Question

A positive charge of 4.70 μC is fixed in place. From a distance of 4.30 cm a from the fixed positive charge, a particle of mass 5.40 g and charge +3.50 μC is released from rest. What is the speed of the +3.50 μC when it is 8.10 cm away from the fixed positive charge?

Answer 1

The electrostatic force is acting between two charge particles which is conservative in nature.

Therefore , the sum of kinetic energy(K) and potential energy(U) is constant for the system. i.e K + U = constant

Initially, the charge particle is released from rest therefore Kinetic energy would be zero i.e K =0 .
Potential energy between two charged particles (q₁ , q₂ ) separated by distance r is given by

where     

here q₁ = 4.70 μC = 4.7 * 10^-6 C

q₂ = +3.50 μC = 3.5 * 10^-6 C

r = distance between two particles = 4.3 cm = 0.043 m

U = 3.443 J

initially K + U = 3.443 J ___________(since K = 0 )

At r = 8.10 cm = 0.081 m

  U = 1.828 J

  

m = mass of the particle = 5.40 g = 5.4 * 10^-3 kg

v= magnitude of velocity at r= 8.1 cm

Applying K + U = 3.443 J at a distance 8.10 cm

v² = 598.15

v = 24.46 m/s