1. Two charges Q1( + 2.00 μC) and Q2( + 2.00 μC) are placed along the x-axis at x = 3.00 cm and x=-3 cm. Consider a charge Q3 of charge +4.00 μC and mass 10.0 mg moving along the y-axis. If Q3 starts from rest at y = 2.00 cm, what is its speed when it reaches y = 4.00 cm?

The speed of the charge q₃ when it starts from rest at y = 2 cm and reaches y = 4 cm is; v = 14.89 m/s

We are given;

Charge 1; q₁ = 2.00 μC = 2 × 10⁻⁶ C

Charge 2; q₂ = 2.00 μC = 2 × 10⁻⁶ C

Distance of charge 1 along x = 3 cm = 3 × 10⁻² m

Distance of charge 2 along x = -3 cm = -3 × 10⁻² m

Charge 3; q₃ = +4.00 μC  = 4 × 10⁻⁶ C

mass; m = 0.01 g

distance of charge 3 along y = 4 cm = 4 × 10⁻² m

q₃ starts from rest at y = 2 × 10⁻² m and reaches y = 4 × 10⁻² m.

Thus;

Distance of charge 1 from the initial position of q₃;

r₁ = √((3 × 10⁻²)² + ((2 × 10⁻²)²)

r₁ = 0.0361 m

Distance of charge 2 from the final position of q₃;

r₂ = √((3 × 10⁻²)² + ((4 × 10⁻²)²)

r₂ = 0.05 m

Now, formula for electric potential is;

V = kq/r

Where k = 9 × 10⁹ N.m²/s²

Thus,change in potential is;

ΔV = V₁ - V₂

Now, Net V₁ = 2kq₁/r₁

Net V₂ = 2kq₂/r₂

Thus;

ΔV = 2kq₁/r₁ - 2kq₂/r₂

ΔV = (2 × 9 × 10⁹)[(2 × 10⁻⁶/0.0361) - (2 × 10⁻⁶/0.05)]

ΔV = 277229.92 V

Now, from conservation of energy;

½mv² = q₃ΔV

Thus;

½ × 0.01 × v² = 4 × 10⁻⁶ × 277229.92

v² = 2 × 4 × 10⁻⁶ × 277229.92/0.01

v = √(221.783936)

v = 14.89 m/s

Read more about point charges at;link

the answer is 12 as i may remember

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