A charge Q1 = 1.13 μC is at rest and is located 2.40 cm away from another fixed charge Q2 = 1.95 μC. The first charge is then released. Calculate the kinetic energy of charge Q1 when it is 4.90 cm away from charge Q2.

421.6 x 10⁻³J

The change in potential energy () of between two charges is equal to the gain in kinetic energy() of the charge. i.e

= -

Remember that, the potential energy () of between two charges (Q₁ and Q₂) at a given position (r) is given by;

= k x Q₁ x Q₂ ÷ r   ---------------------  (i)

Where;

k = electric constant = 9 x 10⁹ Nm²/C²

Therefore to get the 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

=> = 9 x 10⁹ x 1.13 x 10⁻⁶ x 1.95 x 10⁻⁶ / 0.024

=> = 826.3 x 10⁻³ J

Also, to get the 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

=> = 9 x 10⁹ x 1.13 x 10⁻⁶ x 1.95 x 10⁻⁶ / 0.049

=> = 404.7 x 10⁻³ J

Therefore, the kinetic energy is calculated as follows;

= -

= 826.3 x 10⁻³ - 404.7 x 10⁻³

= 421.6 x 10⁻³J

this is a good question! i'll do some research and come back when i have the answer!