^{-2})}{(9*10^9)(1.6*10^{-19})}$

$q_2 = 3.6531nC$

Therefore the charge on the sphere is 3.6531nC