Two parallel plates separated by a distance of 5,mm are kept at a potential difference of 50,V. A pa

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2. Electric Potential and Capacitance

easy

Two parallel plates separated by a distance of $5\,mm$ are kept at a potential difference of $50\,V.$ A particle of mass ${10^{ - 15}}\,kg$ and charge ${10^{ - 11}}\,C$ enters in it with a velocity ${10^7}\,m/s.$ The acceleration of the particle will be

${10^8}\,m/{s^2}$

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

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

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

Solution

(a) $a = \frac{{qE}}{m} = \frac{q}{m}\left( {\frac{V}{d}} \right)$$ = \frac{{{{10}^{ – 11}}}}{{{{10}^{ – 15}}}} \times \frac{{50}}{{5 \times {{10}^{ – 3}}}} = {10^8}\,m/se{c^2}$

Standard 12

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easy

(JEE MAIN-2022)

Two parallel plates separated by a distance of $5\,mm$ are kept at a potential difference of $50\,V.$ A particle of mass ${10^{ - 15}}\,kg$ and charge ${10^{ - 11}}\,C$ enters in it with a velocity ${10^7}\,m/s.$ The acceleration of the particle will be

Solution

Similar Questions

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