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.
4 years 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 ))(1.602×10−19(C))=4.806×10−15(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 =(4.806×10−15(N))(9.109×10−31(kg)) =5.276×1015(ms2 )