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Step-by-Step Solution
Step 1: Identify the Known Quantities
• Charge on particle A, $q_A = 1\mu \text{C} = 1\times 10^{-6}\,\text{C}$
• Charge on particle B, $q_B = 1\mu \text{C} = 1\times 10^{-6}\,\text{C}$
• Mass of particle B, $m_B = 4\times 10^{-9}\,\text{kg}$
• Initial separation between A and B, $r_i = 1\,\text{mm} = 1\times 10^{-3}\,\text{m}$
• Final separation between A and B, $r_f = 9\,\text{mm} = 9\times 10^{-3}\,\text{m}$
• Coulomb’s constant, $ k = 9\times 10^{9}\,\text{N\,m}^2\text{C}^{-2} $
Step 2: Apply Energy Conservation
When particle B is released, it moves due to the electrostatic repulsion from A. By the principle of conservation of energy, the decrease in electrostatic potential energy from $r_i$ to $r_f$ is converted entirely into the kinetic energy of particle B.
Hence,
$ \frac{1}{2} m_B \, v^2 \;=\; k\, q_A \, q_B \left(\frac{1}{r_i} - \frac{1}{r_f}\right). $
Step 3: Substitute the Known Values
Substituting all values:
$ \frac{1}{2}\,\bigl(4\times 10^{-9}\bigr)\,v^2 \;=\; \bigl(9\times 10^{9}\bigr) \,\bigl(1\times 10^{-6}\bigr) \,\bigl(1\times 10^{-6}\bigr) \left(\frac{1}{1\times 10^{-3}} - \frac{1}{9\times 10^{-3}}\right). $
First, compute the factor inside parentheses:
$ \frac{1}{1\times 10^{-3}} \;=\; 10^{3}, \quad
\frac{1}{9\times 10^{-3}} \;=\; \frac{10^{3}}{9}\;=\;\frac{1000}{9}. $
Therefore,
$ \left(10^{3} - \frac{1000}{9}\right)
= 10^{3}\left(1 - \frac{1}{9}\right)
= 10^{3}\left(\frac{8}{9}\right)
= \frac{8\times 10^{3}}{9}. $
So, the right-hand side becomes:
$ (9\times 10^{9}) \,(1\times 10^{-6}) \,(1\times 10^{-6}) \,\frac{8\times 10^{3}}{9}
= 9\times 10^{9}\times 10^{-12}\times \frac{8\times 10^{3}}{9}
= 9\times 10^{-3}\times \frac{8\times 10^{3}}{9}. $
Simplify:
$ = 9\times \frac{8\times 10^{0}}{9}
= 8. $
Thus, the potential energy difference is $8\,\text{J}$. Now equate this to the left-hand side:
$ \frac{1}{2}\,\bigl(4\times 10^{-9}\bigr)\,v^2 = 8. $
Step 4: Solve for the Velocity
Rewriting:
$ 2\times 10^{-9}\,v^2 = 8
\;\;\Longrightarrow\;\;
v^2 = \frac{8}{2\times 10^{-9}}
= 4\times 10^{9}.
$
Hence the speed $v$ is:
$ v = \sqrt{4\times 10^{9}}
= 2\times 10^{4.5}
= 2\times 10^{4}\,\sqrt{10}
\approx 6.32\times 10^{4}\,\text{m/s}.
$
Final Answer
$\displaystyle 6.32\times 10^{4}\,\text{m/s}$
