Two tiny metal spheres A and B, mass mA $ 5.00 g and mB $ 10.0 g, have equal positive charge q $ 5.00 mC. The spheres are connected by a massless nonconducting string of length d $ 1.00 m, which is much greater than the radii of the spheres. (a) What is the electric potential energy of the system? (b) Suppose you cut the string. At that instant, what is the acceleration of each sphere? (c) A long time after you cut the string, what is the speed of each sphere?

Given Information:  

Mass of sphere A = m1 = 5 g

Mass of sphere B = m2 = 10 g

Charge = q1 = q2 = q = 5 mC = 5x10⁻³ C

Length = d = 1 m

Required Information:  

Potential energy = ?  

Acceleration = ?

Speed = ?

U = 2.247x10⁵ j

a1 = 4.49x10⁷ m/s²

a2 = 2.247x10⁷ m/s²

v1 = -7740.8 m/s

v2 = 3870.4 m/s

Explanation:  

(a) What is the electric potential energy of the system?

The electric potential energy can be found using

U = q²/4πe₀d

Where 1/4πe₀ is equal to 8.99x10⁹

U = ( 5x10⁻³)²*8.99x10⁹/1

U = 2.247x10⁵ j

(b) Suppose you cut the string.At that instant, what is the acceleration of each sphere?

Sphere A:

As we know from Newton's second law of motion

F1 = m1a1

We need the force, that can be found using Coulomb's law

F1 = q²/4πe₀d²

F1 = (5x10⁻³)²*8.99x10⁹/1²

F1 = 2.247x10⁵ N

a1 = F1/m1

a1 = 2.247x10⁵/0.005

a1 = 4.49x10⁷ m/s²

Sphere B:

F2 = q²/4πe₀d²

F2 = (5x10⁻³)²*8.99x10⁹/1²

F2 = 2.247x10⁵ N

a2 = F2/m2

a2 = 2.247x10⁵/0.010

a2 = 2.247x10⁷ m/s²

(c) A long time after you cut the string, what is the speed of each sphere?

The potential energy would be converted into kinetic energy

0.5m1v1² + 0.5m2v2² = 2.247x10⁵

According to the equation of conservation of linear momentum

m1v1 + m2v2 = 0

v1 = -m2/m1(v2)

v1² = m2²/m1²(v2²)

0.5m1(m2²/m1²(v2²))

so

0.5v2²((m2²/m1)+m2) = 0.5v2²((0.01²/0.005)+0.01) = 0.5v2²(0.03)

0.5v2²(0.03) = 2.247x10⁵

v2² = 2.247x10⁵/0.015

v2 = √1.49x10⁷

v2 = 3870.4 m/s

v1 = -m2/m1(v2)

v1 = -0.01/0.005(3870.4)

v1 = -7740.8 m/s

The negative sign means that both spheres move in opposite directions.

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