Question

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?

Answer 1

Answer:

speed when it reaches y = 4.00cm is

v = 14.9 g.m/s

Explanation:

given

q₁=q₂ =2.00 ×10⁻⁶

distance along x = 3.00cm= 3×10⁻²

q₃= 4×10⁻⁶C

mass= 10×10 ⁻³g

distance along y = 4×10⁻²m

r₁ = $\sqrt{3^{2} +2^{2} }$ = $\sqrt{13}$ = 3.61cm = 0.036m

r₂ = $\sqrt{4^{2} + 3^{2} }$ = $\sqrt{25}$ = 5cm = 0.05m

electric potential V = $\frac{kq}{r}$

change in potential ΔV = $V_{1} - V_{2}$

ΔV = $\frac{2kq_{1} }{r_{1}} - \frac{2kq_{2} }{r_{2} }$ , where $q_{1} = q_{2}=$2.00μC

ΔV = $2kq(\frac{1}{r_{1}} - \frac{1}{r_{2} })$

ΔV = 2 × 9×10⁹ × 2×10⁻⁶ × $(\frac{1}{0.036} - \frac{1}{0.05} )$

ΔV= 2.789×10⁵

$\frac{1}{2}mv^{2}$ = ΔV × q₃

$\frac{1}{2}$ ˣ 10×10⁻³ ×v² = 2.789×10⁵× 4 ×10⁻⁶

v² = 223.12 g.m/s

v = 14.9 g.m/s