Question

Two protons are either side of an elecron the electron is 30 micrometer away from the proton on its left and 10 micrometer away from the proton on its rightwhat is the magnitude and direction of the net electric force

Answer 1

p+ ----30μm------- e- ---10μm-----p+

$Q_{1} =1.6*10^{-19} \\Q_{2} =-1.6*10^{-19} \\Q_{3} =1.6*10^{-19} \\\\r_{1} =30*10^{-6}m\\r_{2} =10*10^{-6}m \\\\F_{21} =\frac{kQ_{1}Q_{2}}{r_{1}^{2} } } =\frac{9*10^{9}*1.6*10^{-19}*1.6*10^{-19} }{900*10^{-12} } =2.56*10^{9-19-19-2+12} =2.56*10^{-19} N\\F_{23}=\frac{kQ_{3}Q_{2}}{r_{2}^{2} } } =\frac{9*10^{9}*1.6*10^{-19}*1.6*10^{-19} }{100*10^{-12} } =23.04*10^{9-19-19-2+12} =23.04*10^{-19} \\Therefore the net force on the e- will be \\=F_{23}-F_{21}=(23.04-2.56)*10^{-19} =20.48*10^{-19} N$

Here we are subtracting the forces as bothe the protons will attract the electron towards each other ie, in the opposite direction

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