a)1/2mv^2=q(stopping potential)
m=1.67*10^-27kg
stopping potential=263077.1875 V
b)The direction of electric field is opposite to motion of
proton
i.e -x
c)Electric field points in direction of decreasing potential
Potential at final point=220000+263077.1875=483kV
d)The electron having lesser mass will have lesser KE and hence
lesser stopping potential
The electric force on electron is opposite to electric field
acceleration=g-qE/m
g=9.8m/s^s
q=1.6*10^-19C
E=4.4*10^-11N/C
m=9.1*10^-31kg
(Substituting values)
a=2.1m/s^2
v^2=u^2+2as
0=36^2-2*2.1*s
a)s=308.57 m(height which electron reaches)
b)change in gravitational potential energy+change in electric
potential energy=change in kinetic energy
9.1*10^-31*10*308.57+change in electric potential
energy=1/2(9.1*10^-31)(36^2)
change in electric potential energy=-2.16*10^-27J
c)Work done by electric force=-qEs
=-1.6*10^-19*4.4*10^-11*308.57
=-2172.3*10^-30J
d)Voltage=Es
4.4*10^-11*308.57
=1357.7*10^-11V