# A positive point charge q1 = +5.00 × 10−4C is held at a fixed position. A small object with mass 4.00×10−3kg and charge q2 = −3.00×10−4C is

Question

A positive point charge q1 = +5.00 × 10−4C is held at a fixed position. A small object with mass 4.00×10−3kg and charge q2 = −3.00×10−4C is projected directly at q1. Ignore gravity. When q2 is 4.00m away, its speed is 800m/s. What is its speed when it is 0.200m from q1?

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1 year 2021-09-03T16:21:35+00:00 1 Answers 50 views 0

## Answers ( )

Therefore the speed of q₂ is 1961.19 m/s when it is 0.200 m from from q₁.

Explanation:

Energy conservation law: In isolated system the amount of total energy remains constant.

The types of energy are

1. Kinetic energy.
2. Potential energy.

Kinetic energy $$=\frac{1}{2} mv^2$$

Potential energy =$$\frac{Kq_1q_2}{d}$$

Here, q₁= +5.00×10⁻⁴C

q₂=-3.00×10⁻⁴C

d= distance = 4.00 m

V = velocity = 800 m/s

Total energy(E) =Kinetic energy+Potential energy

$$=\frac{1}{2} mv^2$$+ $$\frac{Kq_1q_2}{d}$$

$$=\frac{1}{2} \times 4.00\times 10^{-3}\times(800)^2 +\frac{9\times10^9\times 5\times10^{-4}\times(-3\times10^{-4})}{4}$$

=(1280-337.5)J

=942.5 J

Total energy of a system remains constant.

Therefore,

E $$=\frac{1}{2} mv^2$$ + $$\frac{Kq_1q_2}{d}$$

$$\Rightarrow 942.5 = \frac{1}{2} \times 4 \times10^{-3} \times V^2 +\frac{9\times10^{9}\times5\times 10^{-4}\times(-3\times 10^{-4})}{0.2}$$

$$\Rightarrow 942.5 = 2\times10^{-3}v^2 -6750$$

$$\Rightarrow 2 \times10^{-3}\times v^2= 942.5+6750$$

$$\Rightarrow v^2 = \frac{7692.5}{2\times 10^{-3}}$$

$$\Rightarrow v= 1961.19$$   m/s

Therefore the speed of q₂ is 1961.19 m/s when it is 0.200 m from from q₁.