Question

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.

Answer 1

421.6 x 10⁻³J

Step-by-step explanation:

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

`K_(E)` =
`P_(E(at position 1))` -
`P_(E(at position 2))`

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

`P_(E)` = k x Q₁ x Q₂ ÷ r --------------------- (i)

Where;

k = electric constant = 9 x 10⁹ Nm²/C²

Therefore to get the
`P_(E)` 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

=>
`P_(E)` = 9 x 10⁹ x 1.13 x 10⁻⁶ x 1.95 x 10⁻⁶ / 0.024

=>
`P_(E)` = 826.3 x 10⁻³ J

Also, to get the
`P_(E)` 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

=>
`P_(E)` = 9 x 10⁹ x 1.13 x 10⁻⁶ x 1.95 x 10⁻⁶ / 0.049

=>
`P_(E)` = 404.7 x 10⁻³ J

Therefore, the kinetic energy
`K_(E)` is calculated as follows;

`K_(E)` =
`P_(E(at position 1))` -
`P_(E(at position 2))`

`K_(E)` = 826.3 x 10⁻³ - 404.7 x 10⁻³

`K_(E)` = 421.6 x 10⁻³J