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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^-11 C enters in it with a velocity 10⁷ m/sec. The acceleration of the particle will be

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$2 \times 10^{3} m / s^{2}$

$5 \times 10^{5} m / s^{2}$

$10^{8} m / \sec^{2}$

$10^{5} m / s^{2}$

answer is A.

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Detailed Solution

Electric field between the plates is

$\begin{array}{l} E = \frac{V}{d} = \frac{50}{5 \times 10^{- 3}} = 10^{4} \end{array}$

Acceleration of the charge will be

$a = \frac{Eq}{m} = 10^{8} m / s^{2}$

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