Question

Suppose you wanted to hold up an electron against the force of gravity by the attraction of a fixed proton some distance above it. How far above the electron would the proton have to be? (k = 1/4πε0 = 9.0 × 109 N ∙ m2/C2, e = 1.6 × 10-19 C, mproton = 1.67 × 10-27 kg, melectron = 9.11 × 10-31 kg)

Answer 1

Answer:

The value is  $r = 5.077 \ m$

Explanation:

From the question we are told that

   The  Coulomb constant is  $k = 9.0 *10^{9} \ N\cdot m^2 /C^2$

   The  charge on the electron/proton  is  $e = 1.6*10^{-19} \ C$

    The  mass of proton $m_{proton} = 1.67*10^{-27} \ kg$

    The  mass of  electron is  $m_{electron } = 9.11 *10^{-31} \ kg$

Generally for the electron to be held up by the force gravity

   Then    

       Electric force on the electron  =  The  gravitational Force

i.e  

            $m_{electron} * g = \frac{ k * e^2 }{r^2 }$

         $\frac{9*10^9 * (1.60 *10^{-19})^2 }{r^2 } = 9.11 *10^{-31 } * 9.81$

         $r = \sqrt{25.78}$

         $r = 5.077 \ m$