Question

Two identical particles, each with a charge of 2.0 × 10−4 C and mass of 10 g, are kept at a separation of 10 cm and then released. What would be the speed of the particles when the separation becomes large?

Answer 1

600 m/s is the speed of the particles when the separation becomes large.

Explanation:

Step 1:

The total initial Coulomb potential energy is  

$v = \frac{Q^{2}}{4 \pi \varepsilon_{0} r}$ joules

As r→∞r→∞ the potential energy approaches zero, so by conservation of energy the total kinetic energy of the two particles must become equal to the total initial potential energy, with each particle having half that energy.

Step 2:

Neglecting relativistic effects the final kinetic energy of each particle is 1/2 $mv^2$=V/2 from  

Which

$\begin{aligned}&V=\sqrt{\frac{Q^{2}}{4 \pi \varepsilon_{0} r m}}\\&V=\sqrt{\frac{Q^{2}}{4 \pi \varepsilon_{0} r m}}\end{aligned}$

Given values

Q = 2.0 × $10^{-4\\}$ C

Mass (m) = 10 g =0.01 kg

r= 10 cm = 0.1 m

ε0=8.854 × $10^{-12}$

Step 3:

On substituting the values, we get

$\begin{array}{c}{V=\sqrt{\frac{\left(2 \times 10^{-4}\right)^{2}}{4 \times 3.14 \times 8.854 \times 10^{-12} \times 0.01 \times 0.1}}} \\\\\\{V=\sqrt{\frac{4 \times 10^{-8}}{12.56 \times 8.854 \times 10^{-15}}}}\end{array}$

$\begin{aligned}&V =\sqrt{\frac{4 \times 10^{-8}}{111.20 \times 10^{-15}}} \\& V=\sqrt{0.0359 \times 10^{7}} \\&V=\sqrt{359692.0461} \\&V =599.74 \approx 600 m / s\end{aligned}$

This speed, faster than sound and more like the speed of a bullet, is small compared to c, so it was ok to neglect relativistic effects.

Answer 2

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600 m/s speed.