• # question_answer Two metal pieces having a potential difference of 800V are 0.02 m apart horizontally. A particle of mass$1.96\times {{10}^{-15}}\,kg$is suspended equilibrium between the plates. If e is the elementary charge, then charge on the particle is : A)  8e                                          B)  6e C)  3e                                          D)  e

Solution :

For equilibrium  weight of particle = electrostatic force $mg=qE$ $mg=q\left( \frac{V}{d} \right)$ Putting the given values we get $1.96\times {{10}^{-15}}\times 9.8=q\left( \frac{800}{0.02} \right)$ $q=\frac{1.96\times {{10}^{-15}}\times 9.8\times 0.02}{800\times 1.6\times {{10}^{-19}}}e$$=3e$

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