2. Solved Papers
  3. EAMCET Medical

EAMCET Medical EAMCET Medical Solved Paper-1998

  • question_answer
    A dust particle of mass \[{{10}^{-13}}g\]is stationary between plates of a horizontal parallel plate capacitor of 0.016 m separation which is connected to a voltage of 100 V. How many fundamental charges \[(e=1.6\times {{10}^{-19}}C)\] the dust particle carries : \[(g=10\text{ }m/{{s}^{2}})\]

    A)  1                                            

    B)  10

    C)  100                                       

    D)  1000

    Correct Answer: A

    Solution :

                     From the relation, \[q\,E=mg\] \[\left( \text{Since,}\,E=\frac{V}{d} \right)\] \[\left( E=\frac{100}{0.016} \right)\]                 \[\therefore \]  \[q=\frac{mg}{E}\]                                          ?(i) \[\left( \begin{align}   & \text{Given: m =1}{{\text{0}}^{-13}}\,g \\  & ={{10}^{-16}}\,\text{kg} \\  & d=0.016 \\  & V=100\,\text{volt} \\  & e\text{ }=\text{ }1.6\times {{10}^{-19}}\text{coulomb} \\  & g\text{ }=\text{ }10\text{ }m/{{s}^{2}} \\ \end{align} \right)\] Now, putting the given value and value of E then form (ii) in Eq. (i) \[q=\frac{{{10}^{-6}}\times 10\times 0.016}{100}\,\text{coulomb}\] Number of charges are given as \[n=\frac{q}{e}\] \[=\frac{{{10}^{-16}}\times 10\times 0.016}{100\times 1.6\times {{10}^{-19}}}\] = 1


You need to login to perform this action.
You will be redirected in 3 sec spinner