A proton is launched from the surface of a 2.0 cm diameter glass sphere that is charged to -25.0 nC. What is the speed of the proton when it is launched so that it just escapes the influence of the charge of the glass sphere?
To find the speed of the proton when it just escapes the influence of the charged glass sphere, we need to consider the balance between the electrical force and the centripetal force acting on the proton.
The electrical force between the proton and the negatively charged glass sphere is given by Coulomb's law:
F_electric = (k * q1 * q2) / r^2
Where k is the electrostatic constant (9.0 x 10^9 N m^2/C^2), q1 and q2 are the charges of the two objects (proton and glass sphere), and r is the distance between them.
The centripetal force acting on the proton is given by:
F_centripetal = (m * v^2) / r
Where m is the mass of the proton and v is its velocity.
Since we're looking for the minimum speed at which the proton escapes the influence of the charged glass sphere, the electrical force and the centripetal force must cancel each other out. Therefore:
F_electric = F_centripetal
(k * |Q| * |q|) / r^2 = (m * v^2) / r
|Q| = absolute value of the charge on the glass sphere (25.0 nC = 25.0 x 10^-9 C)
|q| = absolute value of the charge on the proton (1.6 x 10^-19 C)
m = mass of the proton (1.67 x 10^-27 kg)
r = radius of the glass sphere (1.0 cm = 0.01 m)
Plugging in these values, the equation becomes:
(9.0 x 10^9 N m^2/C^2) * (25.0 x 10^-9 C) * (1.6 x 10^-19 C) / (0.01 m)^2 = (1.67 x 10^-27 kg) * v^2 / (0.01 m)
Simplifying further:
225 * 10^(-18) / 10^(-4) = (1.67 x 10^-27) * v^2
225 * 10^(-14) = 1.67 x 10^-27 * v^2
v^2 = (225 * 10^(-14)) / (1.67 x 10^-27)
v^2 = (225 / 1.67) * (10^(-14) / 10^-27)
v^2 ≈ 134.731 * 10^13
v ≈ sqrt(134.731) * sqrt(10^13)
v ≈ 11.60 * 10^6 m/s
Therefore, the speed of the proton when it is launched is approximately 11.60 x 10^6 m/s in order to just escape the influence of the charge of the glass sphere.