Electric Potential and Potential Energy (Uniform Fields) The potential difference ∆V needed to stop a moving...

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

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An electron is initially at ground level, and the electricpotential at that point is assumed to be exactly zero. The electronis immersed in a uniform electric field that points down with amagnitude 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 haveany effect in this problem. a) If the electron is given an initialupward velocity of 36 m/s, to what maximum height above groundlevel does it reach? b) What is the change in electric potentialenergy (final - initial) during this motion? c) How much work isdone by the electric force during this motion? d) What is thevoltage at the electron’s maximum height?

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Answer & Explanation Solved by verified expert
4.2 Ratings (723 Votes)

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


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