Question

Instruction:
1. Illustrate the situation described in the problem below. Label it with the given quantities and the
quantity being asked.
2. Make a Conceptual Reasoning diagram involving the quantities in the problem.
3. Solve the problem.
Problem:
Two particles are held 1.0 mm apart. One particle has a mass of 1.0 x 10-5
kg and a net charge
of + 3.2 x 10-10 C. The other particle has a mass of 2.0 x 10-5
kg and a net charge
of + 4.8 x10-10 C. If the two particles are released, what will their speeds be when they are 2.0
mm apart?

Answer 1

Answer on Question #69622, Physics / Electric Circuits

1. Illustrate the situation described in the problem below. Label it with the given quantities and the quantity being asked.

2. Make a Conceptual Reasoning diagram involving the quantities in the problem.

3. Solve the problem.

Problem:

Two particles are held $1.0 \, \text{mm}$ apart. One particle has a mass of $1.0 \cdot 10^{-5} \, \text{kg}$ and a net charge of $+3.2 \cdot 10^{-10} \, \text{C}$. The other particle has a mass of $2.0 \cdot 10^{-5} \, \text{kg}$ and a net charge of $+4.8 \cdot 10^{-10} \, \text{C}$. If the two particles are released, what will their speeds be when they are $2.0 \, \text{mm}$ apart?

Solution.

The force between two particles:

$$F = k \frac{q_1 q_2}{r^2}$$

$q_1, q_2$ are charges, $r$ is distance between particles

$$k = 9 \cdot 10^9 \, N \cdot m^2 / C^2$$ $$F = m a$$ $$a = \frac{dv}{dt}; \, v = \frac{dr}{dt}; \, dt = \frac{dr}{v}; \, a = v \frac{dv}{dr}$$ $$m a = k \frac{q_1 q_2}{r^2}$$ $$m v \frac{dv}{dr} = k \frac{q_1 q_2}{r^2}$$ $$v d v = k \frac{q_1 q_2}{m r^2} \, dr$$ $$\int_{v_0}^{v} v \, dv = \int_{r_0}^{r} k \frac{q_1 q_2}{m r^2} \, dr$$ $$v_0 = 0; \, r_0 = 1 \, \text{mm}; \, r = 2 \, \text{mm}$$

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Answer on Question #69622, Physics / Electric Circuits

$$\frac {v ^ {2}}{2} = k \frac {q _ {1} q _ {2}}{m r _ {0}} - k \frac {q _ {1} q _ {2}}{m r}$$ $$v = \sqrt {2 k \frac {q _ {1} q _ {2}}{m} \left(\frac {1}{r _ {0}} - \frac {1}{r}\right)}$$

For the first particle:

$$v _ {1} = \sqrt {2 k \frac {q _ {1} q _ {2}}{m _ {1}} \left(\frac {1}{r _ {0}} - \frac {1}{r}\right)} = \sqrt {2 \cdot 9 \cdot 1 0 ^ {9} \cdot \frac {3 . 2 \cdot 4 . 8 \cdot 1 0 ^ {- 2 0}}{1 \cdot 1 0 ^ {- 5} \cdot 1 0 ^ {- 3}} \left(1 - \frac {1}{2}\right)} = 0. 3 7 2 m / s$$ $$v _ {2} = \sqrt {2 k \frac {q _ {1} q _ {2}}{m _ {2}} \left(\frac {1}{r _ {0}} - \frac {1}{r}\right)} = \sqrt {2 \cdot 9 \cdot 1 0 ^ {9} \cdot \frac {3 . 2 \cdot 4 . 8 \cdot 1 0 ^ {- 2 0}}{2 \cdot 1 0 ^ {- 5} \cdot 1 0 ^ {- 3}} \left(1 - \frac {1}{2}\right)} = 0. 2 6 3 m / s$$

The first particle

The second particle

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