Numerical

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?

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#### Solution

Given:

Magnitude of charges, q = 2.0 × 10^{−4} C

Mass of particles, m = 10 g = 0.01 kg

Separation between the charges, r = 10 cm = 0.1 m

Force of repulsion,

\[F = \frac{1}{4\pi \epsilon_0}\frac{q^2}{r^2}\]

\[F = 1 . 8 \times {10}^4 \] N

\[F = m \times a\]

\[ \Rightarrow a = 1 . 8 \times {10}^6 \text{ m/ s}^2\]

Now,

\[v^2 = u^2 + 2\] as

\[ v^2 = 0 + 2 \times 1 . 8 \times {10}^6 \times 10 \times {10}^{- 2} \]

\[v = \sqrt{36 \times {10}^4}\]

\[v = 600 \] m/s

Concept: Coulomb’s Law

Is there an error in this question or solution?

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