A proton is placed between two oppositely charged plates, with a potential diffe

Question

A proton is placed between two oppositely charged plates, with a potential difference of 5.0 volts. If the distance between the plates in 0.10 cm, calculate the following using SI units: 30 points total

a) The acceleration of the proton as it moves from the positive to the negative plate. Hint: What is the electric field between the plates and what force does it produce?

d) What is the change in electrical potential (volts) a proton would experience if moved from exactly between the two plates to the positive plate? Hint: What is the value of the electric field, and does it differ at each position? How is the potential difference related to the electric field in this instance?

A 10.0 cm length of wire carrying a current of -50.0 mA j is placed into a magnetic field of -5.0 i + 7.0 k (T). What is the magnetic force that it experiences? What will be the net acceleration of the wire if its mass is 20.0 g? Remember, acceleration is a vector, so be complete in your answer. 25 points total for the problem, 15 points for the force calculation and 10 points for the acceleration vector calculatio

Explanation / Answer

E=V/x=5/.001= 5000 N/c

a) force on proton=q*E=1.6*10^-19*5000

accelration=1.6*5000*10^-19/(1.673*10^-27) m/s^2= 4.78*10^11 m/s^2

b)work = chaneg in potential energy= 1.6*10^-19*5= 8*10^-19 Joule

c)Change in kinetic energy= decrease in potential energy= v*q= Â 8*10^-19 Joule

d) the electric field would be constant = 5000 N/C Â the potential difference V is relateed to E as:d V=-E*dx Â

E is taken positive in the direction of dx and dx is the distance between two points in direcion of E

Â

thus potential difference from mid way= 5000/2 = 2500 volts

Force= I*L(CROSS)B= -50*10^-3*.1 j (CROSS) [-5i+7k]

Â

Force=5*10^-3[-5k-7i] Â thus magnitude= Â 43.012*10^-3 N

Accelration= Force/mass= 2.15 m/s^2

accelration in vector form=Force/mass=-1.25i-1.75j m/s^2

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