A proton at speed v = 3.00 × 105 m/s orbits at radius r = 1.00 cm outside a charged sphere. Find the sphere’s charge.

q = 1.04×10^-9C

Explanation:

The force on a charge orbiting at the center of the sphere (Fc) is must be equal to the electrostatic force acting on the charge.(Fq)

Fc = Fq

The force that causes a body to bev attracted towards the center of the sphere is the centripetal force.

Centripetal force (Fc) = ma = mv²/r where m is the mass of the body

v is the speed

r is the velocity

The electrostatic force acting on the body is derived according to the Coulomb's law of electrostatic attraction.

Fq = kqe/r²

Where k is the constant

e is the electron orbiting outside the sphere

r is the distance between the charge and the electron

Since fc=fq,

mv²/r = kqe/r²

Making the charge the subject of the formula we have;

mv²r = kqe

q = mv²r/ke

Given the mass of the charge =

8.84×10^-12kg

v = 3.0×10^5

r = 1.0cm = 0.01m

k = 1/4Π£o = 1/4Π×1.67×10^-27

k = 4.77×10^-27

e = 1.6×10^-19C

Substituting the values in the formula given,

q = 8.84×10^-12×(3.0×10^5)²(0.01)/(4.77×10^-27)×1.6×10^-19

q = 1.04×10^-9C

The sphere's charge is 1.04×10^-9

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explanation:

layers of the earth affect the waves and these effects are seen on instruiments such as seimometers