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Home  >>  AIMS  >>  Class12  >>  Physics  >>  Electric Charges and Fields
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Suppose that the particle in Exercise in 1.33 is an electron projected with velocity $V_x= 2.0 × 10^6 m s^{−1}$. If $E$ between the plates separated by $0.5\; cm$ is $9.1 \times 10^2 N/C,$ where will the electron strike the upper plate? $(|e| =1.6 \times 10^{−19} C, m_e = 9.1 \times 10^{−31} kg.)$

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1 Answer

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1.6 cm
Hence A is the correct answer.
answered Jun 4, 2014 by meena.p
 
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