Two parallel plates separated by a distance of 5mm are kept at a potential difference of 50V. A particle of mass 10^–15 kg and charge 10^–11C enters in it with a velocity 10⁷m/s. The acceleration of the particle will be

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10⁸ m/s²

5 × 10⁵ m/s²

2 × 10³ m/s²

10⁵ m/s²

answer is A.

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Detailed Solution

$\begin{array}{l} a = \frac{E_{q}}{m} = \frac{V}{d} (\frac{q}{m}) \\ = \frac{50}{5 \times 10^{- 3}} (\frac{10^{- 11}}{10^{- 15}}) = 10^{8} m / s^{2} \end{array}$

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