Question

Electric Potential and Potential Energy (Uniform Fields) The potential difference ∆V needed to stop a moving charged particle is called the “stopping potential.” Suppose a proton is moving in the +x direction with an initial speed of 7.1 x 106 m/s. Assume the electric force is the only relevant force in this problem. a) Find the stopping potential necessary to bring the proton to rest. b) In which direction is the electric field pointing in order to slow the proton down? c) At the initial position of the proton, the electric potential is 220,000 Volts (220 kilovolts or kV). What is the electric potential at the position where the proton comes to rest? d) Would an electron traveling at the same speed requires greater or lesser magnitude stopping potential? Explain your answer (hint: do an electron and proton moving at the same speed have the same kinetic energy?).

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An electron is initially at ground level, and the electric potential at that point is assumed to be exactly zero. The electron is immersed in a uniform electric field that points down with a magnitude of 4.4 x 10-11 N/C as well as Earth’s gravitational field (9.8 N/kg, pointing down). Only gravity and the electric force have any effect in this problem. a) If the electron is given an initial upward velocity of 36 m/s, to what maximum height above ground level does it reach? b) What is the change in electric potential energy (final - initial) during this motion? c) How much work is done by the electric force during this motion? d) What is the voltage at the electron’s maximum height?

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will vote, please show work and signs

Answer 1

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