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Electricity and Magnetism Level 2

Two identical particles, each having a charge of q = 200 μ C q=200\text{ }\mu\text{C} and mass of 10 gm 10\text{ gm} , are kept at a separation of 10 cm 10\text{ cm} and then released. what would be the speed of the particles when the separation becomes large?

400m/sec 300m/sec 500m/sec 600m/sec

Ranveer Gour by Ranveer Gour , 28, India

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2 solutions

Chew-Seong Cheong
Aug 20, 2014

The potential energy of the two charges is given by:

U = k q 2 r = 9 × 1 0 9 ( 200 × 1 0 − 6 ) 2 0.1 = 3600 J U = \cfrac{kq^2}{r} = \cfrac{9\times 10^9(200\times 10^{-6})^2}{0.1} = 3600J

The potential energy U U is converted to kinetic energy E k E_k of the two moving charges. And when their separation is large, E k = 2 ( 1 2 m v 2 ) = m v 2 = 0.01 v 2 E_k = 2(\frac{1}{2}mv^2) = mv^2 = 0.01v^2 .

Therefore, 0.01 v 2 = 3600 ⇒ v = 360000 = 600 m / s 0.01v^2 = 3600 \quad \Rightarrow v = \sqrt{360000} = 600 m/s .

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Just used conservation of energy : Initial velocity =0

Final kinetic energy of combined system = mv²

And this will be equal to work done by electrostatic force = F.dx Then, just integrate and we are done

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