Let the sphere is having charge Q and radius R
Now if the proton is released from rest
By energy conservation we can say
[tex]U = K[/tex]
[tex]\frac{kQe}{R} = \frac{1}{2}mv^2[/tex]
[tex]\frac{2kQe}{mR} = v^2[/tex]
now take square root of both sides
[tex]v =\sqrt{\frac{2kQe}{mR}}[/tex]
so the proton will move by above speed and
here Q = charge on the sphere
R = radius of sphere
[tex]k = 9 * 10^9 [/tex]
The speed of proton is about 1 × 10⁷ m/s
Electric charge consists of two types namely positive electric charge and negative electric charge.
There was a famous scientist who investigated on this matter. His name was Coulomb and succeeded in formulating a force of attraction or repulsion between two charges.
[tex]\large {\boxed {F = k \frac{Q_1 Q_2}{R^2} } }[/tex]
F = electric force(N)
k = electric constant (N m² / C²)
q = electric charge (C)
r = distance between charges (m)
The value of k in a vacuum = 9 x 10⁹ (N m² / C²)
Let us now tackle the problem!
Let me repeat the question!
A 3.0 cm -diameter isolated metal sphere carries a net charge of 0.90 μC. If a proton were released from rest at the sphere's surface, what would be its speed far from the sphere?
Given:
diameter of sphere = d = 3 cm
charge of sphere = 0.90 μC
mass of proton = 1.67 × 10⁻²⁷ kg
charge of proton = 1.6 × 10⁻¹⁹ C
Unknown:
speed of proton = v = ?
Solution:
We will use the law of conservation of energy to solve this problem.
The potential energy of proton on the surface of the sphere will be the same as the kinetic energy when it is far from the sphere.
[tex]Ek = Ep[/tex]
[tex]\frac{1}{2} m_{proton}v^2 = k \frac{Q_{sphere} ~ q_{proton}}{R}[/tex]
[tex]\frac{1}{2} (1.67 \times 10^{-27})v^2 = 9 \times 10^9 \times \frac{0.9 \times 10^{-6} \times 1.6 \times 10^{-19}}{\frac{1}{2} \times 3 \times 10^{-2}}[/tex]
[tex]\large {\boxed {v \approx 1 \times 10^7 ~ m/s} }[/tex]
Grade: High School
Subject: Physics
Chapter: Static Electricity
Keywords: Series , Parallel , Measurement , Absolute , Error , Combination , Resistor , Resistance , Ohm , Charge
