Question

A proton orbits just at the surface of a charged sphere of radius 3.53 cm. If the speed of the proton is 2.02 X105 m/s, what is the charge on the sphere?

Answer 1

To solve this problem, apply the equilibrium condition given from the electrostatic force and the centripetal force of the body. Said equilibrium condition can be described under the function,

$F_c = F_q$

$\frac{mv^2}{r} = \frac{kQ_{proton}Q_{Sphere}}{r^2}$

Here,

m = Mass of proton

Q = Charge of each object

k = Coulomb's constant

v = Velocity

Our values are given as,

$q= 1.6*10^{-19} C$

$m = 1.67*10^{-27} kg$

$v = 2.02* 10^5m/s$

$r = 3.53cm = 3.53*10^{-2} m$

Rearranging and replacing we have,

$\frac{mv^2}{r} = \frac{kQ_{proton}Q_{Sphere}}{r^2}$

$Q_{sphere}= \frac{mv^2 r }{kQ_{proton}}$

$Q_{sphere} = \frac{(1.67*10^{-27})(2.02*10^5)^2(3.53*10^{-2})}{(9*10^9)(1.6*10^{-19})}$

$Q_{Sphere} = 1.6704*10^{-9}C$

$Q_{Sphere} = 1.67nC$

Therefore the charge on the Sphere is 1.67nC