A particle, having a charge +5 μC, is initially at rest at the point x = 30 cm on the x axis. The particle begins to move due to the presence of a charge Q that is kept fixed at the origin. Find the kinetic energy of the particle at the instant it has moved 15 cm from its initial position if (a) Q =+15μC and (b) Q = -15μC
Answers
Gain in kinetic energy = Difference in potential energy
Potential Energy,
E = k{q_{1}q_{2} /{r_{12}}
Where q₁ and q₂ are the charges
r₁₂ is the distance between the charges
k is Coulomb's constant = 9×10⁹
Given :
q₁ = +5 μC
q₂ = ±15 μC
r₁₂ = 30
To find: K.E. at the instant q₁ moved 15cm from initial position.
Initial potential energy
E = 9 * 10^{9} *{5*10^{-6} * 15*10^{-6} /{30*10^{-2}} =2.25
(a) Since both charges are positive, the charges repel. Hence the charge q₁ moves away by 15 cm.
Hence r₁₂ = 30+15 =45 cm
E' = 9 * 10^{9} *{5*10^{-6} * 15*10^{-6}} /{45*10^{-2}} =1.5
K.E. = E - E' = 2.25 - 1.5 = 0.75
(b) Since both charges are opposite, the charges attract. Hence the charge q₁ moves closer by 15 cm.
Hence r₁₂ = 30-15 =15 cm
E' = 9 * 10^{9} *{5*10^{-6} * 15*10^{-6} /{15*10^{-2}} =4.5
K.E. = E' - E = 4.5 - 2.25 = 2.25
Cheers!!