Answer to Question #184629 in Electric Circuits for Kathryn

Question

A proton was released horizontally from a gun with an acceleration of 3.2 m/s2

to a point where the  distance is about 5.3 cm and stopped because of another proton was stationary positioned on its way. 

The distance of the two proton is 4.5 m. Compute for the electric field intensity 1cm to the left of  the stopped proton is positioned.

Answer 1

Distance between two protons, $r=4.5\space m$

Electric field at 1 cm right of q

$E=E_1+E_2$

$E=\dfrac{kq}{r_1^2}+\dfrac{kq}{r_2^2}$

$E=9\times10^9\times1.6\times10^{-19}\bigg[\dfrac{1}{(4.51)^2}+\dfrac{1}{(0.01)^2}\bigg]$

$E=1.44\times10^{-5}\space N/C$ (Ans)