<h2>
Answer:</h2>
315.7 x 10⁻³ J
<h2>
Explanation:</h2><h2>
</h2>
Given from the question;
Charge q₁ = 2.30μC, which is stationary
Charge q₂ = -5.0μC, which moves from;
initial point r₁ = [0.170m, 0] =
=
= 0.170m
final point r₂ = [0.250m, 0.250m] =
=
=
= 0.3536m
Now, remember that the work done (W) on a moving point charge(q) by an electric force is the change in the potential energy (ΔU) of the charge. i.e
W = ΔU
Where;
ΔU = U₁ - U₂
U₁ = potential energy at point r₁
U₂ = potential energy at point r₂
=> W = U₁ - U₂ ----------------------------(i)
Potential energy (U) of between two point charges at a point r is given by;
U = k x q₁ x q₂ / r
Where;
k = 8.99 x 10⁹ Nm²/C²
Therefore, we can write;
U₁ = k x q₁ x q₂ / r₁ and
U₂ = k x q₁ x q₂ / r₂
Substitute these values of U₁ and U₂ into equation (i) as follows;
=> W = (k x q₁ x q₂ / r₁) - (k x q₁ x q₂ / r₂)
=> W = k x q₁ x q₂ [
-
] --------------------(ii)
Substitute the values of k, q₁, q₂ and r₂ into the equation (ii) as follows;
=> W = 8.99 x 10⁹ x 2.30 x 10⁻⁶ x 5.00 x 10⁻⁶ x [
-
]
=> W = 8.99 x 10⁹ x 2.30 x 10⁻⁶ x 5.00 x 10⁻⁶ x [5.882- 2.828]
=> W = 8.99 x 10⁹ x 2.30 x 10⁻⁶ x 5.00 x 10⁻⁶ x [3.054]
=> W = 315.7 x 10⁻³
=> W = 315.7 mJ
Therefore, the work done on the moving point charge q₂ is 315.7 x 10⁻³ J