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topjm [15]
2 years ago
11

Two particles, with identical positive charges and separation of 2.55 10-2 m, are released from rest. Immediately after the rele

ase, particle 1 has an acceleration vector a1 whose magnitude is 4.65 103 m/s2, while particle 2 has an acceleration vector a2 whose magnitude is 8.55 103 m/s2. Particle 1 has a mass of 6.05 10-6 kg.
a. Find the charge on each particle. q1 = C q2 = C
b. Find the mass of particle 2. kg
Physics
1 answer:
maria [59]2 years ago
7 0

Answer with Explanation:

We are given that

Distance between two charges,d=2.55\times 10^{-2} m

a_1=4.65\times 10^3 m/s^2

a_2=8.55\times 10^3 m/s^2

Mass of particle 1=m_1=6.05\times 10^{-6} kg

a.We have to find the change on each particle .

Let q_1=q_2=q

Force ,F=\frac{kq_1q_2}{d^2}

Where k=9\times 10^9

Using the formula

F=\frac{9\times 10^9\times q^2}{(2.55\times 10^{-2})^2}

For particle 1

F=m_1a_1

F=6.05\times 10^{-6}\times 4.65\times 10^3=0.028 N

Using the value of F

0.028=\frac{9\times 10^9\times q^2}{(2.55\times 10^{-2})^2}

q^2=\frac{(2.55\times 10^{-2})^2\times 0.028}{9\times 10^9}

q=\sqrt{\frac{(2.55\times 10^{-2})^2\times 0.028}{9\times 10^9}}

q=4.5\times 10^{-8} C

q_1=q_2=4.5\times 10^{-8} C

b.m_2=\frac{F}{a_2}

Using the formula

m_2=\frac{0.028}{8.55\times 10^3}

m_2=3.3\times 10^{-6} kg

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