Question

Hi guys, please help me answer this question and show me all the equations, and steps you used to solve it.

d - A proton (m 1.67 x 10-22 kg) enters a region of space with velocity 10%j m/s. In this region an electric field -10%) N/C exists. Find the dis- tance the proton goes before stopping. d=?

Thanks !

Answer 1

mass. Given of the proton (m) = 1:67X10-27 kg velocity of the proton (1) = 100 mg in ydinection electric field (E) 105 / in y direction Since in positive e (proton) is moving charge opposite direction of electric field so. electric exerted force on proton due to electric field reduces the velocity of proton. and magnitude of this electric = qe eq where q- charge on q= + 1.6x 10-19c on force- proton and = ma eq. 14.3 from eq.0 and QE = ma qE a= m Now from third equation of motion- 12=4²- zad ( Negative sign shows that pro ton is decelerating where to final velocity of proton = omy's Cat meest He waitial velocity de distance travelled

Putting the values in above equation 0 = U2 - 2 QE d xlos Now (106)2 - 2 -19 1.6Xlo 7,67 x 10-27 0 = d 0 = 1012 1.916 13 X 10 d ㅋ 1.916 x 1013 d = 1012 d tola T.916 X 1013 m > d = 0.052 m =5.2cm مكو proton 0.052 meter distance goes before stopping