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THe electric feild intensity inhte region between tow deflecting plates of an electronic deflection apparatus within CRT television tube is 3.00 x 10^4 N/C. Determine the force on an electron passing between these plates. What acceleration does the electron experience.

 

2 months ago

Answered By Majid B

   $F=Eq_e=\left(3\times10^4\left(\frac{N}{C}\right)\right)\left(1.602\times10^{-19}\left(C\right)\right)=4.806\times10^{-15}\left(N\right)$F=Eqe=(3×104(NC ))(1.602×1019(C))=4.806×1015(N)   

   $a=\frac{F}{m_e}=\frac{\left(4.806\times10^{-15}\left(N\right)\right)}{\left(9.109\times10^{-31}\left(kg\right)\right)}=5.276\times10^{15}\left(\frac{m}{s^2}\right)$a=Fme =(4.806×1015(N))(9.109×1031(kg)) =5.276×1015(ms2 )