A charge Q1 = 24C is at rest and is located 3 cm away from another fixed charge Q2 = 6 u C.
The first charge is then released. Calculate the kinetic energy of charge Q1 when it is 7 cm away
from charge q2

Respuesta :

The kinetic energy of charge Q1 at the given distance is [tex]3.086 \times 10^{12} \ J[/tex].

Force between the two charges

The force between the two charges is determined by applying Coulomb's law;

[tex]F = \frac{kq_1 q_2}{r^2} \\\\F = \frac{9\times 10^9 \times 24 \times 6\times 10^{-6}}{(0.03)^2} \\\\F = 1.44 \times 10^9 \ N[/tex]

Kinetic energy of the charge Q1

Based on the law of conservation of energy, the kinetic energy of the charge is equal to the potential energy of the charge.

[tex]K.E = U = \frac{kq_1}{r} = \frac{9 \times 10^9 \times 24}{0.07} = 3.086 \times 10^{12} \ J[/tex]

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