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Two parallel plates separated by a disat...

Two parallel plates separated by a disatnce of `5 mm` are kept at a potential difference of `5.0 V`. A particle of mass `10^(15) kg` and change `10^(-11) C` centre in it with a velocity `10^(7) m//s`. The acceleration of the particle will be

A

`10^(8) m//s^(2)`

B

`5 xx 10^(5) m//s^(2)`

C

`10^(5) m//s^(2)`

D

`2 xx 10^(3) m//s^(2)`

Text Solution

Verified by Experts

The correct Answer is:
A

`a = (qE)/(m) = (q)/(m) ((V)/(d)) = (10^(-11))/(10^(-15)) xx (50)/(5 xx 10^(-3)) = 10^(8) m//sec^(2)`
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