Question

A 0.40-kg cart with charge 5.0

Answer 1

Q₁ = 5 x 10^-5 C

Q₂ = 2 x 10^-4 C

m = 0.4 kg

r = 0.6 m

A) Electric Potential energy between cart and the fixed charge is getting converted into kinetic energy as the cart moves away from the fixed charge.

When Cart start from rest :

Kinetic energy of cart = KE_i = 0

Electric Potential energy between cart and fixed charge is given as

U_i = k Q₁ Q₂ /r = (9 x 10⁹) (5 x 10^-5 C) ( 2 x 10^-4 C) / 0.6 m = 150 J

When cart is far away :

Kinetic energy = KE_f = (0.5) m V²

Electric Potential energy = U_f = k Q₁ Q₂ /r = 0

using conservation of energy :

KE_i + U_i = KE_f + U_f

0 + 150 J = (0.5) m V² + 0

(0.5)(0.40 kg ) V² = 150

V = 27.4 m/s

B) Electric Potential energy between cart and fixed charge when distance r = 2 m

U_f = k Q₁ Q₂ /r = (9 x 10⁹) (5 x 10^-5 C) ( 2 x 10^-4 C) / 2 m = 45 J

using conservation of energy :

KE_i + U_i = KE_f + U_f

0 + 150 J = (0.5) m V² + 45

(0.5) m V² = 150 - 45

(0.5) (0.4) V² = 105

V = 22.91 m/s