421.6 x 10⁻³J
The change in potential energy ([tex]P_{E}[/tex]) of between two charges is equal to the gain in kinetic energy([tex]K_{E}[/tex]) of the charge. i.e
[tex]K_{E}[/tex] = [tex]P_{E(at position 1)}[/tex] - [tex]P_{E(at position 2)}[/tex]
Remember that, the potential energy ([tex]P_{E}[/tex]) of between two charges (Q₁ and Q₂) at a given position (r) is given by;
[tex]P_{E}[/tex] = k x Q₁ x Q₂ ÷ r --------------------- (i)
Where;
k = electric constant = 9 x 10⁹ Nm²/C²
Therefore to get the [tex]P_{E}[/tex] at position 1 (where r = 2.40cm), substitute the values of Q₁, Q₂ and r into equation (i);
Where;
Q₁ = 1.13μC = 1.13 x 10⁻⁶C
Q₂ = 1.95μC = 1.95 x 10⁻⁶C
r = 2.40cm = 0.024m
=> [tex]P_{E}[/tex] = 9 x 10⁹ x 1.13 x 10⁻⁶ x 1.95 x 10⁻⁶ / 0.024
=> [tex]P_{E}[/tex] = 826.3 x 10⁻³ J
Also, to get the [tex]P_{E}[/tex] at position 2 (where r = 4.90cm), substitute the values of Q₁, Q₂ and r into equation (i);
Where;
Q₁ = 1.13μC = 1.13 x 10⁻⁶C
Q₂ = 1.95μC = 1.95 x 10⁻⁶C
r = 4.90cm = 0.049m
=> [tex]P_{E}[/tex] = 9 x 10⁹ x 1.13 x 10⁻⁶ x 1.95 x 10⁻⁶ / 0.049
=> [tex]P_{E}[/tex] = 404.7 x 10⁻³ J
Therefore, the kinetic energy [tex]K_{E}[/tex] is calculated as follows;
[tex]K_{E}[/tex] = [tex]P_{E(at position 1)}[/tex] - [tex]P_{E(at position 2)}[/tex]
[tex]K_{E}[/tex] = 826.3 x 10⁻³ - 404.7 x 10⁻³
[tex]K_{E}[/tex] = 421.6 x 10⁻³J
