Question

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

Answer 1

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

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

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