91Ó°ÊÓ

Charges at a distance apply force on each other, which is equal and opposite in direction. If the charges are of the same nature, this force will be repulsive; if not, it will be an attractive force. This force is known as coulomb force.

Let q₂be 0.800 m away from q₁ charge when it is at point a, and q₁is at point b when it is 0.400 m away from q₂. Refer to the image below.

Work is done by only the electric force in this case, So, W_other = 0.

According to the principle of conservation of energy,

${\mathrm{K}}_{\mathrm{a}}+{\mathrm{U}}_{\mathrm{a}}+{\mathrm{W}}_{\mathrm{other}}={\mathrm{K}}_{\mathrm{b}}+{\mathrm{U}}_{\mathrm{b}}$ ….(¾±)

Here, the K- terms denote the kinetic energy, U-terms denote the potential energy given as,

$U=\frac{1}{4{\mathrm{Ï€Î\mu }}_0}\frac{q_1{q}_2}{r}$

Now, the value of kinetic energy of sphere a is,

$$\begin{gathered} K_a=\frac{1}{2}m{v}_a^2 \\ =\frac{1}{2}\left(1.50×{10}^{-3}\mathrm{kg}\right){\left(22.0\mathrm{m}/\mathrm{s}\right)}^2 \\ =0.3630\mathrm{J} \end{gathered}$$

The value of potential energy of sphere a is,

$$\begin{gathered} U_a=\frac{1}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0}\frac{q_1{q}_2}{r_a} \\ =\left(8.988×{10}^9\mathrm{N}.{\mathrm{m}}^2/{\mathrm{C}}^2\right)\frac{\left(-2.80×{10}^{-6}\mathrm{C}\right)\left(-7.80×{10}^{-6}\mathrm{C}\right)}{0.800\mathrm{m}} \\ =+0.2454\mathrm{J} \end{gathered}$$

Similarly, for sphere b,

$K_b=\frac{1}{2}m{v}_b^2$

And,

$$\begin{gathered} U_b=\frac{1}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0}\frac{q_1{q}_2}{r_b} \\ =\left(8.988×{10}^9\mathrm{N}.{\mathrm{m}}^2/{\mathrm{C}}^2\right)\frac{\left(-2.80×{10}^{-6}\mathrm{C}\right)\left(-7.80×{10}^{-6}\mathrm{C}\right)}{0.400\mathrm{m}} \\ =+0.4907\mathrm{J} \end{gathered}$$

From equation (i), Solving for K_b,

$$\begin{gathered} {\mathrm{K}}_{\mathrm{b}}={\mathrm{K}}_{\mathrm{a}}+\left({\mathrm{U}}_{\mathrm{a}}-{\mathrm{U}}_{\mathrm{b}}\right) \\ \frac{1}{2}m{v}_b^2=+0.3630\mathrm{J}+\left(0.2454\mathrm{J}-0.4907\mathrm{J}\right) \\ =0.1177\mathrm{J} \end{gathered}$$

Thus, the speed of the charged particle q₂ is,

$$\begin{gathered} {\mathrm{v}}_{\mathrm{b}}=\sqrt{\frac{2\left(0.1177\mathrm{J}\right)}{\left(1.50×{10}^{-3}\mathrm{kg}\right)}} \\ =12.5\mathrm{m}/\mathrm{s} \end{gathered}$$

Thus, the speed of charge q₂ when the spheres are 4.00 m apart is $12.5\mathrm{m}/\mathrm{s}$.

Take a point c where the charge q₂loses momentum, and its speed becomes zero. Refer to the image below and apply the Law of conservation of energy at points a and c,

The Law of conservation of energy at a and c is,

${\mathrm{K}}_{\mathrm{a}}+{\mathrm{U}}_{\mathrm{a}}={\mathrm{U}}_{\mathrm{c}}$ …(¾±¾±)

Now,$K_a=0.3630J$and $U_a=+0.2454J$ from part (a).

At c, speed is zero. So, K_c is also zero.

The potential energy at c is,

$U_c=\frac{1}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0}\frac{q_1{q}_2}{r_c}$

From equation (ii),

$$\begin{gathered} \frac{1}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0}\frac{q_1{q}_2}{r_c}=+0.3630\mathrm{J}+0.2454\mathrm{J} \\ =0.6084\mathrm{J} \end{gathered}$$

Solve for the distance,

$$\begin{gathered} r_c=\frac{1}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0}\frac{q_1{q}_2}{0.6084} \\ =\left(8.988×{10}^9\mathrm{N}.{\mathrm{m}}^2/{\mathrm{C}}^2\right)\frac{\left(-2.80×{10}^{-6}\mathrm{C}\right)\left(-7.80×{10}^{-6}\mathrm{C}\right)}{0.6084\mathrm{J}} \\ =0.323\mathrm{m} \end{gathered}$$

Thus, the distance of the closest approach is 0.323 m.